A 5 V battery with internal resistance 2 Ω and a 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 (a) 0.27 A, P₂ to P₁ (b) 0.03 A, P₁ to P₂ (c) 0.03 A, P₂ to P₁ (d) 0.27 A, P₁ to P₂

Answer: (c). From the figure the two cells oppose each other. Take P₁ (bottom) at 0 and P₂ at V. Node equation at P₂: (V - 5)/(2) + (V + 2)/(1) + V/10 = 0 1.6V = 0.5 ⇒ V = 0.3125 V I₁₀ = (0.3125)/(10) ≈ 0.03 A, from higher potential P₂ to P₁. So the answer is (c) 0.03 A, P₂ to P₁.

A 5 V battery with internal resistance 2 Ω and a 2 V battery with… | JEE Main 2008 (AIEEE)

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Cells: EMF, Internal Resistance & Combinations

A $5\text{ V}$ battery with internal resistance $2\ \Omega$ and a $2\text{ V}$ battery with internal resistance $1\ \Omega$ are connected to a $10\ \Omega$ resistor as shown in the figure.

The current in the $10\ \Omega$ resistor is

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Answer: (c)

From the figure the two cells oppose each other. Take $P_1$ (bottom) at 0 and $P_2$ at $V$ .
Node equation at $P_2$ : $\frac{V - 5}{2} + \frac{V + 2}{1} + \frac{V}{10} = 0$
$1.6V = 0.5 \Rightarrow V = 0.3125\text{ V}$
$I_{10} = \frac{0.3125}{10} \approx 0.03\text{ A}$ , from higher potential $P_2$ to $P_1$ .
So the answer is (c) 0.03 A, $P_2$ to $P_1$ .

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