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Question

A 5V battery with internal resistance $2\ \Omega$ and a 2V 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:

5V(2Ω) and 2V(1Ω) connected to 10Ω

![5V(2Ω) and 2V(1Ω) connected to 10Ω](https://www.edvaya.com/jee/physics/q26-two-batteries-10-ohm-2008.png) A). $0.27\ 	ext{A}\ P_2\ 	ext{to}\ P_1$ B). $0.03\ 	ext{A}\ P_2\ 	ext{to}\ P_2$ C). $0.03\ 	ext{A}\ P_1\ 	ext{to}\ P_2$ D). $0.27\ 	ext{A}\ P_1\ 	ext{to}\ P_2$

Edvaya · Accepted Answer

The correct answer is $0.03\ 	ext{A}\ P_1\ 	ext{to}\ P_2$. Using mesh analysis with the two batteries in parallel across $10\ \Omega$:
Let current through $5\ 	ext{V}$ battery $= I_1$ and through $2\ 	ext{V}$ battery $= I_2$ (both toward junction $P_2$).
Voltage across $10\ \Omega$: $V = 5 - 2I_1 = 2 + I_2$
Current through $10\ \Omega$: $I = I_1 + I_2 = V/10$
Solving the simultaneous equations gives $I = 0.03\ 	ext{A}$ directed from $P_1$ to $P_2$.

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