# 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

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.

1. 0.25 A
2. 0.50 A
3. 0.75 A
4. 0.10 A

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

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. -1 Amp
2. 5 Amp
3. 2 Amp
4. -2 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

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

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.

1. 0 A
2. 2 A
3. -2e-6 A
4. 3 + e-6 A

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

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

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?

1. i1 – i2 + i3 – i4 = 0
2. i1 + i2 – i3 + i4 = 0
3. i1 – i2 – i3 + i4 = 0
4. i1 – i2 = 0

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

h({});

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.

1. 5 Ω
2. 3 Ω
3. 12 Ω
4. 10 Ω

Given the current through AO is zero,

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

Hence, VBA = VBO …. (1)

Also, VAC = VOC …. (2)

VAC = 4(3I) volts

VOC = IR

From equation  (2),

1 × 2 I = IR

∴ R = 12 Ω

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