A circuit has two nodes, top T and bottom B, joined by three parallel branches. Left branch: a 6 V cell (positive terminal towards T) in series with 2 Ω. Middle branch: a 3 V cell (positive terminal towards T) in series with 1 Ω. Right branch: a 2 Ω resistor only. The cells have negligible internal resistance. What is the current through the right-hand 2 Ω resistor?
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Let V be the potential of T above B. The junction rule gives (6 − V)/2 + (3 − V)/1 = V/2, so V = 3 V and the current in the right branch is 3/2 = 1.5 A. (No current flows in the 3 V branch.)
- A.
3 A is V divided by 1 Ω; the right branch has 2 Ω.
- B.
0.75 A does not satisfy the junction rule at T.
- C.
4.5 A adds the currents each cell alone would drive through its own resistor.
NCERT: Class 12 Physics, Chapter 3