Test: Kirchhoff’s Current & Voltage Law - Electrical Engineering (EE) MCQ

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 since they are connected in parallel.
v = 12V

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 5 A + 10 A = 15A

Assume lower terminal of 20 ohm at 0V and upper terminal at V volt and applying KCL, we get V/10 + V/20 = 1

V = (20 / 3) V

So, current through 20 ohm = V/20 = (20/3) / 20 = 1/3 = 0.33V

 According to KCL, I₁ + I₂ + I₃ = 0

Hence I₃ = - (I₁ + I₂) = - 5A

•  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

Current = Charge/Time

Here charge = 50C and time = 5 seconds

so current = 50/5 = 10A

KCL states that the amount of charge leaving a node is equal to the amount of charge entering it, hence it is applied at nodes.

KCL is applied for different nodes of a network whether it is planar or non-planar.

 At the junction, I - 2 + 3 - 4 - 5 = 0

Hence I = 8A

Using KVL, 12 - V1 - 8 = 0.

V1 = 4V

8 - V2 - 2 = 0

V2 = 6V

KCL states that the amount of charge entering a junction is equal to the amount of charge leaving it, hence it is the conservation of charge.

Using voltage divider rule, V = 10*12/30 = 4V

To explain: Using KVL in loop 1, 10-100i1 = 0. i1 = 0.1A
Using KVL in outer loop, -100i2 + 20 = 0  i2 = 0.2A.
Correct Answer is d)  0.1, 0.2

According to KVL, the sum of the voltage over any closed loop is equal to 0.

Kirchhoff’s laws, namely Kirchhoff’s Current Law and Kirchhoff’s Voltage law are the basic laws in order to analyze a circuit.

According to Kirchhoff’s Voltage Law, Every mesh is a loop but every loop is not a mesh.

V_eq = 10 + 5 -20 = -5u
R_eq = 5 + 2 + 3 = 10Ω
I = V/R = -5/10 = -0.5A.

For branch A: V_AC= 15 * 20 / (25+15) = 7.5V
For branch B: V_BC= 10 * 20 / (10+40) = 4V
Applying KVL to loop ABC
V_AB+V_BC+V_CA = 0
V_AB = 3.5V

Mesh analysis helps us to utilize the different voltages in the circuit as well as the I_R products in the circuit which is nothing but KVL.

Kirchhoff's Current Law

According to Kirchhoff's Current Law, The total current entering a junction or a node is equal to the charge leaving the node. The algebraic sum of every current entering and leaving the node has to be zero.
KCL is based on the conservation of charge.

Hence, statement 4 is correct.
Kirchhoff's Voltage Law
According to Kirchhoff's Voltage Law, the sum of the total voltage drop in a closed loop is equal to zero.
V = V₁ + V₂. . . . . V_n
V = I(R₁+ R₂. .. R_n)