Question

Four resistances P, Q, R, X formed a Wheatstone bridge. The bridge is balanced when $R = 100\,\Omega$. If P and Q are interchanged the bridge balances for $121\,\Omega$. The value of X is
A. $100\,\Omega$
B. $200\,\Omega$
C. $300\,\Omega$
D. $110\,\Omega$

Answer 1

Use the formula for the balanced condition in a Wheatstone’s bridge network. This formula gives the relation between the resistances of the four resistors connected in the Wheatstone’s bridge network. First we need to use this balanced condition for the first case in the question and then use the balanced condition for the second case given. Solve the expressions obtained from these two cases and calculate the value of X.

Formula used:
The balanced condition for the Wheatstone’s bridge is given by
$\dfrac{P}{Q} = \dfrac{R}{S}$ …… (1)
Here, $P$, $Q$, $R$ and $S$ are the resistances of the four resistors connected in the Wheatstone’s bridge.

Complete step by step answer:
We have given that the Resistances P, Q, R and X forms a Wheatstone’s bridge.We have given two values of the resistance R when the Wheatstone’s bridge is in the balanced condition.We have asked to calculate the value of the resistance X.We have given that for the first case the bridge is balanced when the value of the resistance R is $100\,\Omega$.
$R = 100\,\Omega$
Rewrite equation (1) for the balanced condition for the given Wheatstone’s bridge.
$\dfrac{P}{Q} = \dfrac{R}{X}$
Substitute $100\,\Omega$ for $R$ in the above balanced condition.
$\dfrac{P}{Q} = \dfrac{{100\,\Omega }}{X}$ …… (2)
For the second condition, the resistors P and Q are interchanged from their original position. For this new arrangement of resistors, the balanced condition is obtained when the value of the resistance R is $121\,\Omega$.
$R = 121\,\Omega$
Rewrite equation (1) for this new balanced condition for the given Wheatstone’s bridge.
$\dfrac{Q}{P} = \dfrac{R}{X}$
Substitute $121\,\Omega$ for $R$ in the above balanced condition.
$\dfrac{Q}{P} = \dfrac{{121\,\Omega }}{X}$
$\Rightarrow \dfrac{P}{Q} = \dfrac{X}{{121\,\Omega }}$ …… (3)
From the equations (2) and (3), we can write
$\dfrac{{100\,\Omega }}{X} = \dfrac{X}{{121\,\Omega }}$
$\Rightarrow {X^2} = 12100$
$\therefore X = 110\,\Omega$
Therefore, the value of the resistance is $110\,\Omega$.

Hence, the correct option is D.

Note: The important thing while using the balanced condition for Wheatstone’s bridge network is to check the order in which the resistors are connected in the bridge network. So, we should be careful while using this balanced condition. For the second case given in the question, we have given that the two resistors are interchanged. So, the students should also change the balanced condition accordingly.