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: (d). Take the lower middle node at 0 V and the upper middle node at V. Left branch: 20 V with 5 + 2 = 7 Ω; right branch: 10 V (+ at top) with 4 Ω; middle: 10 Ω. KCL at the top node: (20 - V)/(7) + (10 - V)/(4) = V/10 400 + 350 = 69V ⇒ V ≈ 10.87 V Current up through the 10 V branch = (10 - 10.87)/(4) ≈ -0.21 A, so 0.21 A flows down, entering the cell at + and leaving at -. So the answer is (d) 0.21 A from positive to negative terminal.

In the figure shown, the current in the 10 V battery is close to: | JEE Main 2020 (6 Sep, Shift 2)

Q141JEE Main 2020 · 6 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

In the figure shown, the current in the 10 V battery is close to:

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Answer: (d)

Take the lower middle node at $0$ V and the upper middle node at $V$ . Left branch: $20$ V with $5 + 2 = 7\ \Omega$ ; right branch: $10$ V (+ at top) with $4\ \Omega$ ; middle: $10\ \Omega$ .
KCL at the top node: $\frac{20 - V}{7} + \frac{10 - V}{4} = \frac{V}{10}$
$400 + 350 = 69V \Rightarrow V \approx 10.87$ V
Current up through the $10$ V branch $= \frac{10 - 10.87}{4} \approx -0.21$ A, so $0.21$ A flows down, entering the cell at + and leaving at $-$ .
So the answer is (d) 0.21 A from positive to negative terminal.

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