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

Let the resistances of the heaters A and B be $R_A$ and $R_B$, respectively. From the formula for power $P = \frac{V^2}{R}$, the resistance can be expressed as:
$R = \frac{V^2}{P}$.

So:

$R_A = \frac{V^2}{1 \,kW}$ and $R_B = \frac{V^2}{2 \, kW}$.

Case 1: Series Connection
The total resistance in series is:

$R_{series} = R_A + R_B$.

The power output is inversely proportional to the resistance, so:

$P_{series} = \frac{V^2}{R_{series}} = \frac{V^2}{R_A + R_B}$.

Case 2: Parallel Connection
The total resistance in parallel is:

$R_{parallel} = \frac{R_A R_B}{R_A + R_B}$.

The power output for parallel connection is:

$P_{parallel} = \frac{V^2}{R_parallel} = \frac{V^2}{\frac{R_A R_B}{R_A + R_B}} = \frac{V^2 (R_A + R_B)}{R_A R_B}$.

Case 3: Ratio of Power Outputs
The ratio of power outputs for series and parallel connections is:

$\frac{P_{series}}{P_{parallel}} = \frac{\frac{V^2}{R_A + R_B}}{\frac{V^2 (R_A + R_B)}{R_A R_B}} = \frac{R_A R_B}{(R_A + R_B)^2}$.

Substituting $R_A = \frac{V^2}{1}$ and $R_B = \frac{V^2}{2}$:

$\frac{P_{series}}{P_{parallel}} = \frac{(\frac{V^2}{1})(\frac{V^2}{2})}{V^2 + V^2/2} = \frac{V^4/2}{(3V^2)/2} = \frac{V^2}{3}$.

Simplifying:

$\frac{P_{series}}{P_{parallel}} = \frac{1/2}{3/2} = \frac{1}{3} = \frac{2}{9}$.

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