Test: KCL & KVL in AC Circuits - 2 - Electrical Engineering (EE) MCQ

Complex Power:

Considered a circuit in which supply voltage, V ∠α  and current, I ∠β and impedance Z,

Complex Power (S) = VI*

S = (V ∠+α) (I ∠-β) = VI ∠(α - β)

Considered, (α - β) = ϕ 

S = VI (cos ϕ + j sin ϕ) …. (1) 

And, S = P + jQ …. (2)

From equation (1) and (2),

P = VI cos ϕ

Q = VI sin ϕ

The phasor diagram of the circuit can be drawn as,

From the phasor, we can say the load is inductive,

• Inductive load has the property of absorbing reactive power by the load, therefore reactive power (Q) is positive.
• From the phasor it is clear that the voltage leads the current, therefore the phase angle between the voltage and current (α - β) is positive.

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

According to Kirchhoff’s Voltage Law, Every mesh is a loop but every loop is not a mesh. Mesh is a special case of loop which is planar.

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

KVL states that the sum of the potential energy and taken with the right sign is equal to zero, hence it is the conservation of energy since energy doesn’t enter or leave the system.

Using KVL,
12 - V₁ - 8 = 0.
V₁ =  4V.
8 - V₂ - 2=0.
V₂ = 6V.

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

Assuming i₁ and i₂ be the currents in loop 1 and 2 respectively. In loop 1, 4 + 2i1+3(i₁ - 17/3) + 4(i₁ - i₂) + 5 = 0
In loop 2, i₂(4 + 1 + 5)-4i₁ - 5 = 0
⇒ -4i₁ + 10i₂ = 5.
Solving these equations simultaneously i₂ = 1.041A and i₁ = 1.352A
V= i₂ x 5 = 5.21V.

Using KVL in loop 1,
10-  100 i₁ = 0. i₁ = 0.1A
Using KVL in outer loop,
-100i₂ + 20 = 0 i₂ = 0.2A.

Total resistance = 5 + 10 + 15 = 30 ohm. Current in the circuit is 12/30 A.
Voltage across 10 ohm resistor is 10 x (12/30) = 4V.