Kirchhoff's Laws Important Questions and Answers

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Important Questions on Kirchoff’s Law

Ques 1. If R1 = 2Ω, R2 = 4Ω, R3 = 6Ω, determine the electric current flows in the circuit below. (3 marks)

Resistor 1 (R1) = 2Ω

Resistor 2 (R2) = 4Ω

Resistor 3 (R3) = 6Ω

Source of emf 1 (E1) = 9V

Source of emf 2 (E2) = 3V

From Kirchoff’s law, we arrive at the equation,

– I R1 + E1 – I R2 – I R3 – E2 = 0

– 2 I + 9 – 4 I – 6 I – 3 = 0

– 12 I + 6 = 0

– 12 I = – 6

I = -6 / -12

I = 0.5

Thus current flowing through the circuit is 0.5 A.

Ques 2. Determine the electric current that flows in the circuit as shown in the figure below. (3 marks)

Ans. Consider the direction of the current is the same as the direction of clockwise rotation.

- 20 – 5I -5I – 12 – 10I = 0

- 32 – 20I = 0

- 32 = 20I

I = -32 / 20

I = -1.6 A

Since the answer is negative, it means that the direction of the current is actually anti-clockwise.

Ques 3. Determine the electric current that flows in the circuit as shown in the figure below. (3 marks)

Ans. Consider the direction of the current is the same as the direction of clockwise rotation.

– I – 6I + 12 – 2I + 12 = 0

-9I + 24 = 0

-9I = -24

I = 24 / 9

I = 8 / 3 A

Thus, current flowing through the circuit is 2.66 A. 

Ques 4. An electric circuit consists of four resistors, R1 = 12 Ohm, R2 = 12 Ohm, R3 = 3 Ohm and R4 = 6 Ohm, and are connected with a source of emf E1 = 6 Volt, E2 = 12 Volt. Determine the electric current flows in the circuit as shown in the figure below. (3 marks)

Ans. Resistor 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor:

1/R12 = 1/R1 + 1/R2 = 1/12 + 1/12 = 2/12

R12 = 12/2 = 6 Ω

In this solution, the direction of the current is the same as the direction of clockwise rotation.

– I R12 – E1 – I R3 – I R4 + E2 = 0

– 6 I – 6 – 3I – 6I + 12 = 0

– 6I – 3I – 6I = 6 -12

– 15I = – 6

I = -6/-15

I = 2/5 A

Thus, current flowing through the circuit is 0.4 A. 

Ques 5. Determine the electric current that flows in the circuit as shown in the figure below. (3 marks)

Ans: Resistor 1 (R1) = 10 Ω

Resistor 2 (R2) = 6 Ω

Resistor 3 (R3) = 5 Ω

Resistor 4 (R4) = 20 Ω

Source of emf 1 (E1) = 8 Volt

Source of emf 2 (E2) = 12 Volt

Resistor 3 (R3) and resistor 4 (R4) are connected in parallel. The equivalent resistor :

1/R34 = 1/R3 + 1/R4 = 1/5 + 1/20 = 4/20 + 1/20 = 5/20

R34 = 20/5 = 4 Ω

In this solution, the direction of the current is the same as the direction of clockwise rotation.

– I R1 – I R2 – E1 – I R34 + E2 = 0

– 10I – 6I – 8 – 4I + 12 = 0

– 10I – 6I – 4I = 8 – 12

– 20I = – 4

I = -4/-20

I = 1/5 A

I = 0.2 A

Thus, current flowing through the circuit is 0.2 A. 

Ques 6. Determine the electric current that flows in the circuit as shown in the figure below. (5 marks)

Ans. Resistor 1 (R1) = 1 Ω

Resistor 2 (R2) = 6 Ω

Resistor 3 (R3) = 6 Ω

Resistor 4 (R4) = 4 Ω

Source of emf 1 (E1) = 12 Volt

Source of emf 2 (E2) = 6 Volt

The electric current that flows in a circuit.

