Question

A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, what will be the power consumed (approximately)? ____.

• A. 121 W
• B. 200 W
• C. 400 W
• D. 331 W

Hint:

You can also use the ratio method: $\frac{P_2}{P_1} = \left(\frac{V_2}{V_1}\right)^2$. This avoids calculating $R$ explicitly and reduces the chance of rounding errors in the middle of the problem.

Answer 1

The Correct Option is D

Step 1: Understanding the Topic:
This question deals with "Electric Power" in DC and AC circuits. The core concept is that a heating element (like a resistor) has a fixed physical property called resistance. While its power output depends on the voltage applied, its resistance remains constant (assuming temperature changes don't drastically alter the material's resistivity).
Step 2: Key Formulas and Approach:

Power $P = V^2 / R$.
Since $R$ is constant, we can establish the ratio: $P_2 / P_1 = (V_2 / V_1)^2$.
This approach is faster than calculating $R$ explicitly, though both methods are valid.

Step 3: Detailed Explanation:

Method 1 (Finding Resistance): First, determine the heater's resistance using its rated values. \[ R = \frac{V_{rated}^2}{P_{rated}} = \frac{220 \times 220}{400} = \frac{48400}{400} = 121 \Omega \]
Now, calculate the power consumed when the voltage is changed to 200 V: \[ P_{new} = \frac{V_{new}^2}{R} = \frac{200 \times 200}{121} = \frac{40000}{121} \]
Perform the division: $40000 \div 121 \approx 330.57 \text{ W}$. Rounding to the nearest whole number gives 331 W.
Method 2 (Ratio Method): \[ P_{new} = P_{old} \times \left( \frac{V_{new}}{V_{old}} \right)^2 = 400 \times \left( \frac{200}{220} \right)^2 \] \[ P_{new} = 400 \times \left( \frac{10}{11} \right)^2 = 400 \times \frac{100}{121} = \frac{40000}{121} \approx 331 \text{ W} \]
Both methods lead to the same conclusion: the reduction in voltage significantly reduces the heat output.
Step 4: Final Answer:
The power consumed at 200 V is approximately 331 W.