Question 1.P.6: For the circuit shown in figure below, find the voltage acro......

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