Electricity: Class 10 Science answers, notes

Summary

Electricity is a very useful form of energy in our daily lives, powering our homes, schools, and hospitals. It involves the flow of tiny particles called electric charges. When these charges move through a conductor, like a metal wire, we say there is an electric current. Think of it like water flowing in a river; flowing water makes a water current, and flowing charges make an electric current. For current to flow continuously, it needs a complete, unbroken path called an electric circuit. A switch is a device that can open or close this path, turning devices on or off.

Electric current is measured by how much charge passes a point in a certain amount of time. The unit for electric charge is the coulomb, and the unit for electric current is the ampere. An instrument called an ammeter measures the current in a circuit and is always connected in a line with other components. Conventionally, the direction of electric current is considered to be from the positive end to the negative end of a power source, which is opposite to the direction in which tiny negative charges called electrons actually move.

For charges to flow, there needs to be a difference in electric pressure, similar to how water flows from a higher level to a lower one due to a pressure difference. This electric pressure difference is called potential difference, and it is the work done to move a unit charge from one point to another. Its unit is the volt. A voltmeter measures potential difference and is connected across the points where the difference is to be measured. Cells or batteries provide this potential difference in a circuit.

Ohm’s law describes the relationship between potential difference, current, and resistance. It states that the current flowing through a conductor is directly related to the potential difference across its ends, as long as the temperature stays the same. The constant in this relationship is called resistance, which is the property of a conductor to oppose the flow of charges. Its unit is the ohm. Resistance of a conductor depends on its length, its cross-sectional area, and the material it is made from. A property called resistivity is unique to each material.

Resistors can be connected in two main ways. In a series connection, resistors are joined end-to-end, and the same current flows through each. The total resistance is the sum of individual resistances. If one part breaks, the whole circuit stops working. In a parallel connection, resistors are connected across the same two points. The potential difference is the same across each, but the current divides among them. The total resistance in parallel is less than the smallest individual resistance. This is how most household appliances are wired, allowing them to operate independently.

When current flows through a resistor, some electrical energy is converted into heat. This is called the heating effect of electric current. The amount of heat produced depends on the current, the resistance, and the time for which the current flows. This effect is used in devices like electric heaters and irons. Electric bulbs also use this effect; their filaments, often made of tungsten, get very hot and glow. Fuses are safety devices that use this heating effect to melt and break a circuit if the current becomes too high, protecting appliances. Electric power is the rate at which electrical energy is used. Its unit is the watt. The energy we use is often measured in kilowatt-hours.

Textbook solutions

Intext Questions and Answers I

1. What does an electric circuit mean?

Answer: An electric circuit means a continuous and closed path of an electric current.

2. Define the unit of current.

Answer: The unit of current is the ampere (A). One ampere is constituted by the flow of one coulomb of charge per second, that is, 1 A = 1 C/1 s.

3. Calculate the number of electrons constituting one coulomb of charge.

Answer: To calculate the number of electrons constituting one coulomb of charge, we use the information that an electron possesses a negative charge of 1.6 × 10⁻¹⁹ C.
Let Q be the total charge, which is 1 coulomb (1 C).

Let e be the magnitude of the charge on one electron, which is 1.6 × 10⁻¹⁹ C.
The number of electrons (n) constituting one coulomb of charge is calculated as:

n = Q / e
⇒ n = 1 C / (1.6 × 10⁻¹⁹ C)
⇒ n = 0.625 × 10¹⁹
⇒ n = 6.25 × 10¹⁸ electrons.

The SI unit of electric charge, the coulomb (C), is equivalent to the charge contained in nearly 6 × 10¹⁸ electrons, and our calculation provides a more precise number based on the given charge of an electron.

Intext Questions and Answers II

1. Name a device that helps to maintain a potential difference across a conductor.

Answer: A battery, consisting of one or more electric cells, is a device that helps to maintain a potential difference across a conductor. The chemical action within a cell generates the potential difference across the terminals of the cell. In order to maintain the current in a given electric circuit, the cell has to expend its chemical energy stored in it.

2. What is meant by saying that the potential difference between two points is 1 V?

Answer: Saying that the potential difference between two points is 1 V means that one volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer: The electric potential difference (V) between two points in an electric circuit carrying some current is defined as the work done (W) to move a unit charge (Q) from one point to the other, so Potential difference (V) = Work done (W)/Charge (Q). From this, the work done W, which represents the energy given, is equal to VQ. For a 6 V battery, the potential difference V is 6 V. For each coulomb of charge, Q is 1 C. Thus, the amount of energy given to each coulomb of charge is 6 V multiplied by 1 C, which equals 6 Joules.

Intext Questions and Answers III

1. On what factors does the resistance of a conductor depend?

Answer: The resistance of a conductor depends directly on its length, inversely on its area of cross-section, and also on the material of the conductor.

2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer: Current will flow more easily through a thick wire of the same material when connected to the same source. This is because the resistance of a uniform metallic conductor is inversely proportional to its area of cross-section. A thicker wire has a larger cross-sectional area, which means it offers less resistance to the flow of current.

3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer: If the resistance of an electrical component remains constant while the potential difference across its two ends decreases to half of its former value, the current through it will also decrease to half of its former value. This is because, according to Ohm’s law, current (I) is directly proportional to the potential difference (V) when resistance (R) is constant (I = V/R).

4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer: Coils of electric toasters and electric irons are made of an alloy rather than a pure metal because the resistivity of an alloy is generally higher than that of its constituent metals. Furthermore, alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices.

5. Use the data in Table 11.2 to answer the following –

Category	Material	Resistivity (Ω m)
Conductors	Silver	1.60 × 10⁻⁸
	Copper	1.62 × 10⁻⁸
	Aluminium	2.63 × 10⁻⁸
	Tungsten	5.20 × 10⁻⁸
	Nickel	6.84 × 10⁻⁸
	Iron	10.0 × 10⁻⁸
	Chromium	12.9 × 10⁻⁸
	Mercury	94.0 × 10⁻⁸
	Manganese	1.84 × 10⁻⁶
Alloys	Constantan (alloy of Cu and Ni)	49 × 10⁻⁶
	Manganin (alloy of Cu, Mn and Ni)	44 × 10⁻⁶
	Nichrome (alloy of Ni, Cr, Mn and Fe)	100 × 10⁻⁶
Insulators	Glass	10¹⁰ – 10¹⁴
	Hard rubber	10¹³ – 10¹⁶
	Ebonite	10¹⁵ – 10¹⁷
	Diamond	10¹² – 10¹³
	Paper (dry)	10¹²

(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Answer: (a) Among iron and mercury, iron is a better conductor. This is because iron has a resistivity of 10.0 × 10⁻⁸ Ω m, while mercury has a resistivity of 94.0 × 10⁻⁸ Ω m. A material with lower resistivity is a better conductor.

