Question

In wheatstone bridge, 4 resistors $P = 10\,\Omega $, $Q = 5\,\Omega $, $R = 4\,\Omega $ and $S = 4\,\Omega $ are connected in cyclic order, To ensure no current through galvanometer.
A) $5\,\Omega $ resistance is connected in series with $Q$
B) $4\,\Omega $ resistance is connected in parallel with $S$
C) $10\,\Omega $ resistance is connected in series with $P$
D) $4\,\Omega $ resistance is connected in series with $R$

Answer 1

Hint: The solution of this problem can be determined by using the resistance formula of the wheatstone bridge. If the both sides of the resistance equation are balanced, then the wheat stone bridge is working with no current and the galvanometer shows no current.

Formula Used:
The resistance equation of the wheatstone bridge with four resistors are given by,
$\dfrac{P}{Q} = \dfrac{S}{R}$
Where, $P$, $Q$, $R$ and $S$ are the four resistance of the wheatstone bridge.

Complete step by step solution:
Given that,
The resistance of the resistor $P$ is, $P = 10\,\Omega $,
The resistance of the resistor $Q$ is, $Q = 5\,\Omega $,
The resistance of the resistor $R$ is, $R = 4\,\Omega $,
The resistance of the resistor $S$ is, $S = 4\,\Omega $.
Now,
The resistance equation of the wheatstone bridge with four resistors are given by,
$\dfrac{P}{Q} = \dfrac{S}{R}\,..................\left( 1 \right)$
1. $5\,\Omega $ resistance is connected in series with $Q$:
So, the net resistance of the $Q$ is the sum of the resistance of the $Q$ which is given in the question and the $5\,\Omega $, then
$Q = 10\,\Omega $
Now, the LHS of the equation (1) is calculated by substituting the resistance values of the resistors $P$ and $Q$, then
$\dfrac{P}{Q} = \dfrac{{10}}{{10}}$
By dividing the terms in the above equation, then the above equation is written as,
$\dfrac{P}{Q} = 1\,\Omega \,.....................\left( 2 \right)$
Now, the RHS of the equation (1) is calculated by substituting the resistance values of the resistors $S$ and $R$, then
$\dfrac{S}{R} = \dfrac{4}{4}$
By dividing the terms in the above equation, then the above equation is written as,
$\dfrac{S}{R} = 1\,\Omega \,.....................\left( 3 \right)$
By the equation (2) and equation (3), it is clear that the LHS and RHS of the equation (1) is equal.
Hence, there is no current flow through the galvanometer.

Hence, the option (A) is the correct answer.

Note: Both the sides of the equation (1) having the same resistance, so that the equation is balanced, so the current does not flow across the galvanometer. If there is the change in the resistance, then the current will flow across the galvanometer.