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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 \(.........\,V\)

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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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