Question

A balanced Wheatstone bridge consists of four resistances $P = 10\,\Omega$, $Q = 20\,\Omega$, $R = 15\,\Omega$, and $S$. If the positions of the galvanometer and the battery are interchanged, the balancing condition will remain valid. What is the value of the unknown resistance $S$ under the balanced condition?

• A. \( 15\,\Omega \)
• B. \( 30\,\Omega \)
• C. \( 45\,\Omega \)
• D. \( 60\,\Omega \)

Hint:

To avoid basic algebraic cross-multiplication errors under time pressure, remember that if the bottom resistor on the left side (\(Q\)) is double the top resistor (\(P\)), then the bottom resistor on the right side (\(S\)) must simply be double its corresponding top resistor (\(R\)).

Answer 1

The Correct Option is B

Concept: A Wheatstone bridge is an electrical circuit configuration used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. The network consists of four resistors arranged in a closed loop (a diamond shape) with a voltage source connected across one pair of opposite vertices and a sensitive galvanometer connected across the remaining two vertices. When the bridge is perfectly balanced, the electrical potential at both terminals of the galvanometer becomes exactly equal. Because there is zero potential difference across it, no current flows through the galvanometer branch (\(I_g = 0\)). The mathematical condition governing this state of balance is given by the ratio of adjacent arms: \[ \frac{P}{Q} = \frac{R}{S} \] An interesting property of this network is its conjugate nature. If you interchange the positions of the input battery source and the null-detecting galvanometer, the fundamental ratio that determines the balance remains totally unaffected. The condition for zero deflection stays exactly the same.

Step 1: Substitute the given values into the balancing ratio. We are given the values for three of the circuit arms:
• Resistor \(P = 10\,\Omega\)
• Resistor \(Q = 20\,\Omega\)
• Resistor \(R = 15\,\Omega\) Plugging these parameters directly into our null-balance equation: \[ \frac{10}{20} = \frac{15}{S} \]

Step 2: Isolate and solve for the unknown resistance \(S\). First, simplify the fraction on the left side of the equation: \[ \frac{1}{2} = \frac{15}{S} \] Now, cross-multiply to solve for the value of \(S\): \[ S \cdot 1 = 15 \cdot 2 \] \[ S = 30\,\Omega \] Thus, the value of the unknown resistor must be exactly \(30\,\Omega\) to maintain a state of zero current through the central galvanometer branch.