Question

In the network shown below, remove terminal C and obtain the Norton equivalent seen at terminals a and b.

• A. -0.3 A parallel with 6.67 $\Omega$
• B. -0.21 A in parallel with 0.67 $\Omega$
• C. 0.4 A parallel with 3.4 $\Omega$
• D. 0.42 A parallel with 6.67 $\Omega$

Answer 1

The correct option is A -0.3 A parallel with 6.67 $\Omega$

Removing terminal C the circuit can be rearranged as,


$V_{Th}$ is the voltage dropped by 12 $\Omega$ resistor
$V_{Th}=V_a-V_b=2-V_b$
Apply KCL, at node b

$\frac{V_b-Va}{12}+0.1+\frac{V_b-5}{15}=0$

By solving, we get $V_b=4V$

$V_{Th}=-2V$

$R_{Th}=\frac{12\times 15}{12+15}=6.667\Omega$

and $I_N=\frac{V_{Th}}{R_{Th}}=-0.3A$