JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2026
- Easy
- MCQ
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.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation

Current in circuit.
\(I=\frac{E}{\operatorname{Req}}\)
Req \(=100+\frac{400 \times 100}{400+100}=180 \Omega\)
\(\therefore I =\frac{9}{180}=\frac{1}{20} A\)
Reading of voltmeter \(= V = I \times 80=\frac{1}{20} \times 80=4 V\)
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