(b) Based on the data in Table 11.2, silver is the best conductor, as it has the lowest resistivity of 1.60 × 10⁻⁸ Ω m.

Intext Questions and Answers IV

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Answer:

2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

Net potential = 6 V

Using Ohm’s law V = IR, we have,
6 = I × 25
⟹ I = 6/25
⟹ I = 0.24 Ampere

Now for the 12 Ω resistor, current = 0.24 A

So, using Ohm’s law V = 0.24 × 12 V
⟹ V = 2.88 V

Hence, the reading in the ammeter is 0.24 and voltmeter is 2.88.

Intext Questions and Answers V

1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω, and 10⁶ Ω.

Answer: The formula for the equivalent resistance (Rp) of resistors connected in parallel is: 1/Rp = 1/R1 + 1/R2 + 1/R3 + …

(a) For 1 Ω and 10⁶ Ω in parallel:

Given, R1 = 1 Ω
R2 = 10⁶ Ω

We know,

1/Rp = 1/R1 + 1/R2
⟹ 1/Rp = 1/1 + 1/10⁶
⟹ 1/Rp = (10⁶ + 1) / 10⁶
⟹ Rp = 10⁶ / (10⁶ + 1)
⟹ Rp = 1000000 / 1000001
⟹ Rp = 0.999999 Ω or 1 Ω

(b) For 1 Ω, 10³ Ω, and 10⁶ Ω in parallel:

Give,
R1 = 1 Ω
R2 = 10³ Ω
R3 = 10⁶ Ω

We know,

1/Rp = 1/R1 + 1/R2 + 1/R3
⟹ 1/Rp = 1/1 + 1/10³ + 1/10⁶
⟹ 1/Rp = 1/1 + 1/1000 + 1/1000000
⟹ 1/Rp = (1000000 + 1000 + 1) / 1000000
⟹ 1/Rp = 1001001 / 1000000
⟹ Rp = 1000000 / 1001001
⟹ Rp = 0.999 Ω or 1 Ω

2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer: Given:
Resistance of lamp, R_lamp = 100 Ω
Resistance of toaster, R_toaster = 50 Ω
Resistance of water filter, R_filter = 500 Ω
Voltage of the source, V = 220 V

First, we calculate the total current drawn by the three appliances connected in parallel. The total current is the sum of the currents through each appliance.

Current through lamp, I_lamp = V / R_lamp
⟹ I_lamp = 220 V / 100 Ω
⟹ I_lamp = 2.2 A

Current through toaster, I_toaster = V / R_toaster
⟹ I_toaster = 220 V / 50 Ω
⟹ I_toaster = 4.4 A

Current through water filter, I_filter = V / R_filter
⟹ I_filter = 220 V / 500 Ω
⟹ I_filter = 0.44 A

Total current,
I_total = I_lamp + I_toaster + I_filter
I_total = 2.2 A + 4.4 A + 0.44 A = 7.04 A

The electric iron takes the same amount of current.
Current through the iron, I_iron = 7.04 A.

Now, we find the resistance of the electric iron using Ohm’s law (R = V/I).
Resistance of the iron, R_iron = V / I_iron
⟹ R_iron = 220 V / 7.04 A
⟹ R_iron = 31.25 Ω

The resistance of the electric iron is 31.25 Ω, and the current through it is 7.04 A.

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer: Connecting electrical devices in parallel with a battery instead of in series has several advantages:

One advantage is that a parallel circuit divides the current through the electrical gadgets. This is helpful particularly when each gadget has different resistance and requires different current to operate properly. In a series circuit, the current is constant throughout, which is impracticable if devices need widely different currents.

Another advantage is related to component failure. In a series circuit, when one component fails, the circuit is broken and none of the components works. In a parallel circuit, if one device fails (opens), the other devices connected in parallel can continue to operate independently as they still have a complete path for current.

Additionally, the total resistance in a parallel circuit is decreased. Each device in a parallel circuit gets the full battery voltage, allowing them to operate at their intended power.

4. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Answer: Given resistors: R1 = 2 Ω, R2 = 3 Ω, R3 = 6 Ω.

(a) To get a total resistance of 4 Ω:

This requires a mixed combination. Let’s connect the 3 Ω and 6 Ω resistors in parallel and then connect this combination in series with the 2 Ω resistor.

First, calculate the equivalent resistance of the parallel combination (Rp) of 3 Ω and 6 Ω:

1/Rp = 1/3 + 1/6
⟹ 1/Rp = (2 + 1) / 6
⟹ 1/Rp = 3/6
⟹ 1/Rp = 1/2
⟹ Rp = 2 Ω

Now, add the resistance of the 2 Ω resistor in series with this combination:

Total Resistance, R_total = Rp + 2 Ω
⟹ R_total = 2 Ω + 2 Ω
⟹ R_total = 4 Ω

So, to get 4 Ω, connect the 3 Ω and 6 Ω resistors in parallel, and connect this combination in series with the 2 Ω resistor.

(b) To get a total resistance of 1 Ω:

Let’s try connecting all three resistors in parallel.
⟹ 1/R_total = 1/2 + 1/3 + 1/6
⟹ 1/R_total = (3 + 2 + 1) / 6
⟹ 1/R_total = 6/6
⟹ 1/R_total = 1
⟹ R_total = 1 Ω

So, to get 1 Ω, connect all three resistors (2 Ω, 3 Ω, and 6 Ω) in parallel.

5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer: Given resistors: R1 = 4 Ω, R2 = 8 Ω, R3 = 12 Ω, R4 = 24 Ω.

(a) The highest total resistance:

To obtain the highest resistance, connect the resistors in series.
R_highest = R1 + R2 + R3 + R4
⟹ R_highest = 4 Ω + 8 Ω + 12 Ω + 24 Ω
⟹ R_highest = 48 Ω

(b) The lowest total resistance:

To obtain the lowest resistance, connect the resistors in parallel.
1/R_lowest = 1/R1 + 1/R2 + 1/R3 + 1/R4
⟹ 1/R_lowest = 1/4 + 1/8 + 1/12 + 1/24
⟹ 1/R_lowest = (6 + 3 + 2 + 1) / 24
⟹ 1/R_lowest = 12 / 24 = 1/2
⟹ R_lowest = 2 Ω

Intext Questions and Answers VI

1. Why does the cord of an electric heater not glow while the heating element does?

Answer: The cord of an electric heater does not glow while the heating element does because the heat produced in a resistor is directly proportional to its resistance for a given current, according to Joule’s law of heating. The heating element is made of a material with high resistance, so it becomes very hot and glows when current passes through it; this heating effect is utilised in devices such as electric heaters. The cord, however, is made of a material with very low resistance, such as copper (which has low resistivity and is generally used for electrical transmission lines), and therefore, it generates significantly less heat for the same current and does not glow.

