300+ Network Theorem MCQ – Objective Question Answer for Network Theorem Quiz

Short-circuiting terminals A and B,

20 − 10(I1) = 0

I1 = 2A. 10 − 5(I2)

I2 = 2A.

Current flowing through terminals A and B = 2 + 2 = 4A.

The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 5Ω resistor

= > R = (10×5)/(10 + 5) = 3.33Ω.

To solve for Norton’s current we have to find the current passing through the terminals A and B. Short-circuiting the terminals a and b,

I = 100/(6×10)/(6 + 10) + (15×8)/(15 + 8)) = 11.16 ≅ 11A.

The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 6Ω resistor and parallel combination of the 15Ω resistor and the 8Ω resistor

= > R = (10×6)/(10 + 6) + (15×8)/(15 + 8) = 8.96≅9Ω.

To solve for Norton’s current we have to find the current passing through the terminals A and B. Short-circuiting the terminals a and b

I = 11.16×8.96/(5 + 8.96) = 7.16A.

The voltage drop across 5Ω resistor in the circuit is the product of current and resistance

= > V = 5×7.16 = 35.8 ≅ 36V.

Applying KCL in the given circuit, we get

\(\frac{V}{4} + \frac{V − 2I}{2}\) = I

Or

3V − 4I/4 = I

Or, 3V = 8I

∴ V/I = 8/3 Ω.

By writing loop equations for the circuit, we get,

V_S  = V_X, I_S  = I_X

V_S  = 600(I₁ – I₂) + 300(I₁ – I₂) + 900 I₁

= (600 + 300 + 900) I₁ – 600I₂ – 300I₃

= 1800I₁ – 600I₂ – 300I₃

I₁  = I_S, I₂  = 0.3 V_S

I₃  = 3I_S  + 0.2V_S

V_S  = 1800IS – 600(0.01V_S) – 300(3I_S  + 0.01V_S)

= 1800I_S – 6V_S – 900I_S – 3V_S

10V_S  = 900I_S

For Voltage, V_S  = R_N I_S  + V_OC

Here V_OC  = 0

So, Resistance R_N  = 90Ω.

I_X  = 1 A, V_X  = V_test

V_test  = 100(1 − 2I_X) + 300(1 − 2I_X – 0.01V_S) + 800

V_test  = 1200 – 800I_X – 3V_test

4V_test  = 1200 – 800 = 400

V_test  = 100V

∴ R_N = V/1 = 100 Ω.

R_N = V_OC/I_SC

V_N  = V_OC

Applying KCL at node A

\(\frac{2I − V_N}{1} + 2 = I + V_N/2

Or, I = V_N/1

Putting, 2V_N – V_N  + 2 = V_N + V_N/2

Or, V_N  = 4 V.

∴ R_N  = 4/2 = 2Ω.

\(\frac{V_P − 100}{10} + \frac{V_P}{10}\) + 2 = 0

2V_P – 100 + 20 = 0

∴ V_P  = 80/2 = 40V

∴ R = 20Ω (By Norton’s Theorem).