Kirchhoff’s Laws & Network Solutions - Free MCQ Practice Test with solutions,

Sum of voltage drops + Sum of voltage rise = 0

Σ Voltage rise = Σ Voltage drop

12 V = 5 V + 3 V + V

V = 12 – 5 – 3 = 4 V

According to Kirchhoff's Voltage Law:

V₃ + V₄ - V₅ = 0

Hence,

V₃ + V₄ = V₅

Resistor color coding: Red Yellow Brown Gray = 240 Ω ±10%

For resistor V = IR = 50 mA * 240 ohms = 12 V

Applying Kirchhoff’s voltage law:

Σ Voltage rise = Σ Voltage drop

V1 + 12 V = 12 V + 3 V + 5 V

V1 = 20 V – 12 V = 8 V

Kirchhoffs Voltage Law (KVL):

• Kirchhoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero.
• In other words, the algebraic sum of all voltages within the loop must be equal to zero i.e. ∑V = 0
• KVL is based on the law of conservation of energy.

Consider the figure shown below:

∴ According to KVL,

V1 + V2 + V3 + V4 + V5 = 0

Calculation:

The circuit diagram given in the question can be modified as:

Now by applying KVL in the above circuit, we get

5 + 7 - Vx = 0

∴ - 5 - 7 + Vx = 0

Correct option: C. The calculated values are I1 = 1.5 A, I2 = -0.188 A and I3 = -1.19 A (rounded).

Writing mesh KVL equations for the three clockwise mesh currents gives the linear system:

3I1 - 2I2 = 5

-2I1 + 9I2 - 4I3 = 0

-4I2 + 9I3 = -10

In the first equation the coefficient 3 is 1 Ω + 2 Ω (top and shared bottom resistor) and -2 is the shared 2 Ω to the middle mesh. The third equation has 9 = 5 Ω + 4 Ω and the RHS is -10 V from the right-hand source orientation.

From the first equation I1 = (5 + 2I2)/3.

Substituting into the second and simplifying yields 23I2 - 12I3 = 10.

From the third equation I3 = (4I2 - 10)/9.

Substituting this into 23I2 - 12I3 = 10 and solving gives I2 = -10/53 ≈ -0.188 A.

Then I1 = (5 + 2I2)/3 = 245/159 ≈ 1.54 A, which rounds to 1.5 A as given in the options.

Finally I3 = -190/159 ≈ -1.195 A, rounded to -1.19 A.

Therefore, after standard rounding used in the options, I1 = 1.5 A, I2 = -0.188 A, I3 = -1.19 A, so option C is the correct choice.

5 -3 (I1) + 2 (I2) = 0

9 (I2) - 2(I1) = 0

-4 (I2) + V = 0
On solving,V = 1.739 V.

To find the value of I:

V*I = P

=> 100*I = 1000

=> I = 10A

• Voltage across the 2 ohm resistor
  = 20V
• Voltage across the R resistor
  = 100 - 20 = 80 V

R = V / I

=> R = 80 / 10 = 8 ohm.

• The 4 ohm resistor gets shorted since current always prefers the low resistance path.
• All the current flows to the branch which is connected in parallel to the 4 ohm branch, hence no current flows in the 4 ohm resistance.

Nodal analysis uses Kirchhoff’s Current Law to find all the node voltages.

Hence it is a method used to determine voltage.

Mesh analysis uses Kirchhoff’s Voltage Law to find all the mesh currents.

Hence it is a method used to determine current.

If we move in the clockwise direction, we get the total voltage to be equal to:

-10 - 20 + 30 = 0V

Since I = V / R = 0 / 4 = 0

I = 0 A

 The 15A current source has a lower resistance path associated with it and hence it keeps moving in that particular loop.

It does not leave that loop and enter the circuit, hence the circuit is not affected by it.

KVL employs mesh analysis to find the different mesh currents by finding the I*R products in each mesh.

Concept of Kirchoff's Laws:

• Kirchhoff’s laws are used for voltage and current calculations in electrical circuits.
• These laws can be understood from the results of the Maxwell equations in the low-frequency limit.
• They are applicable for DC and AC circuits at low frequencies where the electromagnetic radiation wavelengths are very large when we compare with other circuits. So they are only applicable for lumped parameter networks.

​Kirchhoff's laws (KCL and KVL) is applicable to networks that are:

1. Unilateral or bilateral 
2. Active or passive 
3. Linear or non-linear
4. Lumped network