## Q. 1.P.47

Norton equivalence of the network `X’ to the left of AB is a current source $I_{ N }=$ 5 A from B to A, $R_{ N }=$ 3 Ω and current through R when it is connected across AB = 2 A. What is the value of resistance R?

## Verified Solution

The resistance is given by

$\begin{gathered} \frac{5 \times 3}{3+R}=2 \Rightarrow \frac{15}{3+R}=2 \\ 5+6+2 R \Rightarrow 2 R=9 \Rightarrow R=\frac{9}{2}=4.5 \Omega \end{gathered}$

Ans. (4.5)