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.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): While a storage battery is being charged, the potential difference across its terminals is greater than its emf.
Reason (R): During charging, the current flows through the battery from its positive terminal to its negative terminal.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both (A) and (R) are correct and (R) is the correct explanation of (A).
Assertion (A) While a storage battery is being charged, the potential difference across its terminals is greater than its emf. — True. During charging the terminal potential difference is \( V=\varepsilon+Ir \), which is more than \( \varepsilon \).
Reason (R) During charging, the current flows through the battery from its positive terminal to its negative terminal. — True. The charging supply pushes current into the positive terminal. This is opposite to the direction of the current when the battery itself drives a circuit.
Why Reason (R) explains Assertion (A)?
When a cell drives a circuit, the current inside it flows from N to P. Then \( V=\varepsilon-Ir \).
When the current flows through the cell from P to N, the sign of the \( Ir \) term changes. So \( V=\varepsilon+Ir \), which is greater than the emf.
NCERT (2026-27):
Cells, Emf, Internal Resistance – “If however R is finite, I is not zero. In that case the potential difference between P and N is V = V++ V– – I r = ε – I r …”
Kirchhoff’s Rules – “… If, while labelling the current I through the cell one goes from P to N, then of course V = ε + I r”