Two heating elements each of resistance, \( R_{1} \) and \( R_{2} \), when operated at a constant supply voltage \( V \), consume power \( P_{1} \) and \( P_{2} \), respectively. What will be the power consumed when both resistors are connected in a series combination with supply voltage \( V \) in a circuit?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( P = (P_{1} \times P_{2})(P_{1} + P_{2})^{-1} \)
At a fixed supply voltage, the power of a resistor is \( P = V^{2}/R \), so each resistance can be written in terms of its power.
\( R_{1} = V^{2}/P_{1} \)
\( R_{2} = V^{2}/P_{2} \)
In series the resistances add:
\( \displaystyle P = \frac{V^{2}}{R_{1}+R_{2}} = \frac{V^{2}}{\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}} = \frac{1}{\frac{1}{P_{1}}+\frac{1}{P_{2}}} \)
\( \displaystyle P = \frac{P_{1}P_{2}}{P_{1}+P_{2}} = (P_{1}\times P_{2})(P_{1}+P_{2})^{-1} \)
Shortcut: Limiting case: take two identical elements, \( P_{1} = P_{2} = P_{0} \). In series the resistance doubles, so the power must be \( P_{0}/2 \). Only Option (2) gives this: \( P_{0}^{2}/2P_{0} = P_{0}/2 \).