Question:
In the figure shown, the current in the 10 V battery is close to :
In the figure shown, the current in the 10 V battery is close to :
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Take the bottom node as 0 V and the top node as V; the 20 V branch has 5 + 2 = 7 ohm in series, the 10 V branch has 4 ohm, and apply KCL at the top node.
Updated On: Nov 5, 2026
- 0.36 A from negative to positive terminal
- 0.42 A from positive to negative terminal
- 0.71 A from positive to negative terminal
- 0.21 A from positive to negative terminal
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The Correct Option is D
Solution and Explanation
From the figure: the 20 V battery has its positive plate on the left, connected through the 5 \(\Omega\) resistor to the top node T; its negative plate connects through the 2 \(\Omega\) resistor to the bottom node B. The 10 V battery has its positive (long) plate at T and its negative plate connected to B through the 4 \(\Omega\) resistor. The 10 \(\Omega\) resistor joins T and B.
Let \(V_B=0\), \(V_T=V\).
Current into T from the 20 V branch: \(\frac{20-V}{5+2}\). Current from T to B through the 10 \(\Omega\): \(\frac{V}{10}\). Current from T to B through the 10 V battery branch: \(\frac{V-10}{4}\).
KCL at T: \(\frac{20-V}{7}=\frac{V}{10}+\frac{V-10}{4}\).
Multiply by 140: \(400-20V=14V+35V-350\), so \(69V=750\), \(V=10.87\) V.
Current in the 10 V battery: \(\frac{10.87-10}{4}=0.217\) A, directed from T (positive terminal) to B (negative terminal), i.e. from positive to negative terminal (the cell is being charged).
Answer: (4) 0.21 A from positive to negative terminal
Let \(V_B=0\), \(V_T=V\).
Current into T from the 20 V branch: \(\frac{20-V}{5+2}\). Current from T to B through the 10 \(\Omega\): \(\frac{V}{10}\). Current from T to B through the 10 V battery branch: \(\frac{V-10}{4}\).
KCL at T: \(\frac{20-V}{7}=\frac{V}{10}+\frac{V-10}{4}\).
Multiply by 140: \(400-20V=14V+35V-350\), so \(69V=750\), \(V=10.87\) V.
Current in the 10 V battery: \(\frac{10.87-10}{4}=0.217\) A, directed from T (positive terminal) to B (negative terminal), i.e. from positive to negative terminal (the cell is being charged).
Answer: (4) 0.21 A from positive to negative terminal
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