Find Norton equivalent resistance, RN, and equivalent current source, in, at terminals a and b of the circuit.

Find Norton equivalent resistance, R_N, and equivalent current source, i_N, at terminals a and b of the circuit.

Right Answer is:
R_N = 4 Ω, i_N = 1 A

SOLUTION

For Norton resistance, replace the sources with their internal resistance.

The voltage source is replaced by short-circuited and the current source is replaced by open-circuited.

Then finding the equivalent resistance across terminals ab

Resistance 8Ω, 4Ω & 8Ω are connected in series

R = (8 + 4 + 8) = 20Ω

Finally, 20Ω is parallel with 5Ω

R_eq = R_N = R_th = 20||5

R_eq = (20 × 5)/(20 + 5)

R_eq = 4Ω

For finding the Norton current, a-b terminals are shorted and therefore the 5-ohm resistor is eliminated and converting voltage source into the current source.

Total current = 2A + 3 A = 5A

Applying the current divider rule

I_N = (4 × 5)/(4 + 8 + 8)

I_N = 1 A

Find Norton equivalent resistance, RN, and equivalent current source, in, at terminals a and b of the circuit.

R_N = 5.33 Ω, i_N = 0.71 A

R_N = 4 Ω, i_N = 1 A

R_N = 12 Ω, i_N = 0.71 A

R_N = 5 Ω, i_N = 2 A

Right Answer is:
R_N = 4 Ω, i_N = 1 A

SOLUTION

Find Norton equivalent resistance, RN, and equivalent current source, in, at terminals a and b of the circuit.

RN = 5.33 Ω, iN = 0.71 A

RN = 4 Ω, iN = 1 A

RN = 12 Ω, iN = 0.71 A

RN = 5 Ω, iN = 2 A

Right Answer is: RN = 4 Ω, iN = 1 A

SOLUTION

R_N = 5.33 Ω, i_N = 0.71 A

R_N = 4 Ω, i_N = 1 A

R_N = 12 Ω, i_N = 0.71 A

R_N = 5 Ω, i_N = 2 A

Right Answer is:
R_N = 4 Ω, i_N = 1 A