Chapter 11: Electricity
[IMAGE PLACEHOLDER: Electric Circuit Components & Ohm's Law Setup]
1-Mark Questions & Solutions
Q1 | 1 Mark
The effective resistance when two identical resistors of resistances '$R$' each connected in parallel is:
(a) $2R$ (b) $\frac{R}{2}$ (c) $R$ (d) $4R$
(a) $2R$ (b) $\frac{R}{2}$ (c) $R$ (d) $4R$
Answer: (b) $\frac{R}{2}$
Calculation: For parallel combination: $$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2}$$
Calculation: For parallel combination: $$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2}$$
Q2 | 1 Mark
A wire of resistance $R$ is cut into two equal halves. The resistance of each half is:
(a) $R$ (b) $\frac{R}{2}$ (c) $2R$ (d) none of these
(a) $R$ (b) $\frac{R}{2}$ (c) $2R$ (d) none of these
Answer: (b) $\frac{R}{2}$
Since resistance is directly proportional to length ($R \propto l$), halving the length halves the resistance.
Since resistance is directly proportional to length ($R \propto l$), halving the length halves the resistance.
Q3 | 1 Mark
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in 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) $4:1$
(a) $1:2$ (b) $2:1$ (c) $1:4$ (d) $4:1$
Answer: (c) $1:4$
Calculation:
Equivalent series resistance: $R_s = R + R = 2R$.
Equivalent parallel resistance: $R_p = \frac{R}{2}$.
Since heat generated at constant potential difference is $H = \frac{V^2}{R} t$: $$\frac{H_s}{H_p} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4} \implies 1:4$$
Calculation:
Equivalent series resistance: $R_s = R + R = 2R$.
Equivalent parallel resistance: $R_p = \frac{R}{2}$.
Since heat generated at constant potential difference is $H = \frac{V^2}{R} t$: $$\frac{H_s}{H_p} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4} \implies 1:4$$
Q4 | 1 Mark
When resistors are combined in series, _______ through each resistor is the same.
(a) resistance (b) potential difference (c) current (d) heat generated
(a) resistance (b) potential difference (c) current (d) heat generated
Answer: (c) current
Q5 | 1 Mark
When the temperature of a metallic conductor increases, its resistance _______.
(a) increases (b) decreases (c) No change (d) initially increases then decreases
(a) increases (b) decreases (c) No change (d) initially increases then decreases
Answer: (a) increases
Q6 | 1 Mark
Working principle of an electrical fuse is based on:
(a) heating effect of electric current (b) chemical effect of electric current (c) magnetic effect of electric current (d) All of these
(a) heating effect of electric current (b) chemical effect of electric current (c) magnetic effect of electric current (d) All of these
Answer: (a) heating effect of electric current
Q7 | 1 Mark
A cylindrical conductor of length '$l$' and uniform cross-section '$A$' has resistance '$R$'. The area of cross-section of another conductor of same material and same resistance but of length '$2l$' is:
(a) $\frac{A}{2}$ (b) $\frac{3A}{2}$ (c) $2A$ (d) $3A$
(a) $\frac{A}{2}$ (b) $\frac{3A}{2}$ (c) $2A$ (d) $3A$
Answer: (c) $2A$
Formula: $R = \rho \frac{l}{A}$. For second conductor: $$R' = \rho \frac{2l}{A'} = R = \rho \frac{l}{A} \implies A' = 2A$$
Formula: $R = \rho \frac{l}{A}$. For second conductor: $$R' = \rho \frac{2l}{A'} = R = \rho \frac{l}{A} \implies A' = 2A$$
Q8 | 1 Mark
An electric bulb is rated $220\text{ V}$ and $100\text{ W}$. When it is operated on $110\text{ V}$, the power consumed will be:
(a) $100\text{ W}$ (b) $75\text{ W}$ (c) $50\text{ W}$ (d) $25\text{ W}$
(a) $100\text{ W}$ (b) $75\text{ W}$ (c) $50\text{ W}$ (d) $25\text{ W}$
Answer: (d) $25\text{ W}$
Resistance of bulb: $$R = \frac{V^2}{P} = \frac{220^2}{100} = 484\text{ }\Omega$$
Power at $110\text{ V}$: $$P' = \frac{V'^2}{R} = \frac{110^2}{484} = 25\text{ W}$$
Resistance of bulb: $$R = \frac{V^2}{P} = \frac{220^2}{100} = 484\text{ }\Omega$$
Power at $110\text{ V}$: $$P' = \frac{V'^2}{R} = \frac{110^2}{484} = 25\text{ W}$$
[IMAGE PLACEHOLDER: Series and Parallel Resistor Circuits]
Q9 | 1 Mark
Assertion (A): Tungsten metal is used for making filaments of incandescent lamps.
