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

The problem involves calculating the ratio of power outputs when two heaters are connected in series and then in parallel to a fixed power source. We are given the power ratings of Heater A (1 kW) and Heater B (2 kW).

First, consider the case where heaters are connected in series. When connected in series, the voltage across each heater is different. The total power \( P_s \) in series is given by the formula:

\[ P_s = \frac{V^2}{{R_A + R_B}} \]

Where \( V \) is the voltage of the power source, and \( R_A \) and \( R_B \) are the resistances of heaters A and B, calculated using the formula \( P = \frac{V^2}{R} \). Thus,

\[ V^2 = 1 kW \times R_A = 2 kW \times R_B \]

Solving for resistances, \[ R_A = \frac{V^2}{1 \text{ kW}},\ R_B = \frac{V^2}{2 \text{ kW}} \]

In series:

\[ P_s = \frac{V^2}{{\frac{V^2}{1} + \frac{V^2}{2}}} = \frac{V^2}{\frac{3V^2}{2}} = \frac{2}{3} \text{ kW} \]

Next, consider the parallel connection. In parallel, the voltage across each heater is the same, and the total power \( P_p \) is:

\[ P_p = P_A + P_B = 1 \text{ kW} + 2 \text{ kW} = 3 \text{ kW} \]

The ratio of power outputs is:

\[ \text{Ratio} = \frac{P_s}{P_p} = \frac{\frac{2}{3}}{3} = \frac{2}{9} \]

Thus, the ratio of power outputs is 2:9, which is the correct answer.