Question

Derive an expression for three resistances connected in parallel.

Answer 1

Step 1: Finding the derivations for voltage and current:

In a parallel circuit, the potential difference is uniform throughout the resistors.

Let $I_1,I_2$ and $I_3$ be the current in the resistors.

Let $R_1,R_2$and $R_3$ be the three resistors.

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

Use Ohm’s law:

$V_1=1_1{R}_1,V_2=I_2{R}_2,V_3=I_3{R}_3$

here, $V=V_1=V_2=V_3$

$\therefore V=I_1{R}_1,V=I_2{R}_2,V=I_3{R}_3$

$\Rightarrow I_1=\frac{V}{R_1},I_2=\frac{V}{R_2}$and $I_3=\frac{V}{R_3}$

$\therefore I=I_1+I_2+I_3$

Step 2: Deriving the expression for the three resistance:

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

$V\times \frac{1}{R_{e}q}=X\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right]$

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

Where $R_{e}q$ is equivalent resistance. From this, we know that the resistance in each circuit is more than the equivalent resistance.

Thus, the expression for the three resistance connected in a parallel circuit is $\frac{1}{R_{e}q}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$.