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer: The heat generated can be computed using the relationship where the work done in moving a charge Q through a potential difference V is VQ, and this energy is dissipated as heat.

Given:
Charge, Q = 96000 C
Potential difference, V = 50 V
Time, t = 1 hour = 3600 s (Note: time is not directly needed if using H = VQ)

The heat generated H is given by:
H = VQ
Substituting the given values:
H = 50 V × 96000 C
H = 4,800,000 J

Thus, the heat generated is 4,800,000 joules.

3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Answer: The heat developed can be calculated using Joule’s law of heating, which states that the amount of heat H produced in time t is H = I²Rt.

Given:
Resistance, R = 20 Ω
Current, I = 5 A
Time, t = 30 s

The heat developed H is given by:
H = I²Rt
Substituting the given values:
H = (5 A)² × 20 Ω × 30 s
H = 25 A² × 20 Ω × 30 s
H = 500 × 30 J
H = 15000 J

Thus, the heat developed in 30 s is 15000 joules.

Intext Questions and Answers VII

1. What determines the rate at which energy is delivered by a current?

Answer: The rate at which electric energy is dissipated or consumed in an electric circuit is termed electric power. The power P is given by P = VI, or P = I²R = V²/R.

2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer: The formula for electric power (P) is: P = V × I

Substitute the given values into the formula:
P = 220 V × 5 A
⟹ P = 1100 W

The formula for electrical energy (E) is: E = P × t

t = 2 hours
t = 2 × 60 minutes/hour × 60 seconds/minute
⟹ t = 7200 s

Now,

E = 1100 W × 7200 s
E = 1100 J/s × 7200 s
⟹ E = 7,920,000 J

This can also be expressed in scientific notation: E = 7.92 × 10⁶ J

Exercise Questions and Answers

1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is –

(a) 1/25
(b) 1/5
(c) 5 (d) 25

Answer: Let the initial resistance of the wire be R.

When the wire is cut into five equal parts, the resistance of each part will be R/5, because resistance is directly proportional to length.
Let R₁ = R₂ = R₃ = R₄ = R₅ = R/5.

These five parts are connected in parallel. The equivalent resistance R’ is given by the formula:

1/R’ = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅
⟹ 1/R’ = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)
⟹ 1/R’ = 5/R + 5/R + 5/R + 5/R + 5/R
⟹ 1/R’ = 5 × (5/R)
⟹ 1/R’ = 25/R
⟹ R’ = R/25

We need to find the ratio R/R’.
R/R’ = R / (R/25)
⟹ R/R’ = R × (25/R)
⟹ R/R’ = 25

Therefore, the correct option is (d) 25.

2. Which of the following terms does not represent electrical power in a circuit?

(a) I²R
(b) IR²
(c) VI
(d) V²/R

Answer: The electrical power P is given by P = VI. Alternatively, using Ohm’s Law (V=IR), we can write P = (IR)I = I²R, and P = V(V/R) = V²/R. The term IR² is not included among these expressions for electrical power.

Therefore, the correct option is (b) IR²

3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Answer: First, we calculate the resistance of the bulb, which remains constant.

Rated Voltage (V_rated) = 220 V
Rated Power (P_rated) = 100 W

Using the formula P = V²/R, we find the resistance R:
R = V_rated² / P_rated
⟹ R = (220 V)² / 100 W
⟹ R = 48400 / 100 Ω
⟹ R = 484 Ω

Now, the bulb is operated at a new voltage (V_new) = 110 V.

The new power consumed (P_new) is:
P_new = V_new² / R
⟹ P_new = (110 V)² / 484 Ω
⟹ P_new = 12100 / 484 W
⟹ P_new = 25 W

Therefore, the correct option is (d) 25 W.

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Answer: Let the resistance of each of the two identical wires be R.
The potential difference (V) is the same in both cases.
The formula for heat produced is H = (V²/R_total) × t.

Case 1: Series Combination
The total resistance in series is R_s = R + R = 2R.
The heat produced in series is H_s = (V² / R_s) × t = (V² / 2R) × t.

Case 2: Parallel Combination
The total resistance in parallel is given by 1/R_p = 1/R + 1/R = 2/R ⟹ R_p = R/2.
The heat produced in parallel is H_p = (V² / R_p) × t = (V² / (R/2)) × t = (2V² / R) × t.

Ratio of Heat Produced
We need to find the ratio H_s / H_p.
H_s / H_p = [ (V² / 2R) × t ] / [ (2V² / R) × t ]
H_s / H_p = (V² / 2R) × (R / 2V²)
H_s / H_p = 1 / 4
⟹ H_s : H_p = 1:4

Therefore, the correct option is (c) 1:4.

5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer: The voltmeter is always connected in parallel across the points between which the potential difference is to be measured.

6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer: Part 1: Finding the length of the wire.

Given:
Diameter (d) = 0.5 mm = 0.5 × 10⁻³ m
Radius (r) = d/2 = 0.25 × 10⁻³ m
Resistivity (ρ) = 1.6 × 10⁻⁸ Ω m
Resistance (R) = 10 Ω

First, calculate the cross-sectional area (A):
A = πr² = 3.14 × (0.25 × 10⁻³ m)²
⟹ A = 3.14 × 0.0625 × 10⁻⁶ m²
⟹ A = 0.19625 × 10⁻⁶ m²

Now, use the resistance formula R = ρ(l/A) to find the length (l):
l = (R × A) / ρ
⟹ l = (10 Ω × 0.19625 × 10⁻⁶ m²) / (1.6 × 10⁻⁸ Ω m)
⟹ l = (1.9625 × 10⁻⁶) / (1.6 × 10⁻⁸) m
⟹ l = 122.7 m

Part 2: Change in resistance when the diameter is doubled.

Resistance is inversely proportional to the area of cross-section (R ∝ 1/A), and area is proportional to the square of the diameter (A ∝ d²).
Therefore, R ∝ 1/d².

If the diameter is doubled (d’ = 2d), the new resistance (R’) will be:
R’ ∝ 1/(2d)² ∝ 1/(4d²)

This means R’ = R/4.
New resistance R’ = 10 Ω / 4 = 2.5 Ω.
The resistance changes from 10 Ω to 2.5 Ω, becoming one-fourth of its original value.

7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

I (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between V and I and calculate the resistance of that resistor.

