Two resistors of 100 Ω each are connected in series with a 9 V battery. A voltmeter of 400 Ω resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ______ V. (a) 3 (b) 2 (c) 4 (d) 4.5

Answer: (c). 100 Ω | 400 Ω = (100 × 400)/(500) = 80 Ω Total R = 100 + 80 = 180 Ω, I = 9/180 = 0.05 A. Reading = 0.05 × 80 = 4 V So the answer is (c) 4.

Two resistors of 100 Ω each are connected in series with a 9 V… | JEE Main 2026 (24 Jan, Shift 1)

Q15JEE Main 2026 · 24 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

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

Show answer and solution

Answer: (c)

$100\ \Omega \| 400\ \Omega = \dfrac{100 \times 400}{500} = 80\ \Omega$
Total $R = 100 + 80 = 180\ \Omega$ , $I = \dfrac{9}{180} = 0.05\text{ A}$ .
Reading $= 0.05 \times 80 = 4\text{ V}$
So the answer is (c) 4.

Report a mistake

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2026 (24 Jan, Shift 1)

More questions on Kirchhoff's Laws & Measuring Instruments