Kirchhoff’s Voltage Law (KVL) with Example: Circuit Analysis

KVL, the Kirchhoff’s voltage law, state that:

The voltage produced and voltage drop in a closed loop (a path) of a circuit is always equal.

$\sum{V_{Rise}= }\sum{V_{Drop}}$

Gustav Kirchhoff was a german physicist, who presented two laws; Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL). Ohm law is a very basic one, which may not be sufficient to analyze a complex circuit. The Kirchhoff’s Voltage Law (KVL) provide the basis for Mesh Current Analysis.

KVL Inforgraphics

Example:

Suppose a circuit with two parallel paths (loops) and a single voltage source (DC), as shown in the diagram below. Find the current and voltage of each element of the circuit for the following given circuit parameter using Kirchhoff’s voltage law.

R1 = 5Ω
R2 = 10Ω
R3 = 5Ω

R4 = 10Ω
V = 20Volts

Kirchhoff's Voltage Law KVL Example

Solution:

Assign Loops Name:

There are two loops (closed paths) in the circuit, loop 1 with two resistors and a single voltage source, where in loop 2 there is no voltage source, three resistors only.

Assign sign to circuit elements:

Now, assign a positive sign to the resistor terminal which is closer to voltage source positive terminal and negative sign to whom which are closer to negative terminals. OR you may assign (+ -) sign to each component terminal by going through the loop. But make sure that the sign of voltage rise (- +) is exactly opposite to sign of voltage drop (+ -). Both will give you same results, even same equations.

Assign  current names:

Assign a name to the current of each loop and as discussed in KCL, write equations for current at each node. For the above circuit KCL equations will be:

at node a: $I_{1}= I_{1}$

at node b: $I_{1}= I_{2 }+ I_{3}$

at node c: $ I_{2 }+ I_{3}= I_{1}$

Write Loop Equations:

The next step is to write equations for each loop. Based on the sign and current name assigned, as shown below.

For loop1: $V =I_{1}R_{1}+ I_{2}R_{4}$

For loop2: $0 =I_{3}R_{2}+ I_{3}R_{3}- I_{2}R_{4}$

Notice the negative sign in the second equation, it is because of being opposite in direction of loop arrow i.e. for R2 and R3 it is (+ -) but for R4 it is (- +).

Now, put the value of I1, which will give you two simultaneous equations with unknown variable I2 and I3. The value of these currents can easily find out using cramer’s rule for the matrix.

$V =(R_{1}+R_{4})I_{2}+ R_{1}I_{3}$

$0 = – R_{4}I_{2}+(R_{3}+ R_{2)}I_{3}$

Let’s put the values of resistors and the voltage source, and see what happen

$20=(5+10)I_{2} +5 I_{3} —-(1)$

$0= -10I_{2}+(5+10)I_{3} —–(2)$

Now multiply the equation (1) by -3 and add with the equation (2)

$0= -10I_{2}+(15)I_{3} —-(2)$

$-60= -45I_{2}-15I_{3} —-(1)$

Slove The Equations:

Adding the above two equation will give us the equation (3)

$-60= -55 I_{2}+0 -(3)$ OR $I_{2}= -\frac{60}{-55}=1.09 Amp$

Now putting the current I$_{2}$ in the equation (2) to get current I$_{3}$

$0= -10(1.09)+15I_{3}$

OR

$I_{3}=10\frac{(1.09)}{15}=0.72 Amp$

By applying KCL to the node b, the following equation can be obtained

$I_{1}=I_{2}+I_{3}$

Now putting the values of the current will give us the value of I$_{1}$

$I_{1}=1.09+0.72=1.81 Amp$

Now, it is easy to find any parameter of the circuit, just use the ohm’s  law, as shown below

$V_{3}=I_{3}R_{3}=0.72(5)=3.6 Volts$

And

$V_{2}=I_{3}R_{2}=0.72 (10)=7.2 Volts$
$V_{4}=I_{2}R_{4}=1.09 (10)=10.9 Volts$
$V_{1}=I_{1}R_{1}=1.81 (5)=9.05 Volts$

Kirchhoff's Voltage Law Limitation:

The KVL has some practical limitation besides it is a very useful tool for circuit analysis.

  1. In the high-frequency circuit, the fluctuating magnetic field can link the circuit, which contradicts the assumption of KVL.
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