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Kirchhoffs first rule – problems and solutions

Kirchhoffs first rule – problems and solutions

1. Based on figure as shown below, if I1 = 5A, I2 = 3 A, I3 = 6 A, then I4 = ………

Kirchhoff’s first rule – problems and solutions 1Solution :

Kirchhoff’s first rule states that the sums of all currents entering the junction must equal the sum of all currents leaving the junction.

I1 and I2 are entering the junction, whereas I3 and I4 are leaving.

Apply Kirchhoff’s first rule :

I1 + I2 = I3 + I4

5 + 3 = 6 + I4

8 = 6 + I4

I4 = 8 – 6

I4 = 2 Ampere

2. If I1 = I3 then I4 =……..

Kirchhoff’s first rule – problems and solutions 2Known :

I1 = I3 = 2 Ampere

I2 = 3 Ampere

I5 = 4 Ampere

Wanted: I4

Solution :

I1 and I4 are entering the junction, whereas I2, I3, and I5 are leaving.

Apply Kirchhoff’s first rule :

I1 + I4 = I2 + I3 + I5

2 + I4 = 3 + 2 + 4

2 + I4 = 9

I4 = 9 – 2

I4 = 7 Ampere

3. I1 = …….

Solution :Kirchhoff’s first rule – problems and solutions 3

I am entering the junction, whereas I1, I2, and I3 are leaving.

Apply Kirchhoff’s first rule :

I = I1 + I2 + I3

7 A = I1 + 1 A + 4 A

7 A = I1 + 5 A

7 A – 5 A = I1

I1 = 2 Ampere

4. I2 = ………

Wanted: I2Kirchhoff’s first rule – problems and solutions 4

Solution :

I1 is entering the junction, whereas I2, I3, and I4 are leaving.

Apply Kirchhoff’s first rule :

I1 = I2 + I3 + I4

6 = I2 + 3 + 1

6 = I2 + 4

I2 = 6 – 4

I2 = 2 Ampere

5. If I2 = 1/4 I1 , then I3 = ………

Known :Kirchhoff’s first rule – problems and solutions 5

I1 = 600 milliAmpere

Wanted: I3

Solution :

I2 = 1/4 I1 = 1/4 (600 mA) = 150 mA

I1 is entering the junction, whereas I2 and I3 are leaving.

Apply Kirchhoff’s first rule :

I1 = I2 + I3

600 mA = 150 mA + I3

I3 = 600 mA – 150 mA

I3 = 450 mA

6. Based on figure below, find I!

Solution :

Apply Kirchhoff’s first rule :

3 + 5 + 2 = 7 + IKirchhoff’s first rule – problems and solutions 10

10 = 7 + I

10 – 7 = I

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I = 3 Ampere

7. Based on figure below, if I1 = 4 A, I2 = 3 A, I4 = 7 A, I5 = 4 A, determine I3.

Solution :

Electric current entering the junction = I1 + I3 = 4 A + I3

Electric current leaving the junction = I2 + I4 + I5 = 3 A + 7 A + 4 A = 14 AKirchhoff’s first rule – problems and solutions 11

14 A = 4 A + I3

I3 = 14 A – 4 A

I3 = 10 A

  1. What is the primary principle behind Kirchhoff’s First Rule?
    • Answer: The primary principle behind Kirchhoff’s First Rule is the conservation of electric charge. It implies that no charge is lost at any node or junction in a circuit.
  2. If three currents, , , and , converge on a node and is entering while and are leaving, how is KCL expressed for this node?
    • Answer: According to KCL, .
  3. Why is KCL important in the analysis of electrical circuits?
    • Answer: KCL provides a fundamental constraint that any valid solution to a circuit problem must satisfy, ensuring that the behavior of currents in a network adheres to the conservation of charge principle.
  4. Is KCL valid only for DC circuits or also for AC circuits?
    • Answer: KCL is valid for both DC and AC circuits. It is a reflection of the conservation of charge, which is universal, regardless of how the currents change over time.
  5. In a circuit with several nodes, is it necessary to apply KCL to every node to analyze the circuit fully?
    • Answer: Not always. While you can apply KCL to every node, often analyzing a subset of nodes is sufficient to determine the behavior of the entire circuit.
  6. How does the concept of a “ground” or reference node in a circuit relate to KCL?
    • Answer: A ground or reference node serves as a common point of reference for voltages in the circuit. Applying KCL to this node, like any other node, ensures that the sum of currents is zero. Ground doesn’t affect the applicability or the outcome of KCL.
  7. If a node has only two branches connected to it, and the current in one branch is known, is it necessary to solve for the current in the other branch using KCL?
    • Answer: No, if the current in one branch is known and there are only two branches connected to the node, the current in the other branch is simply the negative of the known current (indicating it’s in the opposite direction).
  8. How does the behavior of capacitors in a circuit relate to KCL?
    • Answer: While capacitors can store charge, the rate of charge entering a capacitor is equal to the rate of charge leaving its other terminal. Therefore, even with capacitors, the net current at a node (including a node at one terminal of the capacitor) is zero, and KCL remains valid.
  9. In practical circuits with real components, does KCL hold perfectly, or are there any exceptions?
    • Answer: In ideal circuit analysis, KCL holds perfectly. In real-world situations, components might have tiny leakage currents or parasitic effects, but for most practical purposes and analyses, KCL is assumed to hold true.
  10. When might one use both Kirchhoff’s First (Current) Rule and Second (Voltage) Rule in analyzing a circuit?
  • Answer: In many complex circuits, especially those with multiple loops and nodes, both KCL (First Rule) and Kirchhoff’s Voltage Law (Second Rule) are used together to set up a system of equations that can then be solved to determine unknown currents and voltages throughout the circuit.
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