Encyclopedia of Electrical Engineering

The law called Kirchhoff's voltage law (KVL) specifies that

*Which way should I go around the closed path?*

*Should I always follow the direction of the current?*

*How do I apply a sign to the various voltages as I
proceed in a clockwise direction?*
To simplify matters,
By selecting a direction, you eliminate the need to think about
which way would be more appropriate. Any direction will work as long
as you get back to the starting point.
**Fig.no.1: **Applying Kirchhoff's voltage law to a series
dc circuit.
In Fig.no.1 as we proceed from point d to point a across the voltage source,
we move from a negative potential (the negative sign) to a positive potential
(the positive sign), so a positive sign is given to the source voltage E. As we
proceed from point a to point b, we encounter a negative sign, so a drop in potential has occurred, and a negative sign is
applied. Continuing from b to c, we encounter another drop in potential, so
another negative sign is applied. We then arrive back at the starting point d,
and the resulting sum is set equal to zero as defined by Eq. (1).
Writing out the sequence with the voltages and the signs results in the
following:
$$ +E - V1 - V2 = 0$$
which can be rewritten as
$$E = V1 + V2$$
The result is particularly interesting because it tells us that
Kirchhoff's voltage law can also be written in the following form:
$$ \bbox[10px,border:1px solid grey]{\sum_{↻} V_{rises} = \sum_{↻} V_{drops}} \tag{2}$$
This equation revealing that
**Example 1:**Use Kirchhoff's voltage law to determine the unknown
voltage for the circuit in Fig. no.2.

**Fig.no.2: **Dc Series Circuit to be examined for Ex.1.
**Solution:** When applying Kirchhoff's voltage law, be sure to concentrate
on the polarities of the voltage rise or drop rather than on the type
of element. In other words, do not treat a voltage drop across a resistive
element differently from a voltage rise (or drop) across a source. If the
polarity dictates that a drop has occurred, that is the important fact, not
whether it is a resistive element or source.

Application of Kirchhoff's voltage law to the circuit in Fig. no.1 in the clockwise direction results in $$ +E_1 - V_1 + V_2 - E_2 = 0$$ $$ V_1 = E_1 - V_2 - E_2 $$ $$ V_1 = 16V - 4.2V - 9 $$ $$ V_1 = 2.8V $$ The result clearly indicates that you do not need to know the values of the resistors or the current to determine the unknown voltage.

In symbolic form it can be written as
$$ \bbox[10px,border:1px solid grey]{\sum_{↻} V = 0} \tag{1}$$
where $ \sum$ represents summation, ↻ the closed loop, and V the potential
drops and rises. The term algebraic simply means paying attention to
the signs that result in the equations as we add and subtract terms.
Some questions that often arises are, Application of Kirchhoff's voltage law to the circuit in Fig. no.1 in the clockwise direction results in $$ +E_1 - V_1 + V_2 - E_2 = 0$$ $$ V_1 = E_1 - V_2 - E_2 $$ $$ V_1 = 16V - 4.2V - 9 $$ $$ V_1 = 2.8V $$ The result clearly indicates that you do not need to know the values of the resistors or the current to determine the unknown voltage.

Series Voltage Sources
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Voltage Division in a Series Circuit