Question: 35 A filament lamp and a 0-10Ω variable resistor are connected in series with a power supply of constant voltage. The graph shows the voltage–current (V–I) characteristic of the filament lamp. V / V 25 20 15 10 5 0 0.0 1.0 2.0 3.0 4.0 I / A When the resistance of the variable resistor is 4.0Ω, the current in the lamp is 2.0A. What is the power dissipated in the lamp when the resistance of the variable resistor is zero? A 12W B 14W C 16W D 28W E 42W F 96W graph: top-right, a curved V-I characteristic rises from near (0.0, 0) through about (1.0, 1.5), (2.0, 5), (3.0, 14), to about (4.0, 24), with axes labeled V / V vertically and I / A horizontally.

1. **State the problem:** We have a filament lamp and a variable resistor connected in series to a constant voltage power supply. The V-I characteristic of the lamp is given. 2. **Given data:** - Resistance of variable resistor $R_v = 4.0\ \Omega$ when current $I = 2.0\ A$. - The lamp's voltage at $I=2.0\ A$ can be read from the graph approximately as $V_{lamp} = 5\ V$ (from the point near (2.0, 5) on the graph). 3. **Find:** Power dissipated in the lamp when the variable resistor resistance is zero ($R_v=0$). 4. **Step 1: Find total voltage of the power supply.** Since the lamp and resistor are in series, total voltage $V_{total} = V_{lamp} + V_{resistor}$. Calculate $V_{resistor} = I \times R_v = 2.0 \times 4.0 = 8.0\ V$. So, $V_{total} = 5 + 8 = 13\ V$. 5. **Step 2: When $R_v=0$, the total voltage is still $13\ V$ (constant voltage supply).** We need to find the current $I$ through the lamp at $V_{total} = 13\ V$ and $R_v=0$. Since $R_v=0$, voltage across lamp $V_{lamp} = 13\ V$. 6. **Step 3: Find current $I$ at $V_{lamp} = 13\ V$ from the graph.** From the graph, at $V=13\ V$, current $I$ is approximately $2.8\ A$ (interpolating between points (3.0,14) and (4.0,24) volts, but since 13V is near 3.0A, we estimate $I \approx 2.8A$). 7. **Step 4: Calculate power dissipated in the lamp:** Power $P = V_{lamp} \times I = 13 \times 2.8 = 36.4\ W$. This is not one of the options, so let's check the graph more carefully. At $I=3.0A$, $V \approx 14V$ (from graph), so at $V=13V$, $I$ is slightly less than 3.0A, say $2.9A$. Calculate power again: $P = 13 \times 2.9 = 37.7W$. Still no match. 8. **Step 5: Reconsider the graph points given:** - At $I=1.0A$, $V \approx 1.5V$ - At $I=2.0A$, $V \approx 5V$ - At $I=3.0A$, $V \approx 14V$ - At $I=4.0A$, $V \approx 24V$ We want $V=13V$, which is between $I=2.0A$ and $I=3.0A$. Interpolate linearly: $$I = 2.0 + \frac{13 - 5}{14 - 5} \times (3.0 - 2.0) = 2.0 + \frac{8}{9} = 2.0 + 0.89 = 2.89A$$ Power: $$P = 13 \times 2.89 = 37.57W$$ Still no match. 9. **Step 6: Check if total voltage is correct.** Total voltage $V_{total} = V_{lamp} + V_{resistor} = 5 + 8 = 13V$. This is correct. 10. **Step 7: Check power options given:** Options are 12, 14, 16, 28, 42, 96 W. Our calculated power is closest to 42W. 11. **Step 8: Calculate power using $P = I^2 R$ for the lamp at $R_v=0$.** Find lamp resistance at $I=2.0A$: $$R_{lamp} = \frac{V}{I} = \frac{5}{2} = 2.5\ \Omega$$ At $V=13V$, $I=2.89A$, lamp resistance: $$R_{lamp} = \frac{13}{2.89} = 4.5\ \Omega$$ Power: $$P = I^2 R = (2.89)^2 \times 4.5 = 8.35 \times 4.5 = 37.57W$$ Again close to 42W. 12. **Step 9: Since 42W is the closest option, the answer is E 42W.** --- **Final answer:** E 42W