Kirchhoff’s voltage law KVL or Kirchhoff’s second law deals with conservation of energy in a closed loop electric circuit. The KVL is fundamental law for in electrical circuit analysis. According to the Kirchhoff Voltage Law KVL **The algebraic sum of voltage produced and the voltage dropped in a closed loop (a closed path) of an electric circuit are always equal. **Mathematically,

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

In other words, the energy is conserved across a closed loop of electric circuit.

Note that the algebraic sum means taking into account the polarities and signs of voltage sources and voltage drops. When applying the KVL to any circuit, it is important that all the signs (+ -) or (- +) must be taken into account correctly otherwise the calculation will be wrong.

Gustav Kirchhoff 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.

## Applying Kirchhoff's Voltage Law

### Step 1

Identify closed loops in electric circuit and assign the loop name to each closed loop. Suppose a loop current current in certain direction.

### step 2

Now, caption each circuit element with polarity sign as following the supposed current direction.

### step 3

Apply Kirchhoff’s Current Law KCL at each node and write current equations for each node.

### Step 4

Now apply KVL at each closed loop and write equation for each loop. All the voltage produce should be equal to all the voltage drop.

### step 5

Now substitute the KCL equations in KVL equations to get final equations. Solve those equations using any of the solver.

## Kirchhoff's Voltage Law 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.

R_{1} = 5Ω

R_{2} = 10Ω

R_{3} = 5Ω

R_{4} = 10Ω

V = 20Volts

## 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 R_{2} and R_{3} it is (+ -) but for R_{4} it is (- +).

Now, put the value of I_{1}, which will give you two simultaneous equations with unknown variable I_{2} and I_{3}. 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.

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