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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

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

  1. A \(40\)
  2. B \(45\)
  3. C \(80\)
  4. D \(90\)
Verified Solution

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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