CBSE Class X Science Chapter 11 Electricity Questions and Answers

Class X Science Solution Guide

Chapter 11: Electricity - Complete In-Text & End-of-Chapter Questions and Answers

In-Text Questions (Page 172)

Q1. What does an electric circuit mean?

An electric circuit is a continuous and closed path along which an electric current flows.

Q2. Define the unit of current.

The SI unit of electric current is the ampere (A). One ampere is defined as the current constituted by the flow of one coulomb of charge per second through a conductor ($1\text{ A} = 1\text{ C/s}$).

Q3. Calculate the number of electrons constituting one coulomb of charge.

The charge on one electron, $e = 1.6 \times 10^{-19}\text{ C}$.

Total charge, $Q = 1\text{ C}$.

Using the formula $Q = n \cdot e$, where $n$ is the number of electrons:

n = Q / e n = 1 C / (1.6 × 10⁻¹⁹ C) n = 6.25 × 10¹⁸ electrons (approximately 6 × 10¹⁸ electrons)

Thus, nearly $6 \times 10^{18}$ electrons constitute one coulomb of charge.

In-Text Questions (Page 173)

Q1. Name a device that helps to maintain a potential difference across a conductor.

A cell or a battery (consisting of one or more electric cells) helps to maintain a potential difference across a conductor.

Q2. What is meant by saying that the potential difference between two points is 1 V?

The potential difference between two points is said to be 1 volt (1 V) when 1 joule of work is done to move a charge of 1 coulomb from one point to the other ($1\text{ V} = 1\text{ J/C}$).

Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Given:

• Potential difference, $V = 6\text{ V}$
• Charge, $Q = 1\text{ C}$

Energy given is equal to the work done ($W$):

W = V × Q W = 6 V × 1 C = 6 J

Therefore, 6 joules of energy is given to each coulomb of charge.

In-Text Questions (Page 181)

Q1. On what factors does the resistance of a conductor depend?

The resistance of a uniform conductor depends on the following factors:

• Its length ($l$) – resistance is directly proportional to length ($R \propto l$).
• Its area of cross-section ($A$) – resistance is inversely proportional to cross-sectional area ($R \propto \frac{1}{A}$).
• The nature of its material.
• Temperature (resistance varies with temperature).

Q2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Current will flow more easily through a thick wire.

Reason: Resistance of a conductor is inversely proportional to its area of cross-section ($R \propto \frac{1}{A}$). A thick wire has a larger area of cross-section and therefore lower resistance compared to a thin wire of the same material and length. Lower resistance allows current to flow more easily.

Q3. 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?

According to Ohm's law, $I = \frac{V}{R}$.

If the resistance $R$ remains constant and the potential difference $V$ decreases to half ($V' = \frac{V}{2}$), the new current $I'$ will be:

I' = V' / R = (V / 2) / R = (1 / 2) × (V / R) = I / 2

Therefore, the current flowing through the component will also decrease to half of its former value.

Q4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Coils of electric toasters and electric irons are made of an alloy rather than a pure metal for the following reasons:

• The resistivity of an alloy is generally higher than that of its constituent metals.
• Alloys do not oxidize (burn) readily at high temperatures.

Q5. Use the data in Table 11.2 to answer the following:

1. Which among iron and mercury is a better conductor?
2. Which material is the best conductor?

From Table 11.2:

• Resistivity of Iron = $10.0 \times 10^{-8}\ \Omega\text{m}$
• Resistivity of Mercury = $94.0 \times 10^{-8}\ \Omega\text{m}$

(a) Iron is a better conductor than mercury because it has a lower resistivity.

(b) Silver is the best conductor because it has the lowest resistivity ($1.60 \times 10^{-8}\ \Omega\text{m}$) among all listed materials.

In-Text Questions (Page 185)

Q1. 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.

A battery of three cells of $2\text{ V}$ each gives a total potential difference of $2\text{ V} + 2\text{ V} + 2\text{ V} = 6\text{ V}$. All three resistors ($5\ \Omega$, $8\ \Omega$, $12\ \Omega$), the battery, and the plug key are connected in a single series loop.

Figure 11.14: Circuit diagram showing a battery of three 2V cells, key, and 5 Ω, 8 Ω, 12 Ω resistors in series.

Q2. 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?

Figure 11.15: Circuit diagram with ammeter connected in series and voltmeter in parallel across the 12 Ω resistor.