Answer:

Calculation of Resistance:

The resistance (R) is the slope of the V-I graph.
Slope = ΔV / ΔI

Let’s take two points from the data, for instance, the first and the last:
R = (13.2 V – 1.6 V) / (4.0 A – 0.5 A)
⟹ R = 11.6 V / 3.5 A
⟹ R = 3.31 Ω

8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer: Given:
Voltage (V) = 12 V
Current (I) = 2.5 mA = 0.0025 A

Using Ohm’s Law, R = V/I:
R = 12 V / 0.0025 A
⟹ R = 4800 Ω or 4.8 kΩ

9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer: In a series circuit, the current is the same through all components.

First, calculate the total equivalent resistance (R_s):
R_s = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
⟹ R_s = 13.4 Ω

Now, calculate the total current (I) using Ohm’s Law, I = V/R_s:
I = 9 V / 13.4 Ω
⟹ I = 0.67 A

Since the current is the same everywhere in a series circuit, the current flowing through the 12 Ω resistor is 0.67 A.

10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer: Let ‘n’ be the number of resistors required.

Given:
Voltage (V) = 220 V
Total Current (I) = 5 A
Resistance of one resistor (R₁) = 176 Ω

First, find the total equivalent resistance (R_eq) required for the circuit:
R_eq = V / I
⟹ R_eq = 220 V / 5 A
⟹ R_eq = 44 Ω

For ‘n’ identical resistors connected in parallel, the equivalent resistance is R_eq = R₁ / n.
n = R₁ / R_eq
⟹ n = 176 Ω / 44 Ω
⟹ n = 4

Four resistors are required.

11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer: Let the three resistors be R₁, R₂, and R₃, each being 6 Ω.

(i) To get a total resistance of 9 Ω:

Connect two resistors (R₁ and R₂) in parallel, and then connect this combination in series with the third resistor (R₃).

Resistance of the parallel part (R_p):
1/R_p = 1/R₁ + 1/R₂
⟹ R_p = 1/6 + 1/6
⟹ R_p = 2/6
⟹ R_p = 1/3
⟹ R_p = 3 Ω.

Total resistance (R_total) = R_p + R₃ = 3 Ω + 6 Ω = 9 Ω.

(ii) To get a total resistance of 4 Ω:

Connect two resistors (R₁ and R₂) in series, and then connect this combination in parallel with the third resistor (R₃).

Resistance of the series part (R_s):
R_s = R₁ + R₂
= 6 Ω + 6 Ω
= 12 Ω.

Total resistance (R_total):
1/R_total = 1/R_s + 1/R₃
⟹ 1/R_total = 1/12 + 1/6
⟹ 1/R_total = (1 + 2)/12
⟹ 1/R_total = 3/12
⟹ 1/R_total = 1/4
⟹ R_total = 4 Ω.

12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Answer: Let ‘n’ be the number of lamps.

Given:
Power of one bulb (P) = 10 W
Voltage (V) = 220 V
Maximum allowable current (I_max) = 5 A

First, find the current drawn by a single bulb (I_bulb):
I_bulb = P / V
⟹ I_bulb = 10 W / 220 V
⟹ I_bulb = 0.04545 A

Now,

n × I_bulb = I_max
⟹ n = I_max / I_bulb
⟹ n = 5 A / 0.04545 A
⟹ n = 110

A maximum of 110 lamps can be connected.

13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer: Given:
V = 220 V
R_A = 24 Ω
R_B = 24 Ω.

Case 1: Used separately

The current will be the same for each coil.
I_separate = V / R_A
⟹ I_separate = 220 V / 24 Ω
⟹ I_separate = 9.17 A

Case 2: Used in series

Total resistance R_s = R_A + R_B
⟹ R_s = 24 Ω + 24 Ω
⟹ R_s = 48 Ω.

Current I_series = V / R_s
⟹ I_series = 220 V / 48 Ω
⟹ I_series = 4.58 A

Case 3: Used in parallel

Total resistance 1/R_p = 1/R_A + 1/R_B
⟹ R_p = 1/24 + 1/24
⟹ R_p = 2/24
⟹ R_p = 1/12
⟹ R_p = 12 Ω.

Current I_parallel = V / R_p
⟹ I_parallel = 220 V / 12 Ω
⟹ I_parallel = 18.33 A

14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer: Circuit (i):

V = 6 V.
Resistors 1 Ω and 2 Ω are in series.

Total resistance R_s = 1 Ω + 2 Ω = 3 Ω.
Current in the circuit I₁ = V / R_s = 6 V / 3 Ω = 2 A.
The current through the 2 Ω resistor is 2 A.

Power used in the 2 Ω resistor is P₁ = I₁² × R
⟹ P₁ = (2 A)² × 2 Ω
⟹ P₁ = 8 W.

Circuit (ii):

V = 4 V.
Resistors 12 Ω and 2 Ω are in parallel.
The voltage across the 2 Ω resistor is the same as the battery voltage, which is 4 V.

Power used in the 2 Ω resistor is P₂ = V² / R
⟹ P₂ = (4 V)² / 2 Ω
⟹ P₂ = 16 / 2 W
⟹ P₂ = 8 W.

Comparison: The power used in the 2 Ω resistor is the same in both circuits (P₁ = P₂ = 8 W). The ratio of power used is 1:1.

Extras

Additional MCQs (Knowledge Based)

1. What is the SI unit of electric charge?

A. Coulomb
B. Ampere
C. Volt
D. Ohm

Answer: A. Coulomb

2. What property of a conductor opposes the flow of electric current?

A. Potential
B. Charge
C. Power
D. Resistance

Answer: D. Resistance

3. What is the commercial unit of electrical energy?

A. Joule
B. Watt-hour
C. Watt
D. Kilowatt-hour

Answer: D. Kilowatt-hour

4. An ammeter is used to measure electric current. Which statement correctly describes its connection in a circuit?

A. In parallel, across component
B. In series, with component
C. Either series or parallel
D. Only with the source

Answer: B. In series, with component

5. Select the material commonly used for the filament of an electric bulb due to its high melting point.

A. Copper
B. Nichrome
C. Tungsten
D. Aluminium

Answer: C. Tungsten

6. Complete the analogy: Current : Ampere :: Potential Difference : __________.

A. Ohm
B. Volt
C. Watt
D. Coulomb

Answer: B. Volt

7. Resistors in Series : Current Same :: Resistors in Parallel : __________.

A. Voltage Same
B. Current Divides
C. Voltage Divides
D. Resistance Increases

Answer: A. Voltage Same

8. A device used to change the resistance in a circuit, thereby regulating the current without changing the voltage source, is called a:

