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


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10 Questions MCQ Test GATE Electrical Engineering (EE) 2024 Mock Test Series - Test: KCL & KVL in AC Circuits - 2

Test: KCL & KVL in AC Circuits - 2 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) 2024 Mock Test Series preparation. The Test: KCL & KVL in AC Circuits - 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: KCL & KVL in AC Circuits - 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: KCL & KVL in AC Circuits - 2 below.
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Test: KCL & KVL in AC Circuits - 2 - Question 1

If the voltage across and the current into a certain load or part of the circuit are expressed by I = |I|∠β, respectively. Reactive power (VI*) will be ___________ when the phase angle ∠(α - β) and current is __________

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 1

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.
Test: KCL & KVL in AC Circuits - 2 - Question 2

KVL is applied in _______

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 2

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.

Test: KCL & KVL in AC Circuits - 2 - Question 3

Every____________ is a ____________ but every __________ is not a __________

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 3

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.

Test: KCL & KVL in AC Circuits - 2 - Question 4

The sum of the voltages over any closed loop is equal to __________

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 4

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

Test: KCL & KVL in AC Circuits - 2 - Question 5

KVL deals with the conservation of?

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 5

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.

Test: KCL & KVL in AC Circuits - 2 - Question 6

Calculate the value of V1 and V2.

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 6

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

Test: KCL & KVL in AC Circuits - 2 - Question 7

What is the basic law that has to be followed in order to analyze the circuit?

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 7

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

Test: KCL & KVL in AC Circuits - 2 - Question 8

What is the voltage across the 5 ohm resistor if current source has current of 17/3 A?

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 8

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.

Test: KCL & KVL in AC Circuits - 2 - Question 9

Find the value of the currents I1 and I2.

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 9

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

Test: KCL & KVL in AC Circuits - 2 - Question 10

Calculate the voltage across the 10 ohm resistor.

Detailed Solution for Test: KCL & KVL in AC Circuits - 2 - Question 10

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.

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