Resistor 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor :

1/R12 = 1/R1 + 1/R2 = 1/1 + 1/6 = 6/6 + 1/6 = 7/6

R12 = 6/7 Ω

The direction of current is the same as the direction of clockwise rotation.

E1 – I R12 – E2 – I R4 – I R3 = 0

12 – (6/7)I – 6 – 4I – 6I = 0

12 – 6 – (6/7)I – 4I – 6I = 0

6 – (6/7)I – 10I = 0

6 = (6/7)I + 10I

6 = (6/7)I + (70/7)I

6 = (76/7)I

(6)(7) = 76I

42 = 76I

I = 42/76

I = 0.5 A

Thus, current flowing through the circuit is 0.5 A. 

Ques 7. From the given circuit find the value of I. (3 marks)

Ans. From Kirchoff’s rule,

The arrows pointing toward P are positive and away from P are negative.

Therefore, 0.2A – 0.4A + 0.6A – 0.5A + 0.7A – I = 0

1.5A – 0.9A – I = 0

0.6A – I = 0

I = 0.6 A

Thus, current flowing through the circuit is 0.6 A. 

Ques 8. The following figure shows a network of conductors. Apply Kirchoff’s voltage rule to the two closed loops like ACE and ABC. (3 marks)

Ans. Thus applying Kirchoff’s second law to the closed loop EACE, 

I₁R₁ + I₂R₂ + I₃R₃ = ξ 

For closed loop ABCA, 

I₄R₄ + I₅R₅ – I₂R₂= 0

Ques 9. Calculate the current that flows in the 1 Ω resistor in the following circuit. (3 marks)

Ans. Consider the loop EFCBE and apply KVR, we get

1I₂ + 3I₁ + 2I₁ = 9

5I₁ + I₂ = 9 …(1)

Applying KVR to the loop EADFE, we get

3 (I₁ – I₂) – 1I₂ = 6

3I₁ – 4I₂ = 6 …(2)

By solving equation (1) and (2), we get

I1 = 1.83 A and I2 = -0.13 A

The negative sign implies that the current in the 1-ohm resistor flows from F to E.

Ques 10. Find the equivalent resistance using Kirchoff's laws. (3 marks)

Ans. From the figure, we can write,

6I+3(I+I₂)=E…(1)

3(3−I)+6(3−(I+I₂))=E

9I+3I₂=E

27−9I−6I₂=E

Applying KVL in loop 1,

6I−3(3−I)=0

2I=3−I

3I=3⇒I=1…(2)

Using 2, 

9I+3I₂ =27−9I−6I₂

9+3I₂=18−6I₂

9I₂=9

I2=1A

From 1, E=12

R=E/3

R=4Ω

Thus, equivalent resistance flowing through the circuit is 4Ω. 

Ques 11. Prove the conservation of energy by Kirchoff’s laws. (3 marks)

Ans. According to Kirchoff’s law, in an electrical circuit, the total power entering the circuit is equal to the total power exiting the circuit.

Consider a circuit consisting of an ideal voltage source with emf ε and a resistance R. 

Power in the emf source is Ps =−VI

Power in the resistance is Pr

By ohm's law, V=IR 

⟹Pr =VI

∴Ps +Pr = 0

Ques 12. How to perform Nodal and Mesh Analysis? (3 marks)

Ans. The following are the steps to perform Nodal Analysis based on Kirchoff’s Current law.

• Assign an arbitrary value of the voltage at each node in the circuit.
• Apply KCL to each node.
• Calculate branch currents.
• Perform mathematical calculations and obtain voltage at the node.

The following are the steps to perform Mesh Analysis based on Kirchoff’s law.

• Every loop is assigned a loop-based current value.
• Kirchhoff’s KVL is implemented for every loop.
• Perform mathematical solvation to obtain loop currents value.

Ques 13. What are the advantages of Kirchhoff’s law in class 12?

Ans. Following are some of the major advantages of Kirchhoff’s law in class 12: 

• In calculating unknown voltages and current in a complete electrical circuit. 
• The flow of current and voltage in a circuit can be easily understood with the help of Kirchhoff’s law. 
• The complex electric circuit can be studied and simplified by this law. 

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