JEE Main 2020 · 6 September · evening · Single correct · +4 / −1
In the figure shown, the current in the 10 V battery is close to :
- (a) 0.36 A from negative to positive terminal
- (b) 0.42 A from positive to negative terminal
- (c) 0.71 A from positive to negative terminal
- (d) 0.21 A from positive to negative terminal
Answer and solution
Answer (d) 0.21 A from positive to negative terminal
Solution
Physics sir · 5 steps
Ground the bottom junction, call the top junction V, and a single KCL equation solves the whole circuit.
- Set up nodes. There are only two junctions: the top one (call its potential V) and the bottom one (take it as 0). The longer plate is the positive terminal, so the 10 V cell has + on top and the 20 V cell has + on the left.
- Branch currents. Take each current flowing up into the top junction. In the left branch the 20 V cell raises the potential, and the current passes through 2 + 5 = 7 Ω, giving (20 - V)/7. In the right branch the 10 V cell acts through 4 Ω, giving (10 - V)/4. The 10 Ω branch carries V/10 downward.
(20 - V)/7 + (10 - V)/4 = V/10
- Solve for V. Multiply by 140: 400 - 20V + 350 - 35V = 14V, so 69V = 750 and V = 10.87 V.
- Current in cell. (10 - V)/4 = (10 - 10.87)/4 = -0.217 A. The minus sign means the current actually flows down the right branch. It enters the 10 V cell at the + terminal and leaves at the - terminal.
- So. About 0.21 A flows from the positive to the negative terminal, so the 10 V cell is being charged. That is option (d).
The idea to keep. With only two junctions, fix one at zero and write KCL at the other. The sign of each branch current then tells you its direction, including whether a cell is charging.
Answer from the NTA final answer key. Solution by PrepWiser.