Question

Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

• A. \((P_1+P_2)\)
• B. \(\dfrac{P_1+P_2}{2}\)
• C. \(\dfrac{P_1P_2}{2(P_1+P_2)}\)
• D. \(\dfrac{P_1P_2}{4(P_1+P_2)}\)

Hint:

For appliances with rating \((P,V)\), \[ R=\frac{V^2}{P}. \] Always convert the ratings into resistance first and then apply series or parallel combinations.

Answer 1

The Correct Option is D

To solve this problem, we need to determine the power consumed by two heaters connected in series, both rated for two different powers and the same voltage, but connected to a voltage source that is half of their rated voltage.

Given:

• Heater 1: Rated Power = P_1, Rated Voltage = V
• Heater 2: Rated Power = P_2, Rated Voltage = V
• Both heaters are connected in series across a dc source of \frac{V}{2} volt.

When resistors are connected in series, the same current flows through each resistor. Let's calculate the resistance of each heater:

• For Heater 1, using the formula P = \frac{V^2}{R}, we get:

R_1 = \frac{V^2}{P_1}

• For Heater 2, similarly:

R_2 = \frac{V^2}{P_2}

When connected in series, the total resistance R_{\text{total}} is:

R_{\text{total}} = R_1 + R_2 = \frac{V^2}{P_1} + \frac{V^2}{P_2}

This simplifies to:

R_{\text{total}} = \frac{V^2(P_1 + P_2)}{P_1 P_2}

The current I flowing through the series combination is given by Ohm's Law, where the applied voltage is \frac{V}{2}:

I = \frac{\text{Voltage}}{\text{Resistance}} = \frac{\frac{V}{2}}{R_{\text{total}}}

I = \frac{\frac{V}{2} \cdot P_1 P_2}{V^2(P_1 + P_2)}

I = \frac{P_1 P_2}{2V(P_1 + P_2)}

The power consumed by the series combination is given by:

P_{\text{consumed}} = I^2 \cdot R_{\text{total}}

P_{\text{consumed}} = \left( \frac{P_1 P_2}{2V(P_1 + P_2)} \right)^2 \cdot \frac{V^2(P_1 + P_2)}{P_1 P_2}

P_{\text{consumed}} = \frac{P_1^2 P_2^2 \cdot V^2(P_1 + P_2)}{4V^2(P_1 + P_2)^2 \cdot P_1 P_2}

P_{\text{consumed}} = \frac{P_1 P_2}{4(P_1 + P_2)}

Thus, the power consumed by the combination of two heaters in series is:

\frac{P_1P_2}{4(P_1+P_2)}

This matches the given correct answer, confirming the solution is accurate.