A `1000 W` heater is designed to operate on a `120 V` line.The line voltage drops to `110 V`. The percentage of heat output drops by

Options

A. `16%`
B. `30%`
C. `9%`
D. `27%`

Answer

To solve the problem, we need to determine the percentage drop in heat output of a heater when the voltage drops from 120 V to 110 V. The heater is rated at 1000 W at 120 V.

### Step-by-Step Solution:

1. **Determine the Resistance of the Heater:**
The resistance (R) of the heater can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the voltage (120 V) and \( P \) is the power (1000 W).
\[ R = \frac{120^2}{1000} = \frac{14400}{1000} = 14.4 \, \Omega \]

2. **Calculate the Power at the Reduced Voltage:**
Now, we need to find the power output when the voltage drops to 110 V. We use the formula for power:
\[ P = \frac{V^2}{R} \]
Substituting \( V = 110 \, V \) and \( R = 14.4 \, \Omega \):
\[ P = \frac{110^2}{14.4} = \frac{12100}{14.4} \approx 840.27 \, W \]
We can round this to 840 W.

3. **Calculate the Percentage Drop in Power:**
The percentage drop in power can be calculated using the formula:
\[ \text{Percentage Drop} = \frac{P_1 - P_2}{P_1} \times 100 \]
where \( P_1 = 1000 \, W \) (original power) and \( P_2 = 840 \, W \) (new power).
\[ \text{Percentage Drop} = \frac{1000 - 840}{1000} \times 100 = \frac{160}{1000} \times 100 = 16\% \]

### Final Answer:
The percentage drop in heat output is **16%**.

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