Kirchhoff First Law

What is Kirchoff's First Law?

In a circuit, at any junction, the sum of the currents entering the junction must equal the sum of the currents leaving the junction. This law is also known as the Junction rule or current law (KCL).

$\sum i=0$

$i_1+i_3=i_2+i_4$

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Solved Examples Based on Kirchoff's First Law

Example 1: In the given figure, which of the following relation is correct?

1) $i_1+i_2-i_3+i_4-i_5-i_6+i_7=0$
2) $i_1+i_2+i_3-i_4-i_5+i_6-i_7=0$
3) $i_1-i_2-i_3+i_4+i_5-i_6-i_7=0$
4) $i_1+i_2-i_3+i_4+i_5-i_6-i_7=0$

Solution:

In a circuit at any junction, the sum of the currents entering the junction must equal the sum of the currents leaving the junction.

This law is also known as the Junction rule or current law

Total incoming current = Total outgoing current

$\begin{array}{rl}& \Rightarrow i_1+i_4+i_5=i_2+i_3+i_6+i_7\\ & \Rightarrow i_1-i_2-i_3+i_4+i_5-i_6-i_7=0\end{array}$

Hence, the answer is option (3).

Example 2: The Kirchhoff's first law $(\sum i=0)$ and second law $(\sum iR=\sum E)$ where the symbols have their usual meanings, are respectively based on-

1) conservation of charge, conservation of energy

2) conservation of charge, conservation of momentum

3) conservation of energy, conservation of charge

4) conservation of momentum, conservation of charge.

Solution:

Kirchoff's first law

$\begin{array}{rl}& \sum i=0\\ & i_1+i_3=i_2+i_4\end{array}$

wherein

Kirchhoff's first law $[\sum i=0]$ is based on conservation of charge
Kirchhoff's second law $[\sum iR=\sum E]$ is based on the conservation of energy.

Hence, the answer is option (1).

Example 3: In the given circuit diagram, the currents, I₁ = -0.3 A, I4 = 0.8 A and I5 = 0.4 A, are flowing as shown. The currents I2 , I3 and I6 ,respectively, are :

1) -0.4 A, 0.4 A, 1.1 A

2) 1.1 A -0.4 A, 0.4 A

3) 0.4 A,1.1 A,0.4 A

4) 1.1 A,0.4 A,0.4 A

Solution:

Apply Kcl at Q

I₆+I₃ = I₁+ I₂ .........(1)

I₁ = 0.3A

I₄ = 0.8A

I₅ = 0.4A

KCL at 5

I₄= I₅ +I₃

I₃ = I₄ - I₅

= 0.8 +0.4

=0.4A

KCL at R

I₁+I₂ = I₄

I₂ = I₄ - I₁

= 0.8 - (-0.3)

=1.1A

From equation (1)

I₆ + I₃ = I₁+ I₂

I₆ = I₁ + I₂ - I₃

= -0.3+1.1 -0.4

=0.4A

I₂ = 1.1A

I₃ = 0.4A

I₆ = 0.4A

Hence, the answer is option (4).

Example 4: When the switch S, in the circuit shown, is closed, then the value of current I (in amperes) will be:

1) 5

2) 3

3) 4

4) 2

Solution:

By KCL

i₁ + i₂ = i

Let voltage at C be V

$\frac{20-V}{2}+\frac{10-V}{4}=\frac{V}{2}$

12.5 = 1.25 V

V = 10 volt

$\mathrm{i}=\frac{V}{2}=\frac{10}{2}=5\mathrm{A}$

Hence, the answer is option (1).

Example 5: The value of current in the $6\; \Omega$ resistance is :

1) 4A

2) 8A

3) 10A

4) 6A

Solution:

Applying KCL at the junction with voltage x (assigned)

By KCL,

$\begin{array}{rl}& {\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3=0\\ & \left(\frac{x-90}{5}\right)+\left(\frac{x-140}{20}\right)+\left(\frac{x-0}{6}\right)=0\\ & 24x-24\times 90+6x-6\times 140+20x=0\\ & 50x=24\times 90+6\times 140\\ & =6[4\times 90+140]\\ & 50x=3000\\ & x=60\mathrm{V}\end{array}$

Current through the resistor is

$I_3=\left(\frac{x-0}{6}\right)=\left(\frac{60-0}{6}\right)=10\mathrm{A}$

Hence, the answer is option (3).

Summary

Kirchhoff's First Law, which is also called the Current Law, says that the sum of all currents going into any point of a circuit (or junction) is equal to the sum of all currents leaving it. It’s founded on the principle of conservation of electric charge - no charge can either be created or destroyed at these points. By applying the principles of Kirchhoff's First Law it is possible to determine unknown currents and analyze circuit electricity flow in circuits that are complex.