Current Electricity 2 Question 6
6. In a Wheatstone bridge (see figure), resistances $P$ and $Q$ are approximately equal. When $R=400 \Omega$, the bridge is balanced. On interchanging $P$ and $Q$, the value of $R$ for balance is $405 \Omega$. The value of $X$ is close to
(2019 Main, 11 Jan I)
(a) $404.5 \Omega$
(b) $401.5 \Omega$
(c) $402.5 \Omega$
(d) $403.5 \Omega$
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Solution:
- For a balanced Wheatstone bridge,
$$ \frac{P}{R}=\frac{Q}{X} $$
In first case when $R=400 \Omega$, the balancing equation will be
$$ \begin{aligned} \frac{P}{R} & =\frac{Q}{X} \Rightarrow \frac{P}{400 \Omega}=\frac{Q}{X} \\ \Rightarrow \quad P & =\frac{400 \times Q}{X} \end{aligned} $$
In second case, $P$ and $Q$ are interchanged and $R=405 \Omega$
$$ \begin{aligned} & \therefore \quad \frac{Q}{R}=\frac{P}{X} \\ & \Rightarrow \quad \frac{Q}{405}=\frac{P}{X} \end{aligned} $$
Substituting the value of $P$ from Eq. (i) in Eq. (ii), we get
$$ \frac{Q}{405}=\frac{Q \times 400}{X^{2}} $$
$$ \begin{array}{ll} \Rightarrow & X^{2}=400 \times 405 \\ \Rightarrow & X=\sqrt{400 \times 405}=402.5 \end{array} $$
The value of $X$ is close to $402.5 \Omega$.
