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:

  1. 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 $402.5 \Omega$.



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