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In the circuit shown in figure each cell has emf 5V and has an internal resistance of 0.2 ohm.
What is the reading of ideal voltmeter V.  

Option: 1

0V


Option: 2

1V


Option: 3

2V


Option: 4

5V


Answers (1)

best_answer

As internal resistance of an ideal voltmeter is infinite, the resistance of the battery across which it is connected will not change by its presence as
\frac{1}{r_1}=\frac{1}{r}+\frac{1}{\infty} \quad \Rightarrow \quad r^{\prime}=r
Now as the given 8 batteries are discharging in series i.e
E_{e q}=8 \times 5=40 \mathrm{~V} \text { and } r_{e q}=8 \times 0.2=1.6 \Omega
so current in the circuit 
 \begin{aligned} I & =\frac{E_{\text {eq }}}{r_{\text {eq }}}=\frac{40}{1.6} \\ & =25 \mathrm{~A} \end{aligned}
Hence potential difference across the required battery
\mathrm{v}=\mathrm{E}-\mathrm{Ir}=5-25 \times 0.2=0 \mathrm{~V}

Posted by

Suraj Bhandari

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