Current Electricity

JEE Main level, single correct, medium

314 of 380

Question 314 of 380

Two resistors of 100Ω100 \Omega each are connected in series with a 9 V battery. A voltmeter of 400Ω400 \Omega resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ____\_\_\_\_ V.

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Answer: (C) 4

When the 400Ω voltmeter is connected in parallel with one 100Ω resistor, the equivalent resistance is
Req=100×400500=80ΩR_{eq} = \frac{100 \times 400}{500} = 80Ω.
Total circuit resistance: Rtotal=100+80=180ΩR_{total} = 100 + 80 = 180Ω.
Circuit current: I=9/180=0.05I = 9/180 = 0.05 A.
Voltmeter reading (voltage across parallel combination): V=0.05×80=4V = 0.05 \times 80 = 4 V

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