### Kirchhoffs Voltage Law

• Explain how to apply Kirchhoff Voltage Law to electric circuits.

• Be able to write KVL for every loop in the circuit.

• Be able to solve the KVL equations, especially for simple circuits.

The voltages around a closed path in a circuit must sum to zero.

Kirchhoff’s Voltage Law (or Kirchhoff’s Loop Rule) is a result of the electrostatic field being conservative. It states that the total voltage around a closed loop must be zero. If this were not the case, then when we travel around a closed loop, the voltages would be indefinite. So $\large \sum V = 0$

## Sign Conventions:

• If a resistance is travelled in the same direction as of current then the voltage drop across that resistance is considered negative.

• If a voltage source is travelled from negative to positive, that voltage value is considered positive.

### Solved Example:

#### 19-2-01

According to Kirchhoff's voltage law, the algebraic sum of all IR drops and e.m.fs. in any closed loop of a network is always:

### Solved Example:

#### 19-2-02

A closed path made by several branches of the network is known as:

### Solved Example:

#### 19-2-03

Kirchhoff's voltage law is related to:

### Solved Example:

#### 19-2-04

Find the current supplied by 7 V source.

Solution:
We will consider three loops with loop currents as I$_1$, I$_2$ and I$_3$ as shown in the following figure.

Applying Kirchhoff's Voltage rule to loop with i$_1$,

$7 - 3(i_1 - i_2) - 1(i_1 - i_3) -5 = 0$ $-4i_1 + 3 i_2 + i_3 = -2$

Applying Kirchhoff's Voltage rule to loop with i$_2$,

$-5 i_2 - 3 (i_2 - i_3) - 3(i_2 - i_1) = 0$ $3i_1 - 11 i_2 + 3i_3 = 0$

Applying Kirchhoff's Voltage rule to loop with i$_3$,

$5 - 1(i_3 - i_1) - 3(i_3 - i_2) - 6 i_3 = 0$ $i_ 1 + 3 i_2 - 10i_3 = -5$

Solving these three equations using determinants,

\begin{align*} D &= \begin{vmatrix} -4 & 3 &1\\ 3&-11 &3 \\ 1& 3 &-10 \end{vmatrix}\\ &= -285 \end{align*} \begin{align*} D_{i_1} &= \begin{vmatrix} -2 & 3 &1\\ 0&-11 &3 \\ -5& 3 &-10 \end{vmatrix}\\ &= -302 \end{align*} $i_1 = \dfrac{D_{i_1}}{D} = \dfrac{-302}{-285} = 1.06\ A$