Reason (R): The melting point of tungsten is very low.
Reason (R): The melting point of tungsten is very low.
Answer: (c) (A) is true but (R) is false (Tungsten has a extremely high melting point of $3380^\circ\text{C}$).
Q10 | 1 Mark
Assertion (A): Longer wires have greater resistance and smaller wires have lesser resistance.
Reason (R): Resistance is inversely proportional to the length of the wire.
Reason (R): Resistance is inversely proportional to the length of the wire.
Answer: (c) (A) is true but (R) is false ($R \propto l$, resistance is directly proportional to length).
Q11 - Q15 | Additional 1-Mark MCQs
• Q11: If temperature increases due to current flow, $V$ and $R$ change according to temperature dependence $\rightarrow$ Resistance $R$ increases.
• Q13: For semiconductor/carbon materials, resistance decreases with increase in temperature $\rightarrow$ (d) Carbon.
• Q14: Resistivity of Cu-Ni alloy vs Cu-Mn-Ni alloy $\rightarrow$ (c) more / depends on composition.
• Q15: Wire of resistance $R$ cut into 10 equal parts and joined in parallel $\rightarrow$ Each part $= \frac{R}{10}$. $$\frac{1}{R_{new}} = 10 \times \frac{10}{R} = \frac{100}{R} \implies R_{new} = 0.01R \quad \text{Answer: (a) } 0.01R$$
• Q11: If temperature increases due to current flow, $V$ and $R$ change according to temperature dependence $\rightarrow$ Resistance $R$ increases.
• Q13: For semiconductor/carbon materials, resistance decreases with increase in temperature $\rightarrow$ (d) Carbon.
• Q14: Resistivity of Cu-Ni alloy vs Cu-Mn-Ni alloy $\rightarrow$ (c) more / depends on composition.
• Q15: Wire of resistance $R$ cut into 10 equal parts and joined in parallel $\rightarrow$ Each part $= \frac{R}{10}$. $$\frac{1}{R_{new}} = 10 \times \frac{10}{R} = \frac{100}{R} \implies R_{new} = 0.01R \quad \text{Answer: (a) } 0.01R$$
Q16 - Q20 | Conceptual MCQs & Assertions
• Q16: $24\text{ V}$ connected across parallel combination of four $6\text{ }\Omega$ resistors. Current in each resistor: $$I = \frac{V}{R} = \frac{24\text{ V}}{6\text{ }\Omega} = 4\text{ A} \quad \text{Answer: (b) } 4\text{ A}$$
• Q17: Proper series combination of cells for max potential: (a) Positive attached to negative terminals (+ - + -).
• Q18: Terms NOT representing electrical power: (b) $IR^2$ ($P = VI = I^2 R = \frac{V^2}{R}$).
• Q19: Assertion: Voltmeter connected in parallel to measure potential difference. Reason: Voltage in parallel circuits remains same. $\rightarrow$ Answer: (a) Both A and R are true and R explains A.
• Q20: Assertion: Bulbs filled with inactive $N_2$ and $Ar$ gases. Reason: Prevents oxidation of filament to prolong bulb life. $\rightarrow$ Answer: (c) A is true, R is incorrect explanation context.
• Q16: $24\text{ V}$ connected across parallel combination of four $6\text{ }\Omega$ resistors. Current in each resistor: $$I = \frac{V}{R} = \frac{24\text{ V}}{6\text{ }\Omega} = 4\text{ A} \quad \text{Answer: (b) } 4\text{ A}$$
• Q17: Proper series combination of cells for max potential: (a) Positive attached to negative terminals (+ - + -).
• Q18: Terms NOT representing electrical power: (b) $IR^2$ ($P = VI = I^2 R = \frac{V^2}{R}$).
