Skip to content

Kirchhoff's Laws

Hardphysics

In a two-loop network, a 10 V cell with negligible internal resistance drives current through a 2 (\Omega) resistor that then splits into parallel arms of 6 (\Omega) and 3 (\Omega). What current is supplied by the cell?

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject
physics
Chapter
current electricity
Topic
kirchhoff's laws
Difficulty
Hard
Year
2025
Tags
Kirchhoff's junction ruleKirchhoff's loop ruletwo-loop networkparallel armscircuit current

Solution

Correct Answer:

2.5 A

Kirchhoff's laws govern multi-loop circuits: the junction rule conserves charge so currents in equal currents out, while the loop rule conserves energy so the algebraic sum of potential changes around any closed loop is zero. The two parallel arms of 6 (\Omega) and 3 (\Omega) combine as (\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}), giving (R_p = 2;\Omega). Applying the loop rule around the source, the total resistance seen by the cell is (2 + 2 = 4;\Omega), so the supplied current is (I = \varepsilon/R_{total} = 10/4 = 2.5) A. The value 1.0 A wrongly treats both parallel resistors as a 8 (\Omega) series chain. The value 5.0 A ignores the parallel section entirely, dividing 10 V by 2 (\Omega). The value 3.3 A uses only one parallel branch. This is the canonical JEE application of Kirchhoff's rules with network reduction. A consistency check confirms it: the 2.5 A through the series resistor must equal the sum of the two branch currents downstream, satisfying the junction rule. Furthermore, the branch carrying the smaller 3 (\Omega) resistance must take twice the current of the 6 (\Omega) branch, since at equal potential difference current divides in inverse proportion to resistance, and adding 1.667 A and 0.833 A indeed returns the 2.5 A supplied by the cell.

This hard difficulty physics question is from the chapter current electricity, covering the topic of kirchhoff's laws. It appeared in the 2025 exam.

Looking for more practice? Explore all physics questions or browse current electricity questions on RankGuru.