Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be :

Current Electricity PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q58 with Solution

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Correct answer: A

Let each resistance be $R$ and the battery voltage be $V$ .
When connected in series:
- Equivalent resistance: $R_s = R + R = 2R$
- Power consumed: $P_s = \frac{V^2}{R_s} = \frac{V^2}{2R}$
Given: $\frac{V^2}{2R} = 60$
When connected in parallel:
- Equivalent resistance: $R_p = \frac{R \cdot R}{R+R} = \frac{R}{2}$
- Power consumed: $P_p = \frac{V^2}{R_p} = \frac{V^2}{R/2} = \frac{2V^2}{R}$
Compare the two powers: From series case, $\frac{V^2}{2R} = 60$ Multiply both sides by $4$ : $\frac{2V^2}{R} = 240$

But $P_p = \frac{2V^2}{R}$ Hence, $P_p = 240\text{ W}$
Option check:
- A: $240\text{ W}$ ✅
- B: $60\text{ W}$ ❌
- C: $30\text{ W}$ ❌
- D: $120\text{ W}$ ❌

Therefore, the correct answer is A: $240\text{ W}$ .