JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2023
- Medium
- Numerical
A cell of emf \(90\,V\) is connected across series combination of two resistors each of \(100\,\Omega\) resistance. A voltmeter of resistance \(400\,\Omega\) is used to measure the potential difference across each resistor. The reading of the voltmeter will be \(.........\,V\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(40\)
Step-by-step Solution
Detailed explanation
\(R _{ eq }=\frac{400 \times 100}{500}+100\)
\(=180\,\Omega\)
\(i =\frac{90}{180}=\frac{1}{2}\,A\)
\(\text { Reading }=\frac{1}{2} \times \frac{400}{500} \times 100\)
\(=40\,volt\)
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