Question

Find the current in the 5 ohm resistance using norton theorem?

Answer 1

I2 = (20-A)/2

I1 = (-10-A)

I5 = -A/5

I4 = (12-A)/4

I1+I2+I4+I5 = 0

10 - A/2 -10 -A -A/5 +3 -A/4 = 0

10 - 10A/20 -10 -4A/20 +3 -5A/20 = 0

60 - 10A -20A -4A -5A = 0

60 = 39A

A = 60/39 = 1.538

Answer 2

0.5A the current in the 5 ohm resistance using norton theorem.

Explanation:

Take Shorting all voltage sources and opening all current sources we have:

RN$=(3||6)+10$

    $= 12$ ohm.

• 5 ohm is the load resistance,we shorted it and find the resistance through the short.

• If we apply source transformation between the 6 ohm resistor and the 1A source,we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. Let us consider I1 flowing in the first mesh and I2 flowing in the second mesh.

• The mesh equations are:

$9I1-6I2=4$

$-6I1+16I2=6$

On solving these equations simultaneously, we get $I2=0.72A$, which is the short circuit current.

• Connecting the current source in parallel to RN which is in turn connected in parallel to the load resistance=5ohm, thn en we get Norton’s equivalent circuit.

Using current divider: $I = \frac{0.72*12}{(12+5)}$

                                       $=\frac{8.64}{17}$

                                       $= 0.5 A.$