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

• A. $0.27A,P_{2}to{P}_1$
• B. $0.03A,P_{1}to{P}_2$
• C. $0.03A,P_{2}to{P}_1$
• D. $0.27A,P_{1}to{P}_2$

Answer 1

The correct option is C $0.03A,P_{2}to{P}_1$

Current through $10\Omega$resister


$=I_1-I_2$
$=\frac{1}{32A}$ Loop (1)
$=0.03Afrom{P}_{2}to{P}_1$ $12I_1–10I_2=5$ ---- (1)
Loop (2)
$-10I_1+11I_2=2$ ----(2)
Solving (1) & (2)
$I_1=\frac{75}{32}A$
$I_2=\frac{74}{32}A$