A 1250 W heater operates at `115V`. What is the resistance of the heating coil?

Options

A. `1.6 Omega`
B. `13. 5 Omega`
C. `1250 Omega`
D. `10.6 Omega`

Answer

To find the resistance of the heating coil in the heater, we can use the formula that relates power (P), voltage (V), and resistance (R). The formula is given by:

\[ P = \frac{V^2}{R} \]

Where:
- \( P \) is the power in watts (W),
- \( V \) is the voltage in volts (V),
- \( R \) is the resistance in ohms (Ω).

### Step-by-Step Solution:

1. **Identify the given values:**
- Power, \( P = 1250 \, W \)
- Voltage, \( V = 115 \, V \)

2. **Rearrange the formula to solve for resistance (R):**
We can rearrange the formula to find \( R \):
\[ R = \frac{V^2}{P} \]

3. **Substitute the known values into the equation:**
\[ R = \frac{(115 \, V)^2}{1250 \, W} \]

4. **Calculate \( V^2 \):**
\[ V^2 = 115^2 = 13225 \, V^2 \]

5. **Substitute \( V^2 \) back into the equation for R:**
\[ R = \frac{13225 \, V^2}{1250 \, W} \]

6. **Perform the division:**
\[ R = \frac{13225}{1250} \approx 10.58 \, \Omega \]

7. **Round the result to an appropriate number of significant figures:**
\[ R \approx 10.6 \, \Omega \]

### Final Answer:
The resistance of the heating coil is approximately \( 10.6 \, \Omega \).

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