Calculations for Readings:

1. Total resistance in series combination ($R_s$):

R_s = R₁ + R₂ + R₃ = 5 Ω + 8 Ω + 12 Ω = 25 Ω

2. Total potential difference ($V$) = $6\text{ V}$.

3. Ammeter reading ($I$): Since components are in series, current remains the same throughout the circuit:

I = V / R_s = 6 V / 25 Ω = 0.24 A

The ammeter reading is 0.24 A.

4. Voltmeter reading ($V_{12}$): Potential difference across the $12\ \Omega$ resistor:

V₁₂ = I × R = 0.24 A × 12 Ω = 2.88 V

The voltmeter reading is 2.88 V.

In-Text Questions (Page 188)

Q1. Judge the equivalent resistance when the following are connected in parallel:

1. 1 Ω and 10⁶ Ω
2. 1 Ω and 10³ Ω, and 10⁶ Ω

When resistors are connected in parallel, the equivalent resistance ($R_p$) is always less than the smallest individual resistance.

(a) Among $1\ \Omega$ and $10^6\ \Omega$, the smallest resistance is $1\ \Omega$. Therefore, the equivalent resistance is less than 1 Ω (approximately equal to $1\ \Omega$).

(b) Among $1\ \Omega$, $10^3\ \Omega$, and $10^6\ \Omega$, the smallest resistance is $1\ \Omega$. Therefore, the equivalent resistance is less than 1 Ω (approximately equal to $0.999\ \Omega$).

Q2. 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?

Given resistances in parallel:

• $R_1 = 100\ \Omega$
• $R_2 = 50\ \Omega$
• $R_3 = 500\ \Omega$
• Supply voltage, $V = 220\text{ V}$

1. Equivalent resistance ($R_p$) of the three appliances:

1/R_p = 1/R₁ + 1/R₂ + 1/R₃ 1/R_p = 1/100 + 1/50 + 1/500 1/R_p = (5 + 10 + 1) / 500 = 16 / 500 R_p = 500 / 16 = 31.25 Ω

2. Total current ($I$) drawn from the source by all three appliances:

I = V / R_p = 220 V / 31.25 Ω = 7.04 A

3. An electric iron connected to the same source that takes as much current as all three appliances must have a resistance equal to the equivalent resistance of the three appliances:

• Resistance of the electric iron = 31.25 Ω
• Current through the electric iron = 7.04 A

Q3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

The advantages of connecting electrical devices in parallel are:

• In a parallel circuit, each electrical device gets the full line voltage (potential difference).
• If one component fails or stops working, the circuit is not broken and other devices continue to work independently.
• Parallel circuit divides the current among gadgets according to their requirement, which is useful when devices have different resistances and require different currents to operate properly.
• The total resistance of the circuit decreases, which keeps the current supply high.

Q4. 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?

(a) To get total resistance of 4 Ω:

Connect $3\ \Omega$ and $6\ \Omega$ resistors in parallel, and then connect this combination in series with the $2\ \Omega$ resistor.

Parallel combination of 3 Ω and 6 Ω: 1/R_p = 1/3 + 1/6 = (2 + 1) / 6 = 3 / 6 = 1 / 2 R_p = 2 Ω Total Resistance = R_p + 2 Ω = 2 Ω + 2 Ω = 4 Ω

(b) To get total resistance of 1 Ω:

Connect all three resistors ($2\ \Omega$, $3\ \Omega$, and $6\ \Omega$) in parallel.

1/R_p = 1/2 + 1/3 + 1/6 1/R_p = (3 + 2 + 1) / 6 = 6 / 6 = 1 R_p = 1 Ω

Q5. 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?

Given resistances: $4\ \Omega$, $8\ \Omega$, $12\ \Omega$, $24\ \Omega$.

(a) Highest resistance: Obtained by connecting all four coils in series:

R_max = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω

(b) Lowest resistance: Obtained by connecting all four coils in parallel:

1/R_min = 1/4 + 1/8 + 1/12 + 1/24 1/R_min = (6 + 3 + 2 + 1) / 24 = 12 / 24 = 1 / 2 R_min = 2 Ω

In-Text Questions (Page 190)

Q1. Why does the cord of an electric heater not glow while the heating element does?

The heating element of an electric heater is made of an alloy (like nichrome) having high resistance. When current flows through it, according to Joule's law of heating ($H = I^2Rt$), a large amount of heat is produced, making it red-hot and glowing.

On the other hand, the cord connecting the heater is made of copper or aluminum, which has very low resistance. Therefore, negligible heat is produced in the cord, and it does not glow.

