A coil develops heat of 800 cal / sec . When 20 volts is applied across its ends. The resistance of the coil is (1 cal = 4.2 joule )

Options

A. `1.2 Omega`
B. `1.4 Omega`
C. `0.12 Omega`
D. `0.14 Omega`

Answer

To find the resistance of the coil, we can follow these steps:

### Step 1: Convert the heat developed from calories to joules.
The coil develops heat at a rate of 800 cal/sec. We know that 1 cal = 4.2 joules.

\[ \text{Power (P)} = 800 \, \text{cal/sec} \times 4.2 \, \text{joules/cal} = 3360 \, \text{joules/sec} \]

### Step 2: Use the formula for electrical power.
The power developed in an electrical circuit can also be expressed as:

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

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

### Step 3: Rearrange the formula to solve for resistance.
Rearranging the formula gives us:

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

### Step 4: Substitute the known values into the equation.
We know \( V = 20 \, \text{volts} \) and \( P = 3360 \, \text{joules/sec} \).

\[ R = \frac{20^2}{3360} \]

Calculating \( 20^2 \):

\[ 20^2 = 400 \]

Now substituting back into the equation:

\[ R = \frac{400}{3360} \]

### Step 5: Simplify the fraction.
To simplify \( \frac{400}{3360} \):

\[ R = \frac{400 \div 80}{3360 \div 80} = \frac{5}{42} \]

### Step 6: Calculate the numerical value.
Now we calculate \( \frac{5}{42} \):

\[ R \approx 0.119 \, \text{ohms} \]

### Step 7: Round to two decimal places.
Rounding \( 0.119 \) gives us:

\[ R \approx 0.12 \, \text{ohms} \]

### Final Answer:
The resistance of the coil is approximately **0.12 ohms**.

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