• Q19: Assertion: Voltmeter connected in parallel to measure potential difference. Reason: Voltage in parallel circuits remains same. $\rightarrow$ Answer: (a) Both A and R are true and R explains A.
• Q20: Assertion: Bulbs filled with inactive $N_2$ and $Ar$ gases. Reason: Prevents oxidation of filament to prolong bulb life. $\rightarrow$ Answer: (c) A is true, R is incorrect explanation context.
2-Mark Questions & Solutions
Q21 | 2 Marks
How is the current flowing in a conductor changed if the resistance of conductor is doubled keeping potential difference across it the same?
Answer: According to Ohm's Law, $I = \frac{V}{R}$. At constant potential difference $V$, current is inversely proportional to resistance ($I \propto \frac{1}{R}$). If $R$ is doubled, current becomes **half** of its initial value ($I' = \frac{I}{2}$).
Q22 | 2 Marks
What happens to resistance and resistivity of a wire when: (i) wire is cut into two equal halves (ii) wire is stretched to double its length?
Answer:
(i) Resistance becomes half ($R' = \frac{R}{2}$); Resistivity remains unchanged (material property).
(ii) On stretching to double length ($l' = 2l$, area $A' = \frac{A}{2}$), resistance becomes 4 times ($R' = 4R$); Resistivity remains unchanged.
(i) Resistance becomes half ($R' = \frac{R}{2}$); Resistivity remains unchanged (material property).
(ii) On stretching to double length ($l' = 2l$, area $A' = \frac{A}{2}$), resistance becomes 4 times ($R' = 4R$); Resistivity remains unchanged.
Q23 | 2 Marks
State Ohm’s law. Draw a labelled circuit diagram to verify this law in the laboratory.
Answer:
• Ohm's Law: The electric current $I$ flowing through a metallic conductor is directly proportional to the potential difference $V$ applied across its ends, provided physical conditions like temperature remain constant: $$V \propto I \implies V = IR$$
• Circuit Components: Battery, Key, Ammeter (series), Voltmeter (parallel across resistor $R$), Rheostat (variable resistor).
• Ohm's Law: The electric current $I$ flowing through a metallic conductor is directly proportional to the potential difference $V$ applied across its ends, provided physical conditions like temperature remain constant: $$V \propto I \implies V = IR$$
• Circuit Components: Battery, Key, Ammeter (series), Voltmeter (parallel across resistor $R$), Rheostat (variable resistor).
Q24 | 2 Marks
Two students plot $V\text{--}I$ graphs for series and parallel combinations of two resistors $R_1$ and $R_2$. Which graph is correctly labeled? Justify.
Answer: Slope of $V\text{--}I$ graph represents resistance ($\text{Slope} = \frac{V}{I} = R$). Since equivalent resistance in series ($R_s = R_1 + R_2$) is greater than in parallel ($R_p = \frac{R_1 R_2}{R_1 + R_2}$), the line with steeper slope corresponds to the **series** combination and smaller slope corresponds to **parallel**.
Q25 | 2 Marks
In a circuit setup, find the ratio of ammeter reading when jockey J is connected to point A ($V_1 = 3\text{ V}$) and then to point B ($V_2 = 10\text{ V}$) across resistance $R = 5\text{ }\Omega$.
Answer:
• Connected to A ($V = 3\text{ V}$): $$I_A = \frac{3\text{ V}}{5\text{ }\Omega} = 0.6\text{ A}$$
• Connected to B ($V = 10\text{ V}$): $$I_B = \frac{10\text{ V}}{5\text{ }\Omega} = 2.0\text{ A}$$
• Ratio $\frac{I_A}{I_B} = \frac{0.6}{2.0} = \frac{3}{10} = 3:10$.
• Connected to A ($V = 3\text{ V}$): $$I_A = \frac{3\text{ V}}{5\text{ }\Omega} = 0.6\text{ A}$$
• Connected to B ($V = 10\text{ V}$): $$I_B = \frac{10\text{ V}}{5\text{ }\Omega} = 2.0\text{ A}$$
• Ratio $\frac{I_A}{I_B} = \frac{0.6}{2.0} = \frac{3}{10} = 3:10$.
