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in Physics by (15 points)
edited by

A cell of emf 90V is connected across series combination of two resistors each of 100Ω resistance. A voltmeter of resistance 400Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

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1 Answer

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\(R_{eq}=\frac{400\times100}{500}+100\)

= 180Ω

\(i=\frac{90}{180}=\frac{1}{2}A\)

\(Reading =\frac{1}{2}\times\frac{400}{500}\times100\)

= 40 volt

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