Kirchoff Law – KCL and KVL | Current and Electricty-1

Kirchoff law, Kirchoff Current Law (KCL), and Kirchoff Voltage Law (KVL) are discussed in this section of Current Electricity -1 for JEE , CBSE, Olympiad Physics, CSIR, BSc Physics, INPhO and other examinations aspirants and the Physics enthusiast.

Here we will be discussing the definitions, concepts, applications, limitations and a practical problem is solved for clarity. Let us consider a simple electric circuit and discuss various concepts Kirchoff Law applications. We try to understand it from basics step by step from the given problem statement and circuit.

Kirchoff Law – Problem Statement

Study and analyze the below electrical circuit to find the currents in the circuit flowing through the resistances. The circuits is given as below. Explain the steps in solving the circuit starting from definitions to application of kirchoff law.

Solution

Understanding the circuit

Let us understand the circuit before proceeding for analyzing and applying the Kirchoff Law.

In the above circuit V = Voltage ( Source of power ) = Battery ( Example)

R_1,R₂,R₃, and R₄ are the resistances in the circuit.

A and B are the nodes. The definition of node is the point where three or more wires join.

The branch is the path between two nodes with one current. Hence A-R₁-R₂ -B = 1st branch,

A-R₃-B = 2nd branch, A-R₄-B = 3rd branch.

Hence there are 3 branches in the above circuit.

The resistances R₁ and R₂ are connected in series as connected one after the other along the same wire.

The resistances R₃ and R₄ are connected in parallel.

Finding the unknowns

The unknowns are ,

1. Current in the battery and through the resistances R₁ and R₂ connected in series will be same . Let it be I.
2. let the current through R₃ be I₃
3. Let the current through R₄ be I₄

Hence number of unknowns = 3.

Number of branches = 3.

Application of Kirchoff Current Law (KCL)

As per Kirchoff Current Law (KCL), the amount of Inward current at a node = Outward Current at that node.

Let us consider the node A.

The Current moving towards node A = I

Current moving outwards of A = I₃+I₄

Hence at node A,

I= I₃+I₄

Application of Kirchoff Voltage Law ( KVL )

There are two Loops, One loop is B-V-R₁-R₂-A-R₃-B and

second loop is A-R₃-B-R₄-A

As per Kirchoff law, The KVL stipulates that the net Voltage in a loop is zero.

Hence in first loop, V-IR₁-IR₂-I₃R₃=0.

From above, V= I(R₁+R₂)+I₃R₃

By considering the second loop we get,

-I₃R₃+I₄R₄=0.

Hence I₃R₃=I₄R₄

Application of Kirchoff Law ( Combining both KCL and KVL )

By combining the eqations obtained from both KCL and KVL as above,

I₄=I₃R₃/R₄

I= I₃+I₄=I₃+I₃R₃/R₄=I₃( 1+R₃/R₄)

Hence by substituting above values in V= I(R₁+R₂)+I₃R₃, we get,

V=I₃( 1+R₃/R₄)(R₁+R₂)+I₃R₃ = I₃( R₄+R₃)(R₁+R₂)/R₄+I₃R₃

From above,

VR₄= I₃[(R₁+R₂)(R₃+R₄) + R₃R₄]

Final Answer

From the above , we get the final answers for unknown currents as below,

I₃=VR₄/(R₁R₃+R₁R₄+R₂R₃+R₂R₄+R₃R₄)

I₄= VR₃/(R₁R₃+R₁R₄+R₂R₃+R₂R₄+R₃R₄)

I=I₃+I₄ = V(R₃+R₄)/(R₁R₃+R₁R₄+R₂R₃+R₂R₄+R₃R₄)

Conclusion

As above using the Kirchoff Law , KCL and KVL , the values of unknowns in the electrical circuit can be calculated from the given known values. In the above practice problem, the Voltage and Resistances are known. We found out the unknown currents in the circuit using the kirchoff Law.

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