Q2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Given:

• Charge, $Q = 96000\text{ C}$
• Time, $t = 1\text{ hour} = 3600\text{ s}$
• Potential difference, $V = 50\text{ V}$

Heat generated ($H$) is given by $H = V \times Q$:

H = 50 V × 96000 C H = 4800000 J = 4.8 × 10⁶ J

The heat generated is $4.8 \times 10^6\text{ J}$.

Q3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Given:

• Resistance, $R = 20\ \Omega$
• Current, $I = 5\text{ A}$
• Time, $t = 30\text{ s}$

Using Joule's law of heating, $H = I^2Rt$:

H = (5 A)² × 20 Ω × 30 s H = 25 × 20 × 30 J H = 15000 J = 1.5 × 10⁴ J

The heat developed in $30\text{ s}$ is $15000\text{ J}$ (or $1.5 \times 10^4\text{ J}$).

In-Text Questions (Page 192)

Q1. What determines the rate at which energy is delivered by a current?

The electric power determines the rate at which energy is delivered by a current. It depends on the potential difference across the device and the current flowing through it ($P = VI$).

Q2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Given:

• Current, $I = 5\text{ A}$
• Potential difference, $V = 220\text{ V}$
• Time, $t = 2\text{ h} = 2 \times 3600\text{ s} = 7200\text{ s}$

1. Power of the motor ($P$):

P = V × I = 220 V × 5 A = 1100 W

The power of the motor is 1100 W.

2. Energy consumed ($E$):

E = P × t = 1100 W × 7200 s = 7,920,000 J = 7.92 × 10⁶ J (or in commercial units: E = 1.1 kW × 2 h = 2.2 kWh)

The energy consumed is $7.92 \times 10^6\text{ J}$ (or $2.2\text{ kWh}$).

Exercises (Pages 193-194)

Q1. 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

Correct Answer: (d) 25

Explanation:

Resistance of a wire is directly proportional to its length. When cut into five equal parts, the resistance of each part becomes $R_1 = \frac{R}{5}$.

When these five parts are connected in parallel, the equivalent resistance $R'$ is given by:

1/R' = 1/R₁ + 1/R₁ + 1/R₁ + 1/R₁ + 1/R₁ 1/R' = 5 / R₁ = 5 / (R/5) = 25 / R R/R' = 25

Q2. Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R

Correct Answer: (b) IR²

Explanation: Electric power is given by $P = VI$. By Ohm's law ($V = IR$), we can substitute $V$ to get $P = I^2R$ and substituting $I = \frac{V}{R}$ gives $P = \frac{V^2}{R}$. Thus, $IR^2$ does not represent electrical power.

Q3. 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

Correct Answer: (d) 25 W

Explanation:

1. Calculate the resistance of the bulb ($R$), which remains constant:

P = V² / R R = V² / P = (220)² / 100 = 48400 / 100 = 484 Ω

2. Calculate power consumed at $110\text{ V}$:

P' = (V')² / R = (110)² / 484 = 12100 / 484 = 25 W

Q4. 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

Correct Answer: (c) 1:4

Explanation:

Let the resistance of each wire be $R$.

• In series: $R_s = R + R = 2R$
• In parallel: $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2}$

Heat produced across constant potential difference $V$ in time $t$ is $H = \frac{V^2}{R} t$:

H_series = (V² / R_s) × t = (V² / 2R) × t H_parallel = (V² / R_p) × t = (V² / (R/2)) × t = (2V² / R) × t Ratio = H_series / H_parallel = (V² / 2R) / (2V² / R) = 1 / 4

Thus, the ratio is 1:4.

Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

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

Q6. 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?

Given:

• Diameter, $d = 0.5\text{ mm} = 0.5 \times 10^{-3}\text{ m}$
• Radius, $r = 0.25 \times 10^{-3}\text{ m}$
• Resistivity, $\rho = 1.6 \times 10^{-8}\ \Omega\text{m}$
• Resistance, $R = 10\ \Omega$

1. Calculation of Length ($l$):

R = ρ × (l / A) = ρ × (l / (π × r²)) l = (R × π × r²) / ρ l = [10 × (22 / 7) × (0.25 × 10⁻³)²] / (1.6 × 10⁻⁸) l = [10 × 3.1416 × 0.0625 × 10⁻⁶] / (1.6 × 10⁻⁸) l = 122.72 m

The length of the wire required is 122.72 m.

2. If the diameter is doubled:

Since $R = \frac{\rho l}{A} = \frac{\rho l}{\frac{\pi d^2}{4}}$, resistance is inversely proportional to the square of diameter ($R \propto \frac{1}{d^2}$).

