Question

A $5\text{V}$ battery and a $2\text{V}$ battery are connected to the resistances as shown. The current in the $10\Omega$ resistor is :

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

Answer 1

The correct option is D $0.03\text{A},{\text{P}}_{2}to{\text{P}}_1$

Let's assume potential at ${\text{P}}_1$ to be zero, and potential at ${\text{P}}_2$ to be $x$.


Now let's take currents in the branches and apply KCL

$i_1+i_2+i_3=0$

$\frac{x-5}{2}+\frac{x-0}{10}+\frac{x+2}{1}=0$

$\Rightarrow x=\frac{5}{16}\text{V}$

So, current through $10\Omega$ resistor will be

$\Rightarrow i_2=\frac{5/16-0}{10}=0.03\text{A}$

So, $0.03\text{A}$ current will flow from ${\text{P}}_2$ to ${\text{P}}_1$.