Starting at any node in a circuit, we form a *loop *by traversing through elements (open-circuits included!) and returning to the starting node, never encountering any other node more than once.

** Figure 1 :** For example, the paths *fabef *and *fdcef *are loops:

** Figure 2 :** whereas the paths *becba *and *fde *are not

Kirchhoff’s Voltage Law (KVL) is essentially the law of conservation of energy. If voltages across elements traversed from – to + are positive in sense, and voltages across elements that are traversed from + to – are negative in sense (or vice versa), then KVL can be stated as follows:

KVL defined as

KVL: Around any loop in a circuit, the voltages algebraically sum to zero.

If there are *n *elements in the loop then, in symbols, KVL is:

KVL can also be stated as: In traversing a loop, the sum of the voltage rises equals the sum of the voltage drops.

**Example 1**

In the circuit shown in Figure 1, we select a traversal from – to + to be positive in sense. Then KVL around the loop *abcefa *gives:

– *v*1 – *v*2 + *v*3 – *v*8 + *v*6 = 0

and around loop *bcdeb*, we have:

– *v*2 – 3 – *v*4 + *v*7 = 0

In this last loop, one of the elements traversed (the element between nodes *b *and *e*) is an open-circuit; however, KVL holds regardless of the nature of the elements in the circuit.

**Example 2**

We want to find the current *i*, in the one-loop circuit shown below:

The polarities of *v*1 , *v*2 , *v*3 and the direction of *i *were chosen arbitrarily (the polarities of the 10 V and 34 V sources are given). Applying KVL we get:

10 – *v*1 – 34 – *v*2 + *v*3 = 0

Thus:

*v*1 + *v*2 – *v*3 = -24

From Ohm’s Law:

*v*1 = 2*i*

*v*2 = 4*i*

*v*3 = -6*i*

Substituting these into the previous equation yields:

(2*i*) + (4*i*) – (- 6*i*) = -24

2*i *+ 4*i *+ 6*i *= -24

12*i *= -24

*i *= -2 A

Having solved for *i*, we now find that:

*v*1 = 2*i *= 2(- 2) = -4 V

*v*2 = 4*i *= 4(- 2) = -8 V

*v*3 = -6*i *= (- 6)(- 2) = 12 V

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- Kirchhoffs Voltage Law Examples