If diameter is doubled ($d' = 2d$), the new resistance $R'$ becomes:

R' = R / (2)² = R / 4 = 10 Ω / 4 = 2.5 Ω

The resistance decreases to one-fourth of its original value (becomes $2.5\ \Omega$).

Q7. 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.

Figure 11.16: V-I graph for the given resistor.

Calculation of Resistance ($R$):

The slope of the $V-I$ graph gives the resistance ($R = \frac{\Delta V}{\Delta I}$).

Taking two points on the graph, for example $(I_1 = 0.5\text{ A}, V_1 = 1.6\text{ V})$ and $(I_2 = 4.0\text{ A}, V_2 = 13.2\text{ V})$:

R = (V₂ - V₁) / (I₂ - I₁) R = (13.2 V - 1.6 V) / (4.0 A - 0.5 A) R = 11.6 / 3.5 = 3.31 Ω

The resistance of the resistor is approximately 3.31 Ω.

Q8. 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.

Given:

• Potential difference, $V = 12\text{ V}$
• Current, $I = 2.5\text{ mA} = 2.5 \times 10^{-3}\text{ A}$

Using Ohm's law, $R = \frac{V}{I}$:

R = 12 V / (2.5 × 10⁻³ A) R = (12 / 2.5) × 10³ Ω R = 4.8 × 10³ Ω = 4800 Ω = 4.8 kΩ

The value of resistance is 4800 Ω (or 4.8 kΩ).

Q9. 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?

Given:

• Potential difference, $V = 9\text{ V}$
• Resistors in series: $0.2\ \Omega$, $0.3\ \Omega$, $0.4\ \Omega$, $0.5\ \Omega$, $12\ \Omega$

1. Equivalent resistance ($R_s$):

R_s = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

2. Total current in the circuit ($I$):

I = V / R_s = 9 V / 13.4 Ω ≈ 0.67 A

In a series circuit, the same current flows through each component. Therefore, the current flowing through the $12\ \Omega$ resistor is 0.67 A.

Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Given:

• Resistance of each resistor, $R = 176\ \Omega$
• Current, $I = 5\text{ A}$
• Voltage, $V = 220\text{ V}$

1. Total required equivalent resistance ($R_p$):

R_p = V / I = 220 V / 5 A = 44 Ω

2. Let $n$ be the number of $176\ \Omega$ resistors connected in parallel:

1 / R_p = n / R 1 / 44 = n / 176 n = 176 / 44 = 4

Therefore, 4 resistors of $176\ \Omega$ in parallel are required.

Q11. 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.

(i) To get equivalent resistance of 9 Ω:

Connect two $6\ \Omega$ resistors in parallel, and connect this parallel group in series with the third $6\ \Omega$ resistor.

Parallel combination of two 6 Ω resistors: 1/R_p = 1/6 + 1/6 = 2/6 = 1/3 => R_p = 3 Ω Total resistance = R_p + 6 Ω = 3 Ω + 6 Ω = 9 Ω

(ii) To get equivalent resistance of 4 Ω:

Connect two $6\ \Omega$ resistors in series, and connect this series group in parallel with the third $6\ \Omega$ resistor.

Series combination of two 6 Ω resistors: R_s = 6 Ω + 6 Ω = 12 Ω Connecting 12 Ω in parallel with third 6 Ω resistor: 1/R_eq = 1/12 + 1/6 = (1 + 2) / 12 = 3 / 12 = 1 / 4 R_eq = 4 Ω

Q12. 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?

Given:

• Power of each bulb, $P = 10\text{ W}$
• Voltage, $V = 220\text{ V}$
• Maximum allowable current, $I = 5\text{ A}$

1. Current drawn by one bulb ($I_1$):

I₁ = P / V = 10 W / 220 V = 1 / 22 A

2. Let $n$ be the total number of bulbs connected in parallel:

Total Current = n × I₁ 5 A = n × (1 / 22) A n = 5 × 22 = 110

Therefore, 110 lamps can be connected in parallel.

Q13. 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?

Given:

• Voltage, $V = 220\text{ V}$
• Resistance of coil A = Resistance of coil B = $24\ \Omega$

Case 1: When coils are used separately:

I = V / R = 220 V / 24 Ω ≈ 9.17 A

Current in each coil is 9.17 A.

Case 2: When coils are used in series:

R_s = 24 Ω + 24 Ω = 48 Ω I = V / R_s = 220 V / 48 Ω ≈ 4.58 A

Current in series combination is 4.58 A.

