## Q. 1.P.6

For the circuit shown in figure below, find the voltage across 10 Ω resistor and current passing through it.

## Verified Solution

Assuming voltage V at the node A,

According to KCL

$I_1+I_2+I_3+I_4+5=10$

Using Ohm’s law

$I_1=\frac{V}{5}, \quad I_2=\frac{V}{10}, I_3=\frac{V}{2}, I_4=\frac{V}{1}$

Therefore,

\begin{aligned} & \frac{V}{5}+\frac{V}{10}+\frac{V}{2}+\frac{V}{1}+5=10 \\ & V\left[\frac{1}{5}+\frac{1}{10}+\frac{1}{2}+1\right]=5 \end{aligned}

V = 2.78 Volts
The voltage across 10 Ω resistor is 2.78 V and current passing through it is

$I_2=\frac{V}{10}=\frac{2.78}{10}=0.278 A$