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A 5 V battery with internal resistance 2 Ω and 2 V battery with internal resistance 1 Ω are connected to a 10 Ω resistor as shown in the figure. The current in the 10 Ω resistor is [Figure: three parallel branches between top node P₂ and bottom node P₁: left, the 5 V, 2 Ω battery with positive terminal toward P₂; middle, the 10 Ω resistor; right, the 2 V, 1 Ω battery with positive terminal toward P₁.]

Asked in JEE Main 2008 · Combination of cells

• (1) 0.27 A P₁ to P₂
• (2) 0.27 A P₂ to P₁
• (3) 0.03 A P₁ to P₂
• (4) 0.03 A P₂ to P₁

Step-by-step solution

Why the other options are wrong

• (1) Treats the cells as aiding (about 0.28 A) and also reverses the direction.
• (2) Treats the cells as aiding: (2.5+2)/(1.6)/10≈0.28 A.
• (3) Right magnitude, wrong direction; P₂ is at the higher potential.

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