Question

Consider the circuit shown in the figure below;
The Norton's equivalent seen across the terminal A and B is

• A. $I_N=0,R_N=100\Omega$
• B. $I_N=3.67A,R_N=24\Omega$
• C. $I_N=8.8A,R_N=24\Omega$
• D. $I_N=1.6A,R_N=100\Omega$

Answer 1

The correct option is B $I_N=3.67A,R_N=24\Omega$

Redrawing the circuit, we get ,

To calculate Norton's equivalent resistance, the circuit can be modified as

$R_{AB}=R_N=40\Omega ||60\Omega =\frac{40\times 60}{100}=24\Omega$

To calculate Norton's equivalent current, the circuit can be modified as

Using nodal analysis, we get ,

$\frac{V_{AB}-120}{40}+\frac{V_{AB}-40}{60}=0$

$100V_{AB}-7200-1600=0$

$100V_{AB}=8800$

or $V_{AB}=88V$

$I_{AB}=I_N=\frac{V_{AB}}{R_{eq}}=\frac{88}{24}=3.67A$