A. Voltmeter
B. Rheostat
C. Ammeter
D. Galvanometer

Answer: B. Rheostat

9. The heating effect of electric current, where heat H = I²Rt, is known as:

A. Ohm’s Law
B. Faraday’s Law
C. Joule’s Law
D. Ampere’s Law

Answer: C. Joule’s Law

10. The SI unit of electric current is named after which scientist?

A. Alessandro Volta
B. Georg Simon Ohm
C. Andre-Marie Ampere
D. Michael Faraday

Answer: C. Andre-Marie Ampere

11. The relationship V/I = constant (for a conductor at constant temperature) is known as:

A. Joule’s Law
B. Ohm’s Law
C. Coulomb’s Law
D. Volta’s Law

Answer: B. Ohm’s Law

12. Identify the material that is generally a good insulator of electricity from the following:

A. Copper
B. Silver
C. Glass
D. Aluminium

Answer: C. Glass

13. Which of the following is NOT a factor on which the resistance of a conductor directly depends?

A. Length
B. Material
C. Current
D. Cross-sectional area

Answer: C. Current

14. Which of the following statements are true for resistors connected in series?
P. The current through each resistor is the same.
Q. The total resistance is the sum of individual resistances.
R. The voltage across each resistor is the same.
S. The total resistance is less than the smallest individual resistance.

A. P and Q
B. R and S
C. P and S
D. Q and R

Answer: A. P and Q

15. Alloys are often used in electrical heating devices because:
P. Their resistivity is generally higher than that of their constituent metals.
Q. They do not oxidise (burn) readily at high temperatures.
R. Their resistivity is generally lower than that of pure metals.
S. They have very low melting points.

A. P and R
B. Q and S
C. P and Q
D. R and S

Answer: C. P and Q

16. Which of the following equations does NOT correctly represent electric power (P)?

A. P = IR
B. P = VI
C. P = I²R
D. P = V²/R

Answer: A. P = IR

17. Which of the following is NOT a characteristic of a parallel circuit arrangement in domestic wiring?

A. If one appliance fails, others continue to work.
B. Different appliances can draw different currents.
C. The total resistance of the circuit is increased.
D. The voltage across each appliance is the same.

Answer: C. The total resistance of the circuit is increased.

18. What is the conventional direction of electric current?

A. Same as electron flow
B. Opposite to electron flow
C. Perpendicular to electron flow
D. No specific direction

Answer: B. Opposite to electron flow

19. One ampere is constituted by the flow of one coulomb of charge per:

A. Minute
B. Hour
C. Second
D. Millisecond

Answer: C. Second

20. The SI unit of electrical resistivity is:

A. Ohm-meter
B. Ohm/meter
C. Volt-meter
D. Ohm

Answer: A. Ohm-meter

21. How much is one kilowatt-hour (kWh) in joules?

A. 3.6 x 10³ J
B. 3.6 x 10⁶ J
C. 3600 J
D. 1000 J

Answer: B. 3.6 x 10⁶ J

22. An electric fuse is connected in a circuit to protect appliances. How is it connected with the device?

A. In parallel
B. In series
C. Either series or parallel
D. After the device

Answer: B. In series

23. If the length of a conducting wire is doubled, and its area of cross-section is halved, its resistance will become:

A. Halved
B. Doubled
C. Four times
D. One-fourth

Answer: C. Four times

24. What is the equivalent resistance when two resistors of 10 Ω and 10 Ω are connected in parallel?

A. 20 Ω
B. 10 Ω
C. 5 Ω
D. 100 Ω

Answer: C. 5 Ω

25. If a current of 0.5 A flows through a bulb for 10 minutes, what is the amount of electric charge that flows?

A. 5 C
B. 50 C
C. 300 C
D. 3000 C

Answer: C. 300 C

26. What is the potential difference across a resistor if 24 J of work is done to move a charge of 2 C across it?

A. 48 V
B. 12 V
C. 0.08 V
D. 26 V

Answer: B. 12 V

27. If the potential difference across a conductor is 1 V and the current through it is 1 A, what is its resistance?

A. 0.5 Ω
B. 2 Ω
C. 1 V/A²
D. 1 Ω

Answer: D. 1 Ω

28. Which material has the lowest resistivity among the following common conductors?

A. Silver
B. Aluminium
C. Tungsten
D. Copper

Answer: A. Silver

29. If three resistors of 2 Ω, 3 Ω, and 4 Ω are connected in series, what is their equivalent resistance?

A. 24 Ω
B. 0.9 Ω
C. 9 Ω
D. Less than 2 Ω

Answer: C. 9 Ω

30. An electric iron is rated 840 W at 220 V. What is the current drawn by it?

A. 0.26 A
B. 3.82 A
C. 184800 A
D. 57.6 A

Answer: B. 3.82 A

31. The heat produced in a resistor is directly proportional to the square of the:

A. Voltage
B. Resistance
C. Current
D. Time

Answer: C. Current

32. What is the power of an electric bulb connected to a 220 V generator if the current is 0.50 A?

A. 440 W
B. 110 W
C. 0.002 W
D. 220.5 W

Answer: B. 110 W

33. A voltmeter is always connected in __________ across the points where potential difference is to be measured.

A. Series
B. Parallel
C. Series-parallel
D. Any way

Answer: B. Parallel

34. The resistivity of alloys is generally __________ than that of their constituent metals.

A. Lower
B. Higher
C. Equal
D. Unrelated

Answer: B. Higher

35. If the resistance of an electrical component is constant and the potential difference across it is halved, the current through it will be:

A. Doubled
B. Halved
C. Unchanged
D. Quadrupled

Answer: B. Halved

36. The potential difference is measured by an instrument called:

A. Rheostat
B. Ohmmeter
C. Ammeter
D. Voltmeter

Answer: D. Voltmeter

37. One milliampere (1 mA) is equal to:

A. 10³ A
B. 10⁻⁶ A
C. 10⁶ A
D. 10⁻³ A

Answer: D. 10⁻³ A

38. If the current through a resistor is doubled, keeping resistance constant, the heat produced will become:

A. Halved
B. Doubled
C. Four times
D. One-fourth

Answer: C. Four times

39. The SI unit of resistance is:

A. Volt
B. Ohm
C. Ampere
D. Watt

Answer: B. Ohm

40. In an electric circuit, the direction of conventional current is taken as the direction of flow of:

A. Electrons
B. Negative charges
C. Neutral particles
D. Positive charges

Answer: D. Positive charges

Additional MCQs (Competency Based)

1. Assertion (A): In an electric circuit, the conventional direction of current is considered from the positive terminal to the negative terminal of the source.
Reason (R): When electricity was first discovered, electrons had not yet been identified, and current was assumed to be the flow of positive charges.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (a) Both A and R are true and R is the correct explanation of A.

