Case: Read the passage and answer the question.
Kirchhoff’s first law applies to currents at a junction in a circuit. It states that at a junction in an electrical circuit, the sum of the currents flowing into the junction is equal to the sum of the currents flowing out of the junction. Kirchhoff’s second law, also known as Kirchhoff’s voltage law, states that the sum of all the voltages around a closed loop in any circuit must be equal to zero. The first law is a consequence of the conservation of charge, and the second law is a consequence of the conservation of energy. The Wheatstone bridge, also known as the resistance bridge, finds an unknown resistance by balancing two legs of the bridge circuit and works on Kirchhoff’s laws.
In a Wheatstone bridge, known resistances \(R_1\) and \(R_2\) are in the first and second arms. The unknown resistance \(R_4\) is in the fourth arm. \(R_3\) is varied until the galvanometer shows a null deflection. The bridge is then balanced, with \(R_2/R_1=R_4/R_3\). Which change, made after this, keeps the galvanometer at null deflection?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → The emf of the battery is doubled.
At balance no current flows through the galvanometer. The balance condition is \( R_2/R_1=R_4/R_3 \). It contains only the four resistances. So any change that keeps the two ratios equal keeps the null deflection.
Option (1) \(R_3\) alone is doubled. — Incorrect. Doubling \( R_3 \) makes \( R_4/R_3 \) half, while \( R_2/R_1 \) stays the same. The galvanometer now deflects.
Option (2) The emf of the battery is doubled. — Correct. The emf does not appear in the balance condition. A larger emf sends larger currents through all four arms, but \( I_g \) stays zero.
Option (3) \(R_1\) and \(R_4\) are both doubled. — Incorrect. \( R_2/R_1 \) becomes half and \( R_4/R_3 \) becomes double. The two ratios are no longer equal.
Option (4) \(R_2\) and \(R_3\) are both halved. — Incorrect. Halving \( R_2 \) halves the left ratio. Halving \( R_3 \) doubles the right ratio. So the bridge goes off balance.
NCERT (2026-27):
Wheatstone Bridge – “Hence, we obtain the condition”
\( \displaystyle \frac{R_2}{R_1}=\frac{R_4}{R_3} \)\( \qquad [3.64(a)] \)
“This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.”