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

Concept: For an electrical appliance rated \((P,V)\), \[ P=\frac{V^2}{R} \] Therefore, \[ R=\frac{V^2}{P}. \] The heaters are connected in series and the supply voltage is \(V/2\). The total power consumed is \[ P_{\text{total}} = \frac{V_{\text{supply}}^2}{R_{\text{eq}}}. \]

Step 1: Determine the resistance of each heater.
For heater 1, \[ R_1=\frac{V^2}{P_1}. \] For heater 2, \[ R_2=\frac{V^2}{P_2}. \]

Step 2: Calculate the equivalent resistance.
Since the heaters are connected in series, \[ R_{\text{eq}} = R_1+R_2 \] \[ = \frac{V^2}{P_1} + \frac{V^2}{P_2} \] \[ = V^2\left(\frac{P_1+P_2}{P_1P_2}\right). \]

Step 3: Calculate the total power consumed.
The applied voltage is \[ V_{\text{supply}}=\frac{V}{2}. \] Hence, \[ P_{\text{total}} = \frac{\left(\frac{V}{2}\right)^2} {V^2\left(\frac{P_1+P_2}{P_1P_2}\right)} \] \[ = \frac{\frac{V^2}{4}} {V^2\left(\frac{P_1+P_2}{P_1P_2}\right)} \] \[ = \frac{P_1P_2} {4(P_1+P_2)}. \] Therefore, \[ \boxed{ P_{\text{total}} = \frac{P_1P_2} {4(P_1+P_2)} } \]