Case 3: When coils are used in parallel:

1/R_p = 1/24 + 1/24 = 2/24 = 1/12 => R_p = 12 Ω I = V / R_p = 220 V / 12 Ω ≈ 18.33 A

Current in parallel combination is 18.33 A.

Q14. Compare the power used in the 2 Ω resistor in each of the following circuits:

1. a 6 V battery in series with 1 Ω and 2 Ω resistors, and
2. a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

(i) Circuit 1: 6 V battery in series with 1 Ω and 2 Ω resistors:

Total resistance, R_s = 1 Ω + 2 Ω = 3 Ω Circuit current, I = V / R_s = 6 V / 3 Ω = 2 A

In series, current through $2\ \Omega$ resistor is $I = 2\text{ A}$. Power used in $2\ \Omega$ resistor ($P_1$):

P₁ = I² × R = (2 A)² × 2 Ω = 4 × 2 = 8 W

(ii) Circuit 2: 4 V battery in parallel with 12 Ω and 2 Ω resistors:

In parallel, the potential difference across $2\ \Omega$ resistor is equal to source voltage = $4\text{ V}$. Power used in $2\ \Omega$ resistor ($P_2$):

P₂ = V² / R = (4 V)² / 2 Ω = 16 / 2 = 8 W

Comparison: The power used in the $2\ \Omega$ resistor is 8 W in both cases (the ratio is 1:1).

Q15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Given:

• Lamp 1: $P_1 = 100\text{ W}, V = 220\text{ V}$
• Lamp 2: $P_2 = 60\text{ W}, V = 220\text{ V}$

1. Current drawn by Lamp 1 ($I_1$):

I₁ = P₁ / V = 100 W / 220 V = 5 / 11 A

2. Current drawn by Lamp 2 ($I_2$):

I₂ = P₂ / V = 60 W / 220 V = 3 / 11 A

3. Total current drawn from the line ($I$) in parallel:

I = I₁ + I₂ = 5/11 + 3/11 = 8/11 A ≈ 0.727 A

The total current drawn from the line is approximately 0.73 A.

Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

1. Energy consumed by TV set:

• Power, $P_{TV} = 250\text{ W} = 0.25\text{ kW}$
• Time, $t_{TV} = 1\text{ h}$

E_TV = P × t = 250 W × 3600 s = 900,000 J = 9 × 10⁵ J (or 0.25 kWh)

2. Energy consumed by Toaster:

• Power, $P_{toaster} = 1200\text{ W} = 1.2\text{ kW}$
• Time, $t_{toaster} = 10\text{ min} = 10 \times 60\text{ s} = 600\text{ s}$

E_toaster = P × t = 1200 W × 600 s = 720,000 J = 7.2 × 10⁵ J (or 0.20 kWh)

Comparing the two energy values, $9 \times 10^5\text{ J} > 7.2 \times 10^5\text{ J}$. Therefore, the 250 W TV set in 1 hr uses more energy than the 1200 W toaster in 10 minutes.

Q17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Given:

• Resistance, $R = 44\ \Omega$
• Current, $I = 5\text{ A}$
• Time, $t = 2\text{ h}$

The "rate at which heat is developed" refers to power ($P = \frac{H}{t} = I^2R$):

P = I² × R P = (5 A)² × 44 Ω P = 25 × 44 = 1100 J/s (or 1100 W)

The rate at which heat is developed is 1100 J/s (or 1100 W).

Q18. Explain the following.

1. Why is the tungsten used almost exclusively for filament of electric lamps?
2. Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
3. Why is the series arrangement not used for domestic circuits?
4. How does the resistance of a wire vary with its area of cross-section?
5. Why are copper and aluminium wires usually employed for electricity transmission?

(a) Tungsten is used almost exclusively for filaments of electric lamps because it has a very high melting point ($3380^\circ\text{C}$) and high resistivity. It can retain heat to get very hot and emit light without melting.

(b) Conductors of electric heating devices are made of an alloy rather than a pure metal because alloys have higher resistivity than constituent metals and do not oxidize (burn) easily at high temperatures.

(c) Series arrangement is not used in domestic circuits because:

• In series, the voltage gets divided among appliances.
• If one appliance fused or fails, the whole circuit breaks and all other appliances stop working.
• All appliances would operate on a single switch, making independent control impossible.

(d) The resistance of a wire is inversely proportional to its area of cross-section ($R \propto \frac{1}{A}$). Thus, as cross-sectional area increases, resistance decreases.

(e) Copper and aluminium wires are usually employed for electricity transmission because they have very low electrical resistivity, making them efficient conductors that minimize energy loss as heat.