Current Electricity 2 Question 26
26. 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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Solution:
- 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 $\frac{20}{3} V$.
