## Kirchhoff’s Voltage Law MCQ (Interview-Exam) Question-Answer

**Q.1** Calculate the value of V1 and V2.

**A.** 5V, 6V

**B.** 7V, 8V

**C.** 4V, 6V

**D.** 6V, 7V

**Ans : **4V, 6V

**Explanation: ** KVL, 12-V1-8=0. V1= 4V.

8-V2-2=0. V2=6V.

**Q.2** KVL deals with the conservation of?

**A.** Momentum

**B.** Energy

**C.** Mass

**D.** Charge

**Ans : **Energy

**Q.3** Find the value of the currents I1 and I2.

**A.** -0.1, -0.3

**B.** 0.1, 0.2

**C.** 0.3, 0.1

**D.** -0.3, -0.1

**Ans : **0.1, 0.2

**Explanation: ** KVL in loop 1, 10-100 i1=0. i1=0.1A

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

**Q.4** Calculate the voltage across the 10 ohm resistor.

**A.** 4V

**B.** 0V

**C.** 12V

**D.** 10V

**Ans : **4V

**Explanation: ** Total resistance = 5+10+15 = 30 ohm. Current in the circuit is 12/30 A.

Voltage across 10 ohm resistor is 10*(12/30) = 4V.

**Q.5** What is the basic law that has to be followed in order to analyze the circuit?

**A.** Faraday’s laws

**B.** Kirchhoff’s law

**C.** Newton’s laws

**D.** Ampere’s laws

**Ans : **Kirchhoff’s law

**Q.6** The sum of the voltages over any closed loop is equal to __________.

**A.** Infinity

**B.** 2V

**C.** 0V

**D.** 1V

**Ans : **0V

**Explanation: ** I=V/R. Total resistance R = 20+40=60ohm. V=120V. I=120/60=2A.

**Q.7** What is the voltage across the 5 ohm resistor if current source has current of 17/3 A?

**A.** 5.21V

**B.** 8.96V

**C.** 2.32V

**D.** 6.67V

**Ans : **5.21V

**Explanation: ** 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*5= 5.21V.

**Q.8** Every____________ is a ____________ but every __________ is not a __________.

**A.** Loop, mesh, mesh, loop

**B.** Mesh, loop, mesh, loop

**C.** Mesh, loop, loop, mesh

**D.** Loop, mesh, loop, mesh

**Ans : **Mesh, loop, loop, mesh

**Q.9** KVL is applied in ____________.

**A.** Nodal analysis

**B.** Neither mesh nor nodal

**C.** Mesh analysis

**D.** Both mesh and nodal

**Ans : **Mesh analysis

**Q.10** Calculate VAB.

**A.** 12V

**B.** 6.5V

**C.** 3.5V

**D.** 9.5V

**Ans : **3.5V

**Explanation: ** For branch A: VAC=15*20/(25+15)=7.5V

For branch B: VBC= 10*20/(10+40)=4V

Applying KVL to loop ABC:

VAB+VBC+VCA=0

VAB=3.5V.

## Kirchhoff’s Voltage Law

**Conservation of Energy:**Kirchhoff’s Voltage Law is based on the principle of conservation of energy, stating that the total energy in a closed loop or a mesh in an electrical circuit is conserved.**Loop Voltage Sum:**The algebraic sum of all the voltage drops in any closed loop of a circuit is equal to the algebraic sum of all the voltage rises in that loop. This implies that the total energy supplied in a closed loop is equal to the total energy consumed.**Mathematical Representation:**Mathematically, KVL is expressed as ΣV = 0, where ΣV represents the sum of all the voltages in a closed loop, and the total sum is equal to zero.**Voltage Polarities:**When applying KVL, it is essential to consider the polarities of voltages. The voltage drops (across resistors, capacitors, or inductors) are considered positive if the assumed current direction matches the actual direction.**Mesh Analysis:**KVL is often used in conjunction with mesh analysis, a circuit analysis technique. Mesh analysis breaks down a complex circuit into individual loops, applying KVL to each loop to find the currents and voltages in the circuit.**Applicability:**Kirchhoff’s Voltage Law is applicable to both DC (direct current) and AC (alternating current) circuits, making it a versatile tool for circuit analysis in various electrical systems.**Mesh Currents:**In the context of mesh analysis, KVL is particularly useful for setting up equations involving mesh currents. The law helps establish relationships between the voltages and currents in different parts of the circuit.**Solving Circuit Problems:**KVL is a valuable tool for solving complex electrical circuit problems, providing a systematic approach to understanding and determining the voltage distribution in a circuit.