2. Assertion (A): An ammeter is always connected in parallel across the component whose current is to be measured.
Reason (R): An ammeter has a very low resistance to ensure it does not significantly alter the current it is measuring.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (d) A is false but R is true.

3. Assertion (A): Alloys like nichrome are commonly used for making heating elements in devices such as electric irons and toasters.
Reason (R): Alloys generally have a higher resistivity compared to their constituent pure metals and do not oxidise readily at high temperatures.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (a) Both A and R are true and R is the correct explanation of A.

4. Assertion (A): When resistors are connected in series, the equivalent resistance is less than the smallest individual resistance.
Reason (R): In a series combination, the current through each resistor is the same.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (d) A is false but R is true.

5. Assertion (A): An electric fuse is connected in parallel with the electrical appliance it is meant to protect.
Reason (R): A fuse wire melts and breaks the circuit if the current exceeds a safe value, thus preventing damage to the appliance.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (d) A is false but R is true.

6. Assertion (A): The commercial unit for electrical energy consumption is the kilowatt-hour (kWh).
Reason (R): One kilowatt-hour is the energy consumed when a device with a power of 1 kilowatt operates for 1 hour.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (a) Both A and R are true and R is the correct explanation of A.

7. A student sets up a circuit with a battery, a switch, an ammeter, and a nichrome wire of length ‘L’ and cross-sectional area ‘A’. The ammeter shows a reading ‘I’. If the student replaces the nichrome wire with another nichrome wire of length ‘2L’ and cross-sectional area ‘A’, what is the most likely new reading on the ammeter, assuming the battery voltage remains constant?

(a) 2I
(b) I
(c) I/2
(d) I/4

Answer: (c) I/2

8. Consider the statement: “One volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.”
Based on this statement, which of the following correctly expresses the relationship between volt (V), joule (J), and coulomb (C)?

(a) 1 V = 1 J × 1 C
(b) 1 C = 1 J × 1 V
(c) 1 J = 1 V / 1 C
(d) 1 V = 1 J / 1 C

Answer: (d) 1 V = 1 J / 1 C

9. The following table shows the resistivity of four materials P, Q, R, and S at 20°C:

Material	Resistivity (Ω m)
P	1.60 × 10⁻⁸
Q	100 × 10⁻⁶
R	10¹⁰
S	5.20 × 10⁻⁸

Which material is most suitable for making the heating element of an electric toaster?

(a) P
(b) Q
(c) R
(d) S

Answer: (b) Q

10. Match the electrical quantities in Column A with their respective SI units in Column B.

Column A (Quantity)	Column B (SI Unit)
(i) Electric Charge	1. Watt (W)
(ii) Electric Current	2. Ohm (Ω)
(iii) Potential Difference	3. Coulomb (C)
(iv) Resistance	4. Ampere (A)
(v) Electric Power	5. Volt (V)

Select the correct code:

(a) (i)–3, (ii)–4, (iii)–5, (iv)–2, (v)–1
(b) (i)–4, (ii)–3, (iii)–2, (iv)–5, (v)–1
(c) (i)–3, (ii)–5, (iii)–4, (iv)–1, (v)–2
(d) (i)–5, (ii)–4, (iii)–3, (iv)–2, (v)–1

Answer: (a) (i)–3, (ii)–4, (iii)–5, (iv)–2, (v)–1

11. Arrange the following steps in the correct sequence to verify Ohm’s Law for a conductor:
(i) Plot a graph between V (potential difference) and I (current).
(ii) Note the readings of the ammeter (for current I) and voltmeter (for potential difference V).
(iii) Set up a circuit containing a battery, a key, a rheostat, a resistor, an ammeter in series, and a voltmeter in parallel across the resistor.
(iv) Vary the current in the circuit by adjusting the rheostat and repeat step (ii) for several different values.
(v) Observe if the V-I graph is a straight line passing through the origin.

(a) (iii) → (ii) → (iv) → (i) → (v)
(b) (ii) → (iii) → (iv) → (v) → (i)
(c) (iii) → (iv) → (ii) → (i) → (v)
(d) (i) → (ii) → (iii) → (iv) → (v)

Answer: (a) (iii) → (ii) → (iv) → (i) → (v)

12. A graph is plotted with potential difference (V) on the y-axis and current (I) on the x-axis for a metallic conductor at a constant temperature. The graph is a straight line passing through the origin, with a slope ‘m’. What does this slope ‘m’ represent?

(a) The resistivity of the conductor.
(b) The resistance of the conductor.
(c) The reciprocal of the resistance of the conductor.
(d) The power dissipated in the conductor.

Answer: (b) The resistance of the conductor.

13. An electric heater is rated 1500 W. If it is used for 2 hours daily, and the cost of electrical energy is Rs 5.00 per kWh, what is the cost of using the heater for 10 days?

(a) Rs 75
(b) Rs 150
(c) Rs 300
(d) Rs 15

Answer: (b) Rs 150

14. A student has three resistors of 2 Ω, 3 Ω, and 6 Ω. If they are connected in parallel, what is the equivalent resistance of the combination?

(a) 11 Ω
(b) 1 Ω
(c) 0.9 Ω
(d) 6 Ω

Answer: (b) 1 Ω

15. Consider the following statements about an electric bulb filament:
Statement 1: The filament is made of a material with a high melting point, like tungsten.
Statement 2: The filament is designed to dissipate most of the electrical energy as light rather than heat.
Which of the following is valid?

(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.

Answer: (c) Statement 1 is true, and Statement 2 is false.

16. An electric circuit contains a cell, a key, an ammeter, and two resistors R₁ and R₂ connected in series. The ammeter reads 0.5 A. If a third resistor R₃ is added in series with R₁ and R₂, how will the ammeter reading likely change, assuming the cell’s voltage remains constant?

(a) It will increase.
(b) It will decrease.
(c) It will remain the same.
(d) It will become zero.

Answer: (b) It will decrease.

17. “The heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor.”
This describes:

(a) Ohm’s Law
(b) Coulomb’s Law
(c) Joule’s Law of heating
(d) Faraday’s Law

Answer: (c) Joule’s Law of heating

18. Match the circuit component in Column A with its primary function in Column B.

Column A (Component)	Column B (Function)
(i) Voltmeter	1. Regulates current without changing voltage source
(ii) Rheostat	2. Measures electric current
(iii) Ammeter	3. Provides potential difference
(iv) Electric Cell	4. Measures potential difference

Select the correct code:

(a) (i)–4, (ii)–1, (iii)–2, (iv)–3
(b) (i)–2, (ii)–4, (iii)–1, (iv)–3
(c) (i)–4, (ii)–3, (iii)–2, (iv)–1
(d) (i)–1, (ii)–2, (iii)–3, (iv)–4

Answer: (a) (i)–4, (ii)–1, (iii)–2, (iv)–3

19. Arrange the following events in the correct order when an electric fuse protects an appliance from an unduly high current:
(i) The temperature of the fuse wire increases.
(ii) The circuit breaks.
(iii) Current larger than the specified value flows through the circuit.
(iv) The fuse wire melts.

