Question

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

• A. $0.03A{P}_1\text{to}{P}_2$
• B. $0.03A{P}_2\text{to}{P}_1$
• C. $0.27A{P}_1\text{to}{P}_2$
• D. $0.27A{P}_2\text{to}{P}_1$

Answer 1

The correct option is B $0.03A{P}_2\text{to}{P}_1$

$\begin{array}{c}ApplyingKirchhof{f}^{\prime }slawfortheloops\\ A{P}_2{P}_{1}CAand{p}_{2}BD{p}_1{p}_2.\\ -10y-2x+5=0\\ \Rightarrow 2x+10y=\mathrm{5..............}\left(1\right)\\ =2-1\left(x-y\right)+10.y=0\\ \Rightarrow x-11y=\mathrm{2.............}\left(2\right)\\ 2x-22y=\mathrm{4..............}\left(3\right)\\ from\left(1\right)\&\left(3\right)gives32y=1\\ \Rightarrow y=\frac{1}{32}A\\ =0.03Afrom{p}_{2}to{p}_1.\end{array}$