Question

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 resistor as shown in the figure.


The current in the resistor is

• A. $0.27\text{A},P_2$ to $P_1$
• B. $0.03\text{A},P_1$ to $P_2$
• C. $0.03\text{A},P_2$ to $P_1$
• D. $0.27\text{A},P_1$ to $P_2$

Answer 1

The correct option is C $0.03\text{A},P_2$ to $P_1$

Let potential at $P_1$ is $0\text{V}$ and potetial at $P_2$ is $V_0$. Now apply KCL at $P_2$.


$\frac{V_0-5}{2}+\frac{V_0-0}{10}+\frac{V_0-(-2)}{1}=0$

$V_0=\frac{5}{16}$

So, current through $10\Omega$ resistor is, $V=IR$ (Ohm's law)

$\Rightarrow I=\frac{\Delta V}{R}=\frac{\frac{5}{16}}{10}=\frac{1}{32}\text{A}$

or $I=0.03125\approx 0.03\text{A}$

(From $P_2$ to $P_1$)