Kirchhoff's Laws
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
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.
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About This Question
- Subject
- physics
- Chapter
- current electricity
- Topic
- kirchhoff's laws
- Difficulty
- Hard
- Year
- 2025
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.
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