Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law is used in the study of series circuits, where it is observed that the voltage applied to a closed circuit is equal to the sum of the voltage drops within that circuit. Another way of expressing this is, the algebraic sum of the voltage rises (applied voltage) and the voltage drops must equal zero.

A voltage rise within a circuit is a voltage source that provides a potential difference or electromotive force (emf). A voltage drop is the voltage across a component within a circuit, like the voltage across a resistor, which uses and diminishes the applied voltage.

One convention often used in circuit diagrams to distinguish clearly between the two is to have voltage sources (voltage rises) labelled with an alphabetic subscript, while voltage drops are labelled with a numerical subscript. This is the case in the equations below.

The Equations of Kirchhoff’s Voltage Law

Using the first definition of Kirchhoff’s Loop Rule — another name for KVL that you will come across — that ‘the voltage applied is equal to the sum of the voltage drops’, we can express Kirchhoff’s Voltage Law mathematically as:

\[ \mbox{Voltage applied = sum of voltage drops} \]

\[ V_A = V_1 + V_2 + V_3 + \cdot \cdot \cdot V_N \]

Where $V_A$ is the voltage applied, and $V_1, V_2, V_3$ all the way to $V_N$ are the voltage drops within the circuit. ($V_N$ just signifies that we can use this equation theoretically for any number of voltage drops in a circuit.)

Voltage drops are the drops in voltage that occur when electrical power is used by components of a circuit.

If we manipulate the above expression with algebra, we can understand the equation in a different way:

\[ V_A – V_1 – V_2 – V_3 = 0 \]

or

\[ V_A – (V_1 + V_2 + V_3) = 0 \]

We say this as the applied voltage minus the individual voltage drops is equal to 0. Or we can say that the algebraic sum of the voltage rises and drops within a circuit equals zero. (Speaking of the ‘algebraic sum’ simply means that we determine some of the values as positive (+) and some of the values as negative (-), and make sure to add/subtract correctly.)

We can simplify the above expression even further using the Greek letter capital sigma, $\Sigma$, which in mathematics means the “sum of”. This gives us

\[ \Sigma \Delta V = V_A – V_1 – V_2 – V_3 = 0 \]

or

\[ \Sigma \Delta V = V_A – \big( V_1 + V_2 + V_3 \big) = 0 \]

We can simplify this equation as

\[ \Sigma \Delta V = 0 \]

The state the above literally as the sum of the changes in voltage equals zero.

Tracing Voltage Rises and Drops

Tracing the voltage rises and drops can be tricky, only because the signs will change depending on which direction through the voltage source we choose to trace the rises and drops. It is a simple matter of perspective. So long as we get the signs right, we will get the same answer regardless of which direction we choose.

To trace voltage drops around a circuit, start at the negative terminal. The pathway from the negative terminal through the voltage source to the positive terminal is a voltage rise. Therefore when we go through $V_A$ from – to + then $V_A = +120V$, which makes the voltage drops across any resistance along the way positive. Using the four resistor series circuit in the diagram below, then we are going along the pathway badcb giving us a voltage rise of $V_A = +120V$, and giving us values $V_1 = 50V$, $V_2 = 10V$, $V_3 = 30V$, and $V_4 = 30V$.

If we perform the calculations for our circuit using any of the equations above, we find that

\begin{equation*} \begin{align} \Sigma \Delta V & = V_A – V_1 – V_2 – V_3 – V_4 \\ & = 120 V – 50V – 10V – 30V – 30V \\ & = 0 \end{align} \end{equation}

If we instead choose to trace the voltage drops in the opposite direction, then the signs of everything are reversed. If we start at the positive terminal and pass through $V_A$ from + to – then everything becomes negative, and we trace our voltage drops along the opposite path as before through abcda. $V_A = -120 V$ and the voltage drop across any resistance also becomes negative (–). This gives us $V_4 = -30V$, $V_3 = -30V$, $V_2 = -10V$ and $V_1 = -50V$.

Perform the calculations using these values and we get the same result:

\begin{equation*} \begin{align} \Sigma \Delta V & = V_A – \big( V_1 + V_2 + V_3 + V_4 \big) \\ & = -120V – \big( (-50V) + (-10V) + (-30V) + (-30V) \big) \\ & = -120V – (-120V) = -120V + 120V \\ & = 0 \end{align} \end{equation}

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