Question

When the value of R in the Balanced Wheatstone bridge shown in figure is increased from 5W to 7W, the value of S has to be increased by 3W in order to maintain the balance. What is the initial value of S ?

(A) 1.5W
(B) 3W
(C) 5W
(D) 7.5W

Answer 1

Hint:In order to solve this question,we use the balancing condition of the Wheatstone bridge which state that if ${\displaystyle \frac{P}{Q}}={\displaystyle \frac{R}{S}}$ then current through the galvanometer is zero.
Complete Step by Step Answer:Wheatstone bridge is an arrangement of 4 resistance and galvanometer with one battery as shown in diagram.

In the balancing condition of Wheatstone bridge, the current flowing in galvanometer is zero and the relation between resistances P, Q, R & S is given as
${\displaystyle \frac{P}{Q}}={\displaystyle \frac{R}{S}}$ …..(1)
Given that initial value of $R=5W$
So, ${\displaystyle \frac{P}{Q}}={\displaystyle \frac{5W}{S}}$ …..(2)
Also given that final value of $R=7W$
Final value of S is $S+3$ and P and Q does not change.
So, ${\displaystyle \frac{P}{q}}={\displaystyle \frac{7W}{S+3}}$ …..(3)
From equation 2 & 3
${\displaystyle \frac{5}{S}}={\displaystyle \frac{7}{S+3}}$
$5(S+3)=7S$
$5S+15=7S$
$7S-SS=15$
$2S=15$
$S={\displaystyle \frac{15}{2}}$
$S=7W$
So, option D is correct answer $7.5W$
Note: If the current flowing in galvanometer is O, then Wheatstone is balanced & relation between P, Q, R & S is given as ${\displaystyle \frac{P}{Q}}={\displaystyle \frac{R}{S}}$
If the current flowing in galvanometer is not zero then Wheatstone bridge is not balanced & ${\displaystyle \frac{P}{Q}}\ne {\displaystyle \frac{R}{S}}$