Question

Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is :

• A. 1 : 1
• B. 2 : 9
• C. 1 : 2
• D. 2 : 3

Answer 1

The Correct Option is B

Step 1: Recall the Power Formula 

The power of a device is given by:

$$ P = \frac{V^2}{R} $$

\( P \) = Power

\( V \) = Voltage across the device

\( R \) = Resistance of the device

The resistance of a device can be calculated using:

$$ R = \frac{V^2}{P} $$

Step 2: Case 1 - Series Connection

When the heaters are connected in series, the total resistance is:

$$ R_{\text{total}} = R_A + R_B $$

Where:

$$ R_A = \frac{V^2}{P_A}, \quad R_B = \frac{V^2}{P_B} $$

For heaters A and B with power ratings:

\( P_A = 1 \) kW

\( P_B = 2 \) kW

Thus:

$$ R_A = \frac{V^2}{1\text{ kW}}, \quad R_B = \frac{V^2}{2\text{ kW}} $$

The total resistance in series:

$$ R_{\text{total, series}} = \frac{V^2}{1\text{ kW}} + \frac{V^2}{2\text{ kW}} = \frac{3V^2}{2\text{ kW}} $$

Step 3: Case 2 - Parallel Connection

For a parallel connection, the total resistance is given by:

$$ \frac{1}{R_{\text{total}}} = \frac{1}{R_A} + \frac{1}{R_B} $$

Substituting the values:

$$ \frac{1}{R_{\text{total}}} = \frac{1}{\frac{V^2}{1\text{ kW}}} + \frac{1}{\frac{V^2}{2\text{ kW}}} $$

Which simplifies to:

$$ \frac{1}{R_{\text{total}}} = \frac{1\text{ kW}}{V^2} + \frac{2\text{ kW}}{V^2} = \frac{3\text{ kW}}{V^2} $$

Thus, the total resistance in parallel is:

$$ R_{\text{total, parallel}} = \frac{V^2}{3\text{ kW}} $$

Step 4: Power Output in Series and Parallel

The power output is inversely proportional to the total resistance. Therefore, for the series connection:

$$ P_{\text{series}} \propto \frac{1}{R_{\text{total, series}}} = \frac{2\text{ kW}}{3} $$

For the parallel connection:

$$ P_{\text{parallel}} \propto \frac{1}{R_{\text{total, parallel}}} = \frac{3\text{ kW}}{2} $$

Step 5: Ratio of Power Outputs

The ratio of power outputs in series and parallel cases is:

$$ \frac{P_{\text{series}}}{P_{\text{parallel}}} = \frac{\frac{2\text{ kW}}{3}}{\frac{3\text{ kW}}{2}} $$

Solving:

$$ = \frac{2}{3} \times \frac{2}{3} = \frac{2}{9} $$

Conclusion

The correct ratio of power outputs is 2:9, which corresponds to option (2).