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

(C) $0.03\mathrm{~A}$ $\mathrm{P}_{2}$ to $\mathrm{P}_{1}$

(A) $0.27\mathrm{~A}$ $\mathrm{P}_{2}$ to $\mathrm{P}_{1}$

(B) $0.03\mathrm{~A}$ $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$

(D) $0.27\mathrm{~A}$ $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$

JEE Main · 2008 · 27 Apr 2008 · Offline · Current Electricity

A 5 V battery with internal resistance

2Ω

and a 2 V battery with internal resistance

1Ω

are connected to a

10Ω

resistor as shown in the figure. The current in the

10Ω

resistor is

Answer: (C)

Apply Millman's theorem to the three parallel branches between the two nodes, taking the two cells with their internal resistances and the

10 Ω

resistor. The node potential is

\frac{\frac{5}{2} - \frac{2}{1}}{\frac{1}{2} + \frac{1}{1} + \frac{1}{10}} = \frac{0.5}{1.6} \approx 0.31 V

, and the current driven through the

10 Ω

resistor is

\frac{0.31}{10} \approx 0.03 A

. The higher potential sits at

P_{2}

, so this current of about

0.03 A

flows through the resistor from

P_{2}

P_{1}

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option