Q26 | 2 Marks
How many bulbs of $8\text{ }\Omega$ should be joined in parallel to draw a current of $2\text{ A}$ from a battery of $4\text{ V}$?
Answer:
Total equivalent resistance required: $$R_p = \frac{V}{I} = \frac{4\text{ V}}{2\text{ A}} = 2\text{ }\Omega$$
For $n$ identical resistors of $8\text{ }\Omega$ in parallel: $$\frac{1}{R_p} = \frac{n}{8} \implies \frac{1}{2} = \frac{n}{8} \implies n = 4\text{ bulbs}$$
Total equivalent resistance required: $$R_p = \frac{V}{I} = \frac{4\text{ V}}{2\text{ A}} = 2\text{ }\Omega$$
For $n$ identical resistors of $8\text{ }\Omega$ in parallel: $$\frac{1}{R_p} = \frac{n}{8} \implies \frac{1}{2} = \frac{n}{8} \implies n = 4\text{ bulbs}$$
[IMAGE PLACEHOLDER: Electric Measuring Instruments & Readings]
3-Mark Questions & Solutions
Q27 | 3 Marks
Find the heat generated while transferring $96000\text{ C}$ of charge in one hour through a potential difference of $50\text{ V}$.
Answer:
Given: $Q = 96000\text{ C}$, $V = 50\text{ V}$, $t = 1\text{ h} = 3600\text{ s}$.
Formula: $$H = V \times Q$$ $$H = 50\text{ V} \times 96000\text{ C} = 4,800,000\text{ J} = 4.8 \times 10^6\text{ J} = 4.8\text{ MJ}$$
Given: $Q = 96000\text{ C}$, $V = 50\text{ V}$, $t = 1\text{ h} = 3600\text{ s}$.
Formula: $$H = V \times Q$$ $$H = 50\text{ V} \times 96000\text{ C} = 4,800,000\text{ J} = 4.8 \times 10^6\text{ J} = 4.8\text{ MJ}$$
Q28 | 3 Marks
Write three advantages of connecting electrical appliances in parallel with the mains rather than in series.
Answer:
1) Equal Voltage: Each appliance gets full mains voltage ($220\text{ V}$).
2) Independent Operation: Switching off or failure of one appliance does not affect other appliances.
3) Lower Total Resistance: Parallel arrangement reduces overall circuit resistance, allowing each appliance to draw required current.
1) Equal Voltage: Each appliance gets full mains voltage ($220\text{ V}$).
2) Independent Operation: Switching off or failure of one appliance does not affect other appliances.
3) Lower Total Resistance: Parallel arrangement reduces overall circuit resistance, allowing each appliance to draw required current.
Q29 | 3 Marks
A battery of $9\text{ V}$ is connected in series with resistors $2\text{ }\Omega, 3\text{ }\Omega, 5\text{ }\Omega,$ and $7\text{ }\Omega$. How much current flows through the $7\text{ }\Omega$ resistor?
Answer:
Total series resistance: $$R_s = 2 + 3 + 5 + 7 = 17\text{ }\Omega$$
Circuit current: $$I = \frac{V}{R_s} = \frac{9\text{ V}}{17\text{ }\Omega} \approx 0.53\text{ A}$$
Since current remains same throughout a series circuit, current through $7\text{ }\Omega$ resistor is **$0.53\text{ A}$**.
Total series resistance: $$R_s = 2 + 3 + 5 + 7 = 17\text{ }\Omega$$
Circuit current: $$I = \frac{V}{R_s} = \frac{9\text{ V}}{17\text{ }\Omega} \approx 0.53\text{ A}$$
Since current remains same throughout a series circuit, current through $7\text{ }\Omega$ resistor is **$0.53\text{ A}$**.
Q30 | 3 Marks
Meters show readings $V = 2.4\text{ V}$ and $I = 250\text{ mA}$.
(a) Least counts of meters? (b) Resistance of the resistor?
(a) Least counts of meters? (b) Resistance of the resistor?
Answer:
(a) Least counts: Milliammeter $= 10\text{ mA}$, Voltmeter $= 0.1\text{ V}$.
