10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is :

1. 1000

2. 10

3. 100

4. 1

Doubtly Asked 30 Oct 2026 NEET UG 2023 Practice test neet Save WhatsApp Report

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Best answer

Answer: 3. 100

Series: \(R_s=10R\), so \(I_1=\dfrac{E}{10R}\).

Parallel: \(R_p=\dfrac{R}{10}\), so \(I_2=\dfrac{10E}{R}\).

\(n=\dfrac{I_2}{I_1}=\dfrac{10E/R}{E/10R}=100\). The answer is 100.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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