KCL And KVL Explained With Solved Numericals In Detail

 Kirchoff’s Current (KCL) and Voltage Laws (KVL)

Ohm’s law alone is not sufficient to analyze circuits unless it is coupled with kirchoff’s two laws:

·         Kirchoff’s Current law (KCL)

·         Kirchoff’s Voltage law (KVL)

KCL

KCL states that the algebraic sum of currents entering a node (or a closed boundary) is zero.

Mathematically

                      

Where ‘N’ is the number of branches connected to the node ‘n’ is the n^th branch; and i_n is the n^th branch current leaving or entering a node
               Convention: current entering a node is positive; while leaving a node is negative

KCL equation: i1 – i5 + i4 + i3 – i2 = 0 i1 + i3 + i4  = i2 – i5

                                                                                         

Alternate KCL: The sum of currents entering a node is equal to the sum of currents leaving the node.

Example: Write KCL on node ‘a’ and find out Ι_T.

Solution:

·         So, an application of KCL is to combine current source in parallel into one equivalent current source.

·         A circuit cannot contain two different currents Ι₁ and Ι₂ in series unless Ι₁=i₂; otherwise KCL will be violated.

KVL:

KVL states that the algebraic sum of all voltage round a closed path (or loop) is zero.

Mathematically,

Where M is the no. of voltages in a loop (or number of branches in a loop),  and v_m is the m^th voltage.

Convention: The sign on each voltage is the polarity of the terminal encountred first as we travel around the loop.

Example:

Alternate KVL: Sum of voltage drops is equal to sum of voltage rises.

Example:  Apply loop in the following circuit and find out Vab:

·                     This is an application of KVL where voltage source in series can be combined into one equivalent source.

·                     Note that a circuit cannot contain two different voltage V1 and V2 in parallel unless V1 = V2; Otherwise KVL would be violated.

Example: Find out V₁ and V₂ using KVL.

Solution:

Example: Find out V₁ and V₂ using KVL.

Solution:

                                       

We observe that answers in both examples are handled well by polarity changes.