(a) (iii) → (i) → (iv) → (ii)
(b) (i) → (iii) → (iv) → (ii)
(c) (iii) → (iv) → (i) → (ii)
(d) (ii) → (iii) → (i) → (iv)

Answer: (a) (iii) → (i) → (iv) → (ii)

20. A wire of resistance R is cut into three equal parts. These parts are then connected in parallel. What is the equivalent resistance of this new combination?

(a) R/3
(b) R/9
(c) 3R
(d) 9R

Answer: (b) R/9

21. A current of 0.2 A flows through a conductor for 5 minutes. The amount of electric charge that flows through any cross-section of the conductor is:

(a) 1 C
(b) 100 C
(c) 60 C
(d) 0.04 C

Answer: (c) 60 C

22. Assertion (A): In a parallel combination of resistors, the potential difference across each resistor is the same.
Reason (R): The total current in a parallel circuit divides itself among the branches according to the resistance of each branch.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (b) Both A and R are true but R does not explain A.

23. A device consumes 100 J of energy in 5 seconds. What is its power?

(a) 500 W
(b) 20 W
(c) 0.05 W
(d) 100 W

Answer: (b) 20 W

24. A student observes that the cord of an electric heater does not glow while its heating element does. This is because:

(a) The cord has a much higher resistance than the heating element.
(b) The heating element has a much higher resistance than the cord.
(c) The current in the cord is less than the current in the heating element.
(d) The cord is better insulated than the heating element.

Answer: (b) The heating element has a much higher resistance than the cord.

Additional Questions and Answers

1. What is electric current?

Answer: If an electric charge flows through a conductor, for example, through a metallic wire, we say that there is an electric current in the conductor. Electric current is expressed by the amount of charge flowing through a particular area in unit time. In other words, it is the rate of flow of electric charges.

2. What is the SI unit of electric charge?

Answer: The SI unit of electric charge is coulomb (C).

3. Which particles constitute the flow of electric charges in metallic conductors?

Answer: In circuits using metallic wires, electrons constitute the flow of charges.

4. Why is the conventional direction of electric current taken opposite to the flow of electrons?

Answer: Electrons were not known at the time when the phenomenon of electricity was first observed. So, electric current was considered to be the flow of positive charges, and the direction of flow of positive charges was taken to be the direction of electric current. Conventionally, in an electric circuit, the direction of electric current is taken as opposite to the direction of the flow of electrons, which are negative charges.

5. What causes electric charges to flow in a conductor?

Answer: For the flow of charges in a conducting metallic wire, electrons move only if there is a difference of electric pressure – called the potential difference – along the conductor. This difference of potential may be produced by a battery, consisting of one or more electric cells. The chemical action within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When the cell is connected to a conducting circuit element, the potential difference sets the charges in motion in the conductor and produces an electric current.

6. After which scientist is the unit of electric current named?

Answer: The electric current is expressed by a unit called ampere (A), named after the French scientist, Andre-Marie Ampere (1775–1836).

7. After which scientist is the unit of electric potential difference named?

Answer: The SI unit of electric potential difference is volt (V), named after Alessandro Volta (1745–1827), an Italian physicist.

8. After which scientist is the unit of electrical resistance named?

Answer: The SI unit of electrical resistance is ohm (Ω), named after the German physicist Georg Simon Ohm (1787-1854) who found out the relationship between the current flowing in a metallic wire and the potential difference across its terminals.

9. How should an ammeter be connected in a circuit?

Answer: An instrument called an ammeter measures electric current in a circuit. It is always connected in series in a circuit through which the current is to be measured.

10. What symbol is used to represent a variable resistor in circuit diagrams?

Answer: A variable resistance or rheostat is represented by a resistor symbol with an arrow diagonally across it, or by a resistor symbol with an arrow pointing to its sliding contact from above.

11. Define electric potential difference between two points.

Answer: The electric potential difference between two points in an electric circuit carrying some current is defined as the work done to move a unit charge from one point to the other. Potential difference (V) between two points = Work done (W)/Charge (Q).

12. What is resistivity and what is its SI unit?

Answer: Resistivity, denoted by ρ (rho), is a constant of proportionality related to the resistance of a conductor, its length, and its area of cross-section. It is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is Ω m. It is a characteristic property of the material.

13. State Ohm’s law for a metallic conductor.

Answer: Ohm’s law states that the potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same.

14. What is meant by the heating effect of electric current?

Answer: If an electric circuit is purely resistive, that is, a configuration of resistors only connected to a battery, the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current.

15. State Joule’s law of heating.

Answer: Joule’s law of heating states that the heat H produced in a resistor is given by H = I²Rt. The law implies that heat produced in a resistor is:

(i) directly proportional to the square of current for a given resistance,
(ii) directly proportional to resistance for a given current, and
(iii) directly proportional to the time for which the current flows through the resistor.

16. Define electric power and give its SI unit.

Answer: Electric power is the rate at which electric energy is dissipated or consumed in an electric circuit. The SI unit of electric power is watt (W).

17. What is the commercial unit of electrical energy and how many joules does one unit represent?

Answer: The commercial unit of electric energy is kilowatt hour (kW h), commonly known as ‘unit’. One kilowatt hour (1 kW h) is equal to 3.6 × 10⁶ joules (J).

18. Why is tungsten used for filaments in electric bulbs?

Answer: In an electric bulb, the filament must retain as much of the heat generated as is possible, so that it gets very hot and emits light. It must not melt at such high temperature. A strong metal with a high melting point such as tungsten (melting point 3380°C) is used for making bulb filaments.

19. What is the function of a fuse in an electric circuit?

Answer: A fuse used in electric circuits protects circuits and appliances by stopping the flow of any unduly high electric current.

20. Why are alloys such as nichrome preferred for heating elements instead of pure metals?

Answer: Alloys are commonly used in electrical heating devices because the resistivity of an alloy is generally higher than that of its constituent metals, and alloys do not oxidise (burn) readily at high temperatures.

