JEE MAIN - Physics (2024 - 1st February Morning Shift - No. 8)
The reading in the ideal voltmeter $(\mathrm{V})$ shown in the given circuit diagram is :
_1st_February_Morning_Shift_en_8_1.png)
$3 \mathrm{~V}$
$10 \mathrm{~V}$
$5 \mathrm{~V}$
$0 \mathrm{~V}$
Explanation
$\begin{aligned} \mathrm{i}= & \frac{\mathrm{E}_{\text {eq }}}{\mathrm{r}_{\mathrm{eq}}}=\frac{8 \times 5}{8 \times 0.2} \\\\ & \mathrm{I}=25 \mathrm{~A} \\\\ & V=\mathrm{E}-\mathrm{ir} \\\\ & =5-0.2 \times 25 \\\\ & =0\end{aligned}$
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