AC Network Theorem MCQ [Free PDF] – Objective Question Answer for AC Network Theorem Quiz

Answer: A

According to KVL, the sum of all voltages of branches in a closed loop is zero.

Kirchhoff’s voltage law is based on the principle of conservation of energy. This requires that the total work done in taking a unit positive charge around a closed path and ending up at the original point is zero.

Answer: B. 12 V

The current through the 10 ohm resistor = v1/10 = 2A.

Applying KCL at node 1:

i5 = i10+i2. i2 = 6-2 = 4A.

Thus the drop in the 2 ohm resistor = 4 × 2 = 8V.

v1 = 20V;

hence v2 = 20-v across 2 ohm resistor = 20-8 = 12V

v2 = v s

ince they are connected in parallel.

v = 12V.

Answer.1. 0.25A

Let voltage across the 8 Ω resistance is ‘V’ volt.

∴ Current across the 8 Ω is given by

I = V/8

Now by applying KCL at the node  we get

\({{V – 5} \over 2}+{{V +3} \over 4}+{{V } \over 8}=0\)

4V – 20 + 2V + 6 + V = 0

V = 14/7

Now current flowing through the 8 Ω resistance is

I = 2/8

I = 0.25 A

Answer: C

KCl states that the total current leaving the junction is equal to the current entering it. In this case, the current entering the junction is 5A+10A = 15A.

Answer.1. -1 Amp

Apply KCL at node V₁, we get:

\(\frac{{{{\rm{V}}_1} – 0}}{1} + \frac{{{{\rm{V}}_1} – 8}}{1} + \frac{{{{\rm{V}}_1} – 0}}{1} + \frac{{{{\rm{V}}_1} – 8}}{1} = 0\)

4V₁ – 16 = 0

V₁ = 4 V

Again, applying KCL, we can write:

\({\rm{i}} + \frac{{\left( {0 – {{\rm{V}}_1}} \right)}}{1} + 5 = 0 \)

i = V1 − 5 = 4 − 5 = −1 Amp

Answer: D

Assume a lower terminal of 20 ohms at 0V and upper terminal at V volt and applying KCL, we get

V/10 +V/20 = 1. V = 20/3V

So current through 20 ohm

= V/20 = (20/3)/20

= 1/3 = 0.33V.

Answer.3.

Electric current, i = Rate of transfer of electric charge.

i(t) = dQ/dt

Calculation:

t = 5 s so, equation 3rd is consider.

\(i = \frac{{dQ}}{{dt}} = \frac{d}{{dt}}\left( {3 + {e^{ – 2\left( {t – 2} \right)}}} \right)\)

\(i = {e^{ – 2\left( {t – 2} \right)}}\frac{d}{{dt}}\left[ { – 2\left( {t – 2} \right)} \right]\)

\(i = {e^{ – 2\left( {t – 2} \right)}}\left( { – 2} \right)\)

\(i = – 2{e^{ – 2\left( {t – 2} \right)}}\)

Put the value of t = 5, then we get,

i = −2e⁻⁶A

Answer: A

According to KCl, I1+I2+I3 = 0.

Hence I3 = -(I1+I2) = -5A.

The three mesh equations are:

− 3I1 + 2I2 − 5 = 0

2I1 − 9I2 + 4I3 = 0

4I2 − 9I3 − 10 = 0

Solving the equations, we get

I1 = 1.54A, I2 = − 0.189 and I3 = − 1.195A.

The two mesh equations are:

5I1 − 3I2 = 10

− 3I1 + 7I2 = − 15

Solving the equations simultaneously, we get

I1 = 0.96A and I2 = − 1.73A.