A potential divider is used to divide an input voltage into a smaller output voltage. Consider a circuit where the input voltage, $V$ is applied across two resistors, $R_1$ and $R_2$ in series:
The current flowing from the battery depends on the input voltage and the total resistance: $$ \begin{equation}\begin{aligned} I&=\frac{V}{R_T}\\ \end{aligned}\end{equation} $$
Because the resistors are in series, the same current flows through $R_1$, with a potential difference of $V_1$ across its ends: $$ \begin{equation}\begin{aligned} I&=\frac{V_1}{R_1}\\ \end{aligned}\end{equation} $$
Thus we can equate these two values for current: $$ \begin{equation}\begin{aligned} \frac{V_1}{R_1}&=\frac{V}{R_T}\\ \therefore V_1&=\frac{R_1}{R_T}\cdot V\\ \end{aligned}\end{equation} $$
Recall that for resistors in series, the total resistance is the sum of the individual resistances: $$ \begin{equation}\begin{aligned} R_T&=R_1+R_2\\ \therefore V_1&=\frac{R_1}{R_1+R_2}\cdot V\\ \end{aligned}\end{equation} $$
Thus the ratio, $\frac{R_1}{R_1+R_2}$ determines what portion of $V$, $V_1$ will be.
Example 1
Find the voltage across a $5\Omega$ resistor placed in series with a $10\Omega$ resistor with an input voltage of $9V$ in the circuit.
Solution We deduce the values:
- $V=9V$
- $R_1=5\Omega$
- $R_2=10\Omega$
- $V_1=?$
Using the formula: $$ \begin{equation}\begin{aligned} V_1&=\frac{R_1}{R_1+R_2}\cdot V\\ &=\frac{5\Omega}{5\Omega+10\Omega}\cdot 9V\\ &=\frac{5\Omega}{15\Omega}\cdot 9V\\ &=3V\\ \end{aligned}\end{equation} $$
Example 2
Determine the resistor across which a voltage of $40V$ will be produced if there is an input voltage of $100V$ and the other resistor in the potential divider is $30\Omega$.
Solution We deduce the values:
- $V_1=40V$
- $V=100V$
- $R_2=30\Omega$
- $R_1=?$
Using the formula: $$ \begin{equation}\begin{aligned} 40V&=\frac{x}{x+30\Omega}\cdot 100V\\ x+30\Omega&=x\cdot \frac{100V}{40V}\\ x+30\Omega&=2.5x\\ 30\Omega&=2.5x-x\\ 30\Omega&=1.5x\\ \therefore x&=\frac{30\Omega}{1.5}\\ &=20\Omega\\ \end{aligned}\end{equation} $$