Question

With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.

Answer 1

Step 1: Diagram and assumption for equivalent resistance:

Using the given diagram,

Let the total current in the circuit be I, the potential difference be V and the current passing through resistance R_1, R₂, and R₃ will be I₁, I₂ and I₃ respectively.

The total current will be,
$I=I_1+I_2+I_3$

Step 2: Applying Ohms Law,

$$\begin{gathered} I_1=\frac{V}{R_1} \\ {I}_2=\frac{V}{R_2} \\ {I}_3=\frac{V}{R_3} \\ I=\frac{V}{R_{eq}} \end{gathered}$$

So, the net current will be,

$\frac{V}{R_{eq}}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}$

Factoring out V, we get

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Hence proved that the summation of the reciprocals of each resistance is equal to the reciprocal of the equivalent resistance of a set of resistances connected in parallel.