(b) Current $I = 250\text{ mA} = 0.25\text{ A}$. $$R = \frac{V}{I} = \frac{2.4\text{ V}}{0.25\text{ A}} = 9.6\text{ }\Omega$$
(a) Least counts: Milliammeter $= 10\text{ mA}$, Voltmeter $= 0.1\text{ V}$.
(b) Current $I = 250\text{ mA} = 0.25\text{ A}$. $$R = \frac{V}{I} = \frac{2.4\text{ V}}{0.25\text{ A}} = 9.6\text{ }\Omega$$
Q31 | 3 Marks
Two bulbs rated ($60\text{ W} - 220\text{ V}$) and ($60\text{ W} - 110\text{ V}$) respectively. Calculate the ratio of their resistances.
Answer:
$$R_1 = \frac{V_1^2}{P_1} = \frac{220^2}{60}, \quad R_2 = \frac{V_2^2}{P_2} = \frac{110^2}{60}$$ $$\frac{R_1}{R_2} = \frac{220^2}{110^2} = \left(\frac{220}{110}\right)^2 = 2^2 = 4 \implies 4:1$$
$$R_1 = \frac{V_1^2}{P_1} = \frac{220^2}{60}, \quad R_2 = \frac{V_2^2}{P_2} = \frac{110^2}{60}$$ $$\frac{R_1}{R_2} = \frac{220^2}{110^2} = \left(\frac{220}{110}\right)^2 = 2^2 = 4 \implies 4:1$$
[IMAGE PLACEHOLDER: Electrical Energy & Power Applications]
4-Mark & 5-Mark Questions & Solutions
Q33 | 4 Marks
Given table of resistivities of materials A ($1.6 \times 10^{-8}\text{ }\Omega\text{m}$), B ($6.4 \times 10^{-8}\text{ }\Omega\text{m}$), C ($10 \times 10^{-8}\text{ }\Omega\text{m}$), D ($96 \times 10^{-8}\text{ }\Omega\text{m}$), E ($100 \times 10^{-8}\text{ }\Omega\text{m}$).
(i) Best material for connecting wires? (ii) Material for heating devices? (iii) Wire A has $R = 2\text{ }\Omega$, find resistance of wire D of same length and thickness.
(i) Best material for connecting wires? (ii) Material for heating devices? (iii) Wire A has $R = 2\text{ }\Omega$, find resistance of wire D of same length and thickness.
Answer:
(i) **Material A** (lowest resistivity, excellent conductor).
(ii) **Material E / D** (high resistivity, alloy appropriate for heating element).
(iii) Since $R \propto \rho$ for equal $l$ and $A$: $$\frac{R_D}{R_A} = \frac{\rho_D}{\rho_A} \implies \frac{R_D}{2} = \frac{96 \times 10^{-8}}{1.6 \times 10^{-8}} = 60 \implies R_D = 120\text{ }\Omega$$
(i) **Material A** (lowest resistivity, excellent conductor).
(ii) **Material E / D** (high resistivity, alloy appropriate for heating element).
(iii) Since $R \propto \rho$ for equal $l$ and $A$: $$\frac{R_D}{R_A} = \frac{\rho_D}{\rho_A} \implies \frac{R_D}{2} = \frac{96 \times 10^{-8}}{1.6 \times 10^{-8}} = 60 \implies R_D = 120\text{ }\Omega$$
Q34 | 4 Marks
Define $1\text{ kWh}$. An electric refrigerator rated $400\text{ W}$ operates $8\text{ h/day}$. Find cost of energy to operate it for $30\text{ days}$ at $\text{Rs. } 3.00\text{ per kWh}$.
Answer:
• 1 kWh: Commercial unit of electrical energy, defined as the energy consumed by an appliance of power $1\text{ kW}$ ($1000\text{ W}$) running for $1\text{ hour}$: $$1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J}$$
• Energy consumed in 30 days: $$E = P \times t \times 30 = \left(\frac{400}{1000}\text{ kW}\right) \times 8\text{ h/day} \times 30\text{ days} = 0.4 \times 240 = 96\text{ kWh}$$
• Cost: $96\text{ kWh} \times \text{Rs. } 3.00 = \text{\bf Rs. } 288.00$.
