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

Thevenin’s voltage is equal to the open-circuit voltage across the terminals AB that is across a 12Ω resistor.

Vth  = 10×12/14 = 8.57V.

The resistance into the open circuit terminals is equal to the Thevenin’s resistance

= > Rth  = (12×2)/14 = 1.71Ω.

The equivalent thevenin’s model of the circuit shown is

I = 8.57/(2.4 + 1.71) = 0.33A.

Let us find the voltage drop across terminals A and B.

50 − 25 = 10I + 5I = > I = 1.67A.

Voltage drop across 10Ω resistor = 10×1.67 = 16.7V.

So, Vth = VAB = 50 − V = 50 − 16.7 = 33.3V.

To find Rth, two voltage sources are removed and replaced with a short circuit. The resistance at terminals AB then is the parallel combination of the 10Ω resistor and 5Ω resistor

= > Rth = (10×5)/15 = 3.33Ω.

Current through 3Ω resistor is 0A. The current through 6Ω resistor

= (50 − 10)/(10 + 6) = 2.5A.

The voltage drop across 6Ω resistor = 25×6 = 15V.

So the voltage across terminals A and B = 0 + 15 + 10 = 25V.

To find Rth, two voltage sources are removed and replaced with short circuit

= > Rth = (10×6)/(10 + 6) + 3 = 6.75Ω.

The voltage at terminal a is

Va = (100×6)/16 = 37.5V,

The voltage at terminal b is

Vb = (100×8)/23 = 34.7V.

So the voltage across the terminals ab is

Vab = Va − Vb = 37.5 − 34.7 = 2.7V.

To find Rth, two voltage sources are removed and replaced with short circuit

= > Rab = (6×10)/(6 + 10) + (8×15)/(8 + 15) = 8.96≅9V.

The Equivalent of Thevenin’s circuit for the circuit shown above is

I = 2.7/(8.96 + 5) = 0.193A≅0.2A.

We, Thevenized the left side of xx’ and source transformed the right side of yy’.

V_xx’  = V_th = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

∴ R_th  = 8 || (16 + 8)

= (8×24)/(8 + 24) = 6 Ω.

We, Thevenized the left side of xx’, and the source transformed the right side of yy’.

Thevenin equivalent as seen from terminal yy’ is

Vxx’ = Vth = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

(0.167 + 1)/(0.04167 + 0.125) = 7 V
∴ R_th  = (8 + 16) || 8

= (24×8}/(24 + 8) = 6 Ω.

The Thevenin equivalent of the circuit is as shown below.

Therefore from the figure, we can infer that R_th  = 2 Ω

The Thevenin equivalent of the circuit is as shown below.

I = 10 A, R_th  = 2 Ω

∴ P_max = (10/2)² × 2

= 5×5×2 = 50 W.

Let V_AB  = 1 V

5 V_AB  = 5

Or, 1 = 1 × I₁ or, I₁  = 1

Also, 1 = − 5 + 1(I – I₁)

∴ I = 7

Hence, R = 0.2 kΩ.

Current through 1 Ω = (5/2) – I₁

Using source transformation to 5 V sources

V_OC  = 1 × I₁

V_OC  = − 5 V_OC + (5/2) – I₁) × 1

Eliminating I1, we get, V_OC  = 0.5 V.

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 Thevenin equivalent, V_S  = R_TH I_S  + V_OC

So, Thevenin voltage V_OC  = 0

Resistance R_TH  = 90Ω.

When R_L  = 10 kΩ and V_AB  = 4 V

Current in the circuit

I = V_AB/R_L = 4/10 = 0.4 mA

Thevenin voltage is given by

V_TH  = I (R_TH  + R_L)

= 0.4(R_TH  + 10)

= 0.4R_TH  + 4

Similarly, for R_L  = 2 kΩ and V_AB  = 1 V

So, I = 1/2 = 0.5 mA

V_TH  = 0.5(R_TH  + 2)

= 0.5 R_TH  + 1

∴ 0.1R_TH  = 3

Or, R_TH  = 30 kΩ

And V_TH  = 12 + 4 = 16 V.

I_X  = 1 A, V_X  = V_test

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

Or, V_test  = 1200 – 800I_X – 3V_test

Or, 4V_test  = 1200 – 800 = 400

Or, V_test  = 100V

∴ R_TH = V_test/1 = 100 Ω.

R_TH = V_oc/I_sc

V_TH  = V_OC

Applying KCL at node A,

\(\frac{2I − V_{TH}}{1} + 2 = I + \frac{V_{TH}}{2}\)

Or, I = V_TH/1

Putting, 2V_TH – V_TH  + 2 = V_TH + VTH/2

Or, V_TH  = 4 V.

∴ R_TH  = 4/2 = 2Ω.