The resistance values of the Wheatstone bridge shown are \(P=2000\ \Omega\), \(Q=500\ \Omega\), \(R=3000\ \Omega\). The battery voltage is \(E=50\ \text{V}\).
The battery has an internal resistance of \(1\ \Omega\) and the Galvanometer (G) has a resistance of \(50\ \Omega\).
The value of the resistance S for balanced condition is \(\Omega\) (Answer in integer)
Show Hint
Balance in a Wheatstone bridge needs the ratio of adjacent arms to match; the battery and galvanometer resistances play no part in that ratio.
Step 1: Set up the two potential dividers.
At balance, no current flows through the galvanometer, so the bridge splits into two separate potential dividers fed by the same battery, one made of $P$ and $R$, the other made of $Q$ and $S$. Since no current crosses the galvanometer branch, the midpoint of the $P$-$R$ divider and the midpoint of the $Q$-$S$ divider must sit at the same potential. Step 2: Turn this into an equation.
Equal midpoint potentials on both dividers, fed from the same source, reduce to the ratio condition
\[ \frac{P}{Q}=\frac{R}{S} \]
This is the same balance rule seen from a divider point of view rather than a loop-current point of view. Step 3: Plug in the known resistances.
\[ \frac{2000}{500}=\frac{3000}{S} \]
\[ 4=\frac{3000}{S} \] Step 4: Solve for S.
\[ S=\frac{3000}{4}=750 \] Step 5: Note the role of the extra data.
The battery's internal resistance of $1\ \Omega$ only reduces the terminal voltage the bridge actually receives, it never enters the balance ratio. The galvanometer resistance of $50\ \Omega$ only affects how sensitive the null detection is, and it also drops out once the current through it is zero. Neither number is needed to find $S$.
\[ \boxed{S=750\ \Omega} \]