• 1 kWh: Commercial unit of electrical energy, defined as the energy consumed by an appliance of power $1\text{ kW}$ ($1000\text{ W}$) running for $1\text{ hour}$: $$1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J}$$
• Energy consumed in 30 days: $$E = P \times t \times 30 = \left(\frac{400}{1000}\text{ kW}\right) \times 8\text{ h/day} \times 30\text{ days} = 0.4 \times 240 = 96\text{ kWh}$$
• Cost: $96\text{ kWh} \times \text{Rs. } 3.00 = \text{\bf Rs. } 288.00$.
Q37 | 5 Marks
An electric geyser consumes $2.2\text{ units}$ ($2.2\text{ kWh}$) of electrical energy per hour on $220\text{ V}$ mains.
a) Power-rating? b) Current flowing? c) Resistance? d) Does resistance remain constant during operation? e) Cost of energy per hour at $\text{Rs. } 5/\text{unit}$.
a) Power-rating? b) Current flowing? c) Resistance? d) Does resistance remain constant during operation? e) Cost of energy per hour at $\text{Rs. } 5/\text{unit}$.
Answer:
a) Power rating: $$P = \frac{E}{t} = \frac{2.2\text{ kWh}}{1\text{ h}} = 2.2\text{ kW} = 2200\text{ W}$$
b) Current: $$I = \frac{P}{V} = \frac{2200\text{ W}}{220\text{ V}} = 10\text{ A}$$
c) Resistance: $$R = \frac{V}{I} = \frac{220\text{ V}}{10\text{ A}} = 22\text{ }\Omega$$
d) No, resistance increases slightly as heating increases temperature.
e) Cost: $2.2\text{ units} \times \text{Rs. } 5 = \text{\bf Rs. } 11.00$.
a) Power rating: $$P = \frac{E}{t} = \frac{2.2\text{ kWh}}{1\text{ h}} = 2.2\text{ kW} = 2200\text{ W}$$
b) Current: $$I = \frac{P}{V} = \frac{2200\text{ W}}{220\text{ V}} = 10\text{ A}$$
c) Resistance: $$R = \frac{V}{I} = \frac{220\text{ V}}{10\text{ A}} = 22\text{ }\Omega$$
d) No, resistance increases slightly as heating increases temperature.
e) Cost: $2.2\text{ units} \times \text{Rs. } 5 = \text{\bf Rs. } 11.00$.
Q38 | 5 Marks
An electric lamp of resistance $20\text{ }\Omega$ and a conductor of $4\text{ }\Omega$ are connected in series to a $6\text{ V}$ battery.
Calculate: (a) total resistance, (b) total current, (c) potential difference across (i) lamp and (ii) conductor, (d) power of lamp.
Calculate: (a) total resistance, (b) total current, (c) potential difference across (i) lamp and (ii) conductor, (d) power of lamp.
Answer:
(a) Total resistance: $$R = R_1 + R_2 = 20 + 4 = 24\text{ }\Omega$$
(b) Current: $$I = \frac{V}{R} = \frac{6\text{ V}}{24\text{ }\Omega} = 0.25\text{ A}$$
(c) Potential differences:
(i) Across lamp: $$V_1 = I \times R_1 = 0.25\text{ A} \times 20\text{ }\Omega = 5\text{ V}$$
(ii) Across conductor: $$V_2 = I \times R_2 = 0.25\text{ A} \times 4\text{ }\Omega = 1\text{ V}$$
(d) Power of lamp: $$P = I^2 R_1 = (0.25)^2 \times 20 = 0.0625 \times 20 = 1.25\text{ W}$$
(a) Total resistance: $$R = R_1 + R_2 = 20 + 4 = 24\text{ }\Omega$$
(b) Current: $$I = \frac{V}{R} = \frac{6\text{ V}}{24\text{ }\Omega} = 0.25\text{ A}$$
(c) Potential differences:
(i) Across lamp: $$V_1 = I \times R_1 = 0.25\text{ A} \times 20\text{ }\Omega = 5\text{ V}$$
(ii) Across conductor: $$V_2 = I \times R_2 = 0.25\text{ A} \times 4\text{ }\Omega = 1\text{ V}$$
(d) Power of lamp: $$P = I^2 R_1 = (0.25)^2 \times 20 = 0.0625 \times 20 = 1.25\text{ W}$$
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