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.
A cell of emf 9 V and internal resistance 1 Ω is joined in series with a second cell of emf 3 V and internal resistance 1 Ω. The negative terminals of the two cells are joined to each other. The free terminals are connected across a 4 Ω resistor. The current in the resistor is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 A
The negative terminals of the two cells are joined. So the cells oppose each other in the loop, and the equivalent emf is \( \varepsilon_1-\varepsilon_2 \). The internal resistances still add in series.
\( \begin{aligned} \varepsilon_{eq} &= \varepsilon_1-\varepsilon_2 \\ &= 9-3 \\ &= 6\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} r_{eq} &= r_1+r_2 \\ &= 1+1 \\ &= 2\ \Omega \end{aligned} \)
The current in the loop is then
\( \begin{aligned} I &= \frac{\varepsilon_{eq}}{R+r_{eq}} \\ &= \frac{6}{4+2} \\ &= 1\ \mathrm{A} \end{aligned} \)
Therefore, a current of 1 A flows through the 4 Ω resistor.
NCERT (2026-27):
Cells in Series and in Parallel – “… If instead we connect the two negatives, … we will get εeq = ε1 – ε2 (ε1 > ε2)”
Cells in Series and in Parallel – “(ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances.”