21. Explain how the resistance of a conductor depends on its length and on its area of cross-section.

Answer: The resistance of a conductor depends on its length and its area of cross-section in the following ways:

Observations from experiments show that the ammeter reading, which indicates the current, decreases to one-half when the length of the wire is doubled, implying that resistance increases with length. The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit, implying that resistance decreases with an increase in the area of cross-section.

Precise measurements have shown that the resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A).

22. Explain the concept of resistivity and show how it relates to the resistance of a conductor.

Answer: Resistivity is a characteristic property of the material of a conductor. Since the resistance (R) of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A), we can write R ∝ l/A.

This can be expressed as R = ρ (l/A), where ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor.

The SI unit of resistivity is Ω m. Metals and alloys have very low resistivity in the range of 10⁻⁸ Ω m to 10⁻⁶ Ω m, making them good conductors of electricity. Insulators like rubber and glass have resistivity of the order of 10¹² to 10¹⁷ Ω m. Both the resistance and resistivity of a material vary with temperature. The resistivity of an alloy is generally higher than that of its constituent metals.

23. Discuss the heating effect of electric current and derive the expression for the heat produced in a resistor.

Answer: The heating effect of electric current occurs when an electric circuit is purely resistive, meaning it consists of a configuration of resistors only connected to a battery. In such a circuit, the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current.

To derive the expression for the heat produced in a resistor, consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V. Let t be the time during which a charge Q flows across. The work done in moving the charge Q through a potential difference V is VQ. Therefore, the source must supply energy equal to VQ in time t.

The power input P to the circuit by the source is P = VQ/t. Since I = Q/t, power P = VI.

The energy supplied to the circuit by the source in time t is P × t, that is, VIt. This energy gets dissipated in the resistor as heat. Thus, for a steady current I, the amount of heat H produced in time t is:
H = VIt

Applying Ohm’s law, V = IR. Substituting this into the equation for H:
H = (IR)It
H = I²Rt

This is known as Joule’s law of heating. The law implies that heat produced in a resistor is:

(i) directly proportional to the square of current for a given resistance,
(ii) directly proportional to resistance for a given current, and
(iii) directly proportional to the time for which the current flows through the resistor.

24. Describe three practical applications of the heating effect of electric current.

Answer: Three practical applications of the heating effect of electric current are:

Electrical Heating Appliances: Devices such as the electric laundry iron, electric toaster, electric oven, electric kettle, and electric heater are familiar devices based on Joule’s heating. Alloys are commonly used in these devices because they do not oxidise (burn) readily at high temperatures.
Electric Bulb: Electric heating is used to produce light, as in an electric bulb. The filament of the bulb must retain as much of the heat generated as is possible, so that it gets very hot and emits light. A strong metal with a high melting point, such as tungsten, is used for making bulb filaments.
Electric Fuse: A fuse used in electric circuits is another common application of Joule’s heating. It protects circuits and appliances by stopping the flow of any unduly high electric current. The fuse consists of a piece of wire made of a metal or an alloy of appropriate melting point. If a current larger than the specified value flows, the fuse wire melts and breaks the circuit.

25. Explain how electric power is related to current and potential difference, and outline how electrical energy consumption is measured commercially.

Answer: Electric power is the rate at which electric energy is dissipated or consumed in an electric circuit. The power P is given by the product of the potential difference V and the current I:
P = VI

Using Ohm’s law (V = IR), electric power can also be expressed as:
P = I²R
Or, P = V²/R

The SI unit of electric power is watt (W). One watt is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.

Commercially, electrical energy consumption is measured in kilowatt-hour (kW h), commonly known as a ‘unit’. One watt-hour (Wh) is the energy consumed when 1 watt of power is used for 1 hour. A kilowatt-hour is the energy consumed when 1 kilowatt of power is used for 1 hour.
1 kW h = 1000 watt × 3600 second
1 kW h = 3.6 × 10⁶ watt second
1 kW h = 3.6 × 10⁶ joule (J)

26. Discuss the role and working principle of a fuse in protecting electrical circuits.

Answer: The role of an electric fuse is to protect circuits and appliances by stopping the flow of any unduly high electric current.

The working principle of a fuse is based on Joule’s heating effect of electric current. The fuse is placed in series with the device it is meant to protect. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example, aluminium, copper, iron, or lead. If a current larger than the specified value (the fuse rating) flows through the circuit, the temperature of the fuse wire increases due to the heating effect (H = I²Rt). This melts the fuse wire, which breaks the circuit and stops the current flow, thereby preventing damage to the appliance or circuit due to overheating. The fuse wire is usually encased in a cartridge of porcelain or similar material with metal ends.

27. Explain how the material and design of a bulb filament affect its temperature, light output and lifespan.

Answer: The material and design of a bulb filament significantly affect its temperature, light output, and lifespan:

Material: The filament must be made of a material with a very high melting point. A strong metal such as tungsten, which has a melting point of 3380°C, is used for making bulb filaments. This allows the filament to reach a very high temperature without melting.
Temperature and Light Output: The filament must retain as much of the heat generated as is possible so that it gets very hot and emits light. To achieve this, the filament should be thermally isolated as much as possible, using insulating supports. Most of the power consumed by the filament appears as heat, but a small part of it is in the form of light radiated when the filament is sufficiently hot.
Lifespan: To prolong the life of the filament, bulbs are usually filled with chemically inactive gases like nitrogen and argon. These gases reduce the rate of evaporation of the tungsten filament at high temperatures, thus extending its operational life.

28. Explain why domestic electrical appliances are connected in parallel rather than in series.

Answer: Domestic electrical appliances are connected in parallel rather than in series for several important reasons:

Independent Operation and Correct Current: In a parallel circuit, each appliance can be operated independently with its own switch. A parallel circuit divides the current through the electrical gadgets. This is helpful, particularly when each gadget has different resistance and requires different current to operate properly. In a series circuit, the current is constant throughout, which is impracticable for appliances needing widely different current values.
Continued Operation if One Fails: If one appliance in a parallel circuit fails or is switched off, the other appliances continue to operate. In a series circuit, if one component fails, the circuit is broken, and none of the components work.
Constant Voltage: Each appliance in a parallel circuit receives the same voltage from the source (e.g., 220 V in domestic supply lines). This ensures that all appliances operate at their rated voltage.
Lower Overall Resistance: The total resistance in a parallel circuit is decreased, which allows for a larger total current to be drawn from the supply if needed, without excessive voltage drop across the connecting wires.

Ron'e Dutta

Ron'e Dutta is a journalist, teacher, aspiring novelist, and blogger who manages Online Free Notes. An avid reader of Victorian literature, his favourite book is Wuthering Heights by Emily Brontë. He dreams of travelling the world. You can connect with him on social media. He does personal writing on ronism.

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