## Q. 1.P.46

Find the equivalent Norton circuit of the circuit shown in the following figure.

## Verified Solution

Replacing 10 Ω resistance by short circuit, we have

\begin{aligned} & \frac{V_{ A }-30}{5}+5+\frac{V_{ A }}{5}=0 \\ & \frac{V_{ A }-30+25+V_{ A }}{5}=0 \\ & 2 V_{ A }=5 \Rightarrow V_{ A }=2.5 V \\ & I_{ sc }=\frac{2.5}{5}=0.5 A \end{aligned}

For $R_{ Th }$ between A and B, we have

$R_{ Th }=\frac{10 \times 10}{20}=5 \Omega$

The equivalent Norton circuit is