Question

Three resistors are connected in parallel such that their total resistance is $3\Omega$. If the resistance of two resistors are $10\Omega$ and $30\Omega$ respectively, find the resistance of the third resistor.

Answer 1

Step 1 - Given data and formula

1. The resistance of the first resistor $R_1=10\Omega$
2. The resistance of the second resistor $R_2=30\Omega$
3. The total resistance of the circuit $R_p=3\Omega$

Consider the resistance of the third resistor is $R_3$.

The formula for the total resistance $R_p$ for the parallel combination is

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Step 2 - Finding the formula for the resistance of the third resistor

Rearrange the above formula.

$$\begin{gathered} \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \\ \frac{1}{R_3}=\frac{1}{R_p}-\frac{1}{R_1}-\frac{1}{R_2} \end{gathered}$$

Step 3 - Finding the resistance of the third resistor

Substitute the value into the above formula to calculate $R_3$.

$$\begin{gathered} \frac{1}{R_3}=\frac{1}{3\Omega }-\frac{1}{10\Omega }-\frac{1}{30\Omega } \\ \frac{1}{R_3}=\frac{10-3-1}{30} \\ \frac{1}{R_3}=\frac{6}{30} \\ {R}_3=\frac{30}{6} \\ {R}_3=5\Omega \end{gathered}$$

Hence, the resistance of the third resistor is $5\Omega$.