Current Electricity Ques 46
- An electrical circuit is shown in figure. Calculate the potential difference across the resistor of $400$ $ \Omega$ as will be measured by the voltmeter $V$ of resistance $400$ $ \Omega$ either by applying Kirchhoff’s rules or otherwise.
(1996, 5M)
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Answer:
Correct Answer: 46.($\frac{20}{3} $ $V$)
Solution:
Formula:
- The given circuit actually forms a balanced Wheatstone’s bridge (including the voltmeter) as shown below
$ R _V=400$ $ \Omega $
Here, we see that $\frac{P}{Q}=\frac{R}{S}$
Therefore, resistance between $A$ and $B$ can be ignored and equivalent simple circuit can be drawn as follows
The voltmeter will read the potential difference across resistance $Q$.
$ \text { Currents } i _1=i _2=\frac{10}{100+200}=\frac{1}{30} $ $A $
$\therefore$ Potential difference across voltmeter
$ =Q i _1=(200) (\frac{1}{30})$ $ V$ $=\frac{20}{3} $ $V $
Therefore, reading of voltmeter will be $\frac{20}{3} $ $V$.
BETA
