Kirchhoff’s Voltage Law (KVL) MCQ

1. Find the value of v if v1 = 20V and the value of the current source are 6A.

A. 10V
B. 12V
C. 14V
D. 16V

Show Explanation

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

Since they are connected in parallel.

v = 12V.

2. In the circuit shown in the figure, find the current flowing through the 8 Ω resistance.

0.25 A
0.50 A
0.75 A
0.10 A
Show Explanation

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

3. Calculate the current A by using Kirchhoff’s current law

A. 5A
B. 10A
C. 15A
D. 20A

Show Explanation

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.

4. In the figure shown, the current 𝑖 (in ampere) is __________

-1 Amp
5 Amp
2 Amp
-2 Amp
Show Explanation

Answer.1. -1 Amp

Apply KCL at node V_{1} , we get:

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

4V_{1} – 16 = 0

V_{1} = 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

5. By using Kirchhoff’s current law calculate the current across the 20-ohm resistor.

A. 20A
B. 1A
C. 0.67A
D. 0.33A

Show Explanation

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.

6. The total charge q(t), in the coulombs, that enters the terminal of an element is:

\(q(t) = \left\{ {\begin{array}{*{20}{c}} {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t < 0}\\ {2t\,\,\,\,\,\,\,\,\,\,\,\,0 \le t \le 2}\\ {3 + {e^{ – 2(t – 2)}}\,\,t > 2} \end{array}} \right.\)

Determine the current at t = 5 s.

0 A
2 A
-2e^{-6} A
3 + e^{-6} A
Show Explanation

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^{−6} A

7. Calculate the value of I3, if I1 = 2A and I2 = 3A by applying Kirchhoff’s current law

A. -5A
B. 5A
C. 1A
D. -1A

Show Explanation

Answer: A

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

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

8. What would be the correct equation representing Kirchhoff’s Current Law (KCL) at node a for the given network?

i_{1} – i_{2} + i_{3} – i_{4} = 0
i_{1} + i_{2} – i_{3} + i_{4} = 0
i_{1} – i_{2} – i_{3} + i_{4} = 0
i_{1} – i_{2} = 0
Show Explanation

By applying KCL, at node a

i1 – i2 – i3 + i4 = 0

9. Find the value of i2, i4, and i5 if i1 = 3A, i3 = 1A and i6 = 1A by applying Kirchhoff’s current law

A. 2,-1,2
B. 4,-2,4
C. 2,1,2
D. 4,2,4

Show Explanation

Answer: A

At junction a: i1-i3-i2 = 0. i2 = 2A.

At junction b: i4+i2-i6 = 0. i4 = -1A.

At junction c: i3-i5-i4 = 0. i5 = 2A.

10. In the circuit shown in the following figure, calculate the value of the unknown resistance R when the current in-branch OA is zero.

5 Ω
3 Ω
12 Ω
10 Ω
Show Explanation

Answer.3. 12 Ω

Given the current through AO is zero,

It means node A and node O has the same potential,

Hence, V_{BA} = V_{BO} …. (1)

Also, V_{AC} = V_{OC} …. (2)

V_{AC} = 4(3I) volts

V_{OC} = IR

From equation (2),

1 × 2 I = IR

∴ R = 12 Ω