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 $.........\,V$
JEE MAIN 2023, Medium
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$R _{ eq }=\frac{400 \times 100}{500}+100$
$=180\,\Omega$
$i =\frac{90}{180}=\frac{1}{2}\,A$
$\text { Reading }=\frac{1}{2} \times \frac{400}{500} \times 100$
$=40\,volt$
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