- 2073 Bhadra · 6 marks
Superposition: branch currents, 6 V and 12 V
Short answer
Using superposition, the branch currents are: left branch (3 Ω with 6 V) 0.4 A into the source (charging), middle 6 Ω branch 1.2 A downward, and right branch (3 Ω with 12 V) 1.6 A from the 12 V source.
Calculate the current flowing in each branch using superposition theorem. [Figure: 6 V source in series with 3 Ω on the left; 6 Ω resistor in the middle branch; 3 Ω in series with 12 V source on the right.]
Assumption: the left branch is the 6 V source with 3 Ω, the middle branch is 6 Ω, and the right branch is 3 Ω with the 12 V source. Both sources have their positive terminals toward the top node. Currents are taken as positive in these directions: left branch up into the top node (I1), right branch up into the top node (I3), and middle branch down through the 6 Ω (I2).
Case 1: 6 V alone (12 V shorted)
6 Ω ∥ 3 Ω = 2 Ω; R_total = 3 + 2 = 5 Ω I1' = 6/5 = 1.2 A (up into the node) Through 6 Ω: I2' = 1.2 × 3/(6 + 3) = 0.4 A (down) Right branch: 1.2 × 6/9 = 0.8 A flows down through its 3 Ω, so I3' = −0.8 A.
Case 2: 12 V alone (6 V shorted)
6 Ω ∥ 3 Ω = 2 Ω; R_total = 3 + 2 = 5 Ω I3'' = 12/5 = 2.4 A (up into the node) Through 6 Ω: I2'' = 2.4 × 3/9 = 0.8 A (down) Left branch: I1'' = −1.6 A (1.6 A flows down into the 6 V branch).
Superposition
| Branch | 6 V alone | 12 V alone | Total |
|---|---|---|---|
| Left (3 Ω, 6 V) | +1.2 | −1.6 | −0.4 A |
| Middle (6 Ω) | 0.4 | 0.8 | 1.2 A |
| Right (3 Ω, 12 V) | −0.8 | +2.4 | +1.6 A |
Check: 1.6 − 0.4 = 1.2 A at the node. The voltage across the 6 Ω is 7.2 V; (12 − 7.2)/3 = 1.6 A and (6 − 7.2)/3 = −0.4 A, which matches.
Answer: left branch 0.4 A (flowing into the 6 V source, i.e. charging it), middle 6 Ω branch 1.2 A downward, right branch 1.6 A delivered by the 12 V source.
✓ Answer checked by Claude Opus 5.5 (high effort)
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