KCL Solved Examples and solution | Electric current 12th Physics Concepts

Before solving the KCL Solved examples, we must know what is KCL Law? There are two existing fundamental law which are used to calculate the current circuit property. The two laws are KCL and KVL. KCL stands for Kirchoff’ Current Law while the KVL stands for Kirchoff’ Voltage Law. We have discussed about the Electric Current and its explanation and effect in our previous post on What is Electric Current?

Now here are some solved problems on KCL and examples on properties of current source and we will also discuss about current division method for calculating current in the circuit.

KCL Solved Examples and solution

Ques:1 Using KCL law find the unknown current in the Figure shown below:

kirchoff's law, KCL Solved Examples

Solution:  Given that                              i_{{1}} = 3A, i_{{2}} = 2A, i_{{3}} = 1A, i_{{5}} = 2A,i_{{6}} = 5A, i_{{8}} = 6A,

to find                                      i_{{4}} = ?, i_{{7}} = ?, i_{{9}} = ?

Apply KCL At Node 3          i_{{8}} = i_{{3}} + i_{{4}} + i_{{5}} ---- (1)

for i4                                        i_{{4}} = i_{{8}} - (i_{{3}} + i_{{5}}) \Rightarrow 6-(2+1) = 3A

for i7                                        i_{{7}} = i_{{2}} + i_{{6}} \Rightarrow 2+5 = 7A

for i9                                       i_{{9}} = i_{{7}} + i_{{8}} \Rightarrow 6+7 = 13A


Ques:2 Find the current ‘i‘ through the resister 1K ohm.

kirchoff's law, KCL Solved Examples

Solution: Let us consider the voltage ‘V1’at node A.

Apply KCL at node A                                     \frac{V_{{1}}-20V}{1K}+\frac{V_{{1}}}{1K}=1mA  ————(1)

Calculate the voltage value at Node         V_{{1}} = 10 V


therefore                                                        i=10mA Ans.


Ques:3 Find the current I through Resistor R When 5 Ampere current source is in:

  • same direction to the 10 Ampere current source
  • opposite direction to the 10 Ampere current source.

kirchoff's law, KCL Solved Examples


In case a: using KCL            I=10+5=15A

In case b: using KCL           I=10-5=5A

Current Division Rule:

Now we are going to discuss current division method for calculating branch current in parallel circuit (divider circuit). In series circuit it is not applicable (voltage division method is applicable).                        When current flows through  more than one impedance path i.e in divider circuit,then each branch shares some portion of total current  depending on the impedance on that branch, the formula is given below:

Current to be calculating= \frac{total current}{sum of impedance}\times  Impedance through which current is not calculate

You can make it clear by solving a question on this method.


Ques:4 Find the current through the resistor 4\Omega  in the following circuit.

kirchoff's law, KCL Solved Examples

Solution: First calculate the total impedance of the circuit then calculate the current delivered by the voltage source

Let R1=1\Omega ,R2=2\Omega ,R3=4\Omega ,R4=6\Omega

R5=\frac{R3*R4}{R3+R4} ;           R6=\frac{R5*R2}{R5+R2};           R(equivalent)=R6+R1 ;         R(equivalent)=\frac{23}{11}\Omega

Now we will calculate Current delivered by the Voltage source using Ohm’s law:


For calculate the current through 4\Omega  Resistor,we have to first calculate the branch current.Let Current through R2 be I1 and through R5 be I2 branches can be calculate by using Current division Rule :


Now calculate the current through 4\Omega  resistor I(required)=\frac{I2*R4}{R3+R4}=\frac{1.304*6}{4+6}=0.7824A

Hope you understand the concept through these KCL solved examples. If you find any problem or any queries or if you also want to discuss some other problems you have  please comment below ,we will surely entertain you. In case, any specific topics needs to be discussed on our website, we welcome & give our audience the first priority. Hit the like button and share the post with your friends who need help with this topic.

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