A 12 V battery of negligible internal resistance is connected across two branches joined in parallel. One branch is a 4 Ω resistor. The other branch is a 2 Ω resistor in series with a 6 V cell of negligible internal resistance. In this branch, the 6 V cell is joined so that it opposes the battery. The current drawn from the 12 V battery is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6 A
Both branches have the full 12 V of the battery across them. In the second branch the 6 V cell acts against the battery. The loop rule for each branch gives its current, and the junction rule adds the two currents.
\( \begin{aligned} I_1 &= \frac{12}{4} \\ &= 3\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} 12-6 &= 2I_2 \\ I_2 &= 3\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} I &= I_1+I_2 \\ &= 3+3 \\ &= 6\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Kirchhoff’s Rules – “(a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction …”
Kirchhoff’s Rules – “(b) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero …”