JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 4)
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 :
40 V
90 V
45 V
80 V
Explanation
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$$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{400 \times 100}{500}+100 \\\\ & =180 \Omega \\\\ & \mathrm{i}=\frac{90}{180}=\frac{1}{2} \mathrm{~A} \\\\ & \text { Reading }=\frac{1}{2} \times \frac{400 \times 100}{500} \\\\ & =40 \text { volt } \end{aligned} $$
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