Answer to Question #170576 in Electric Circuits for Innocent

Question

If the resistor are in parallel,prove also that the equivalent resistance is R where

1/R=1/R1 +1/R2 +1/R3+...1/RN

Answer 1

Explanations & Calculations

• For the array of resistors, the total current by it can be written as follows as the total current is the sum of currents through each branch: resistor.

$\qquad\qquad \begin{aligned} \small I&=\small i_1+i_2+i_3+i_4+.....+i_n \end{aligned}$

• As the potential drop across each resistor is equal in a parallel arrangement, individual currents from the above equation can be replaced by $i=\frac{V}{R}$,

$\qquad\qquad \begin{aligned} \small I&=\small \frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}+\frac{V}{R_4}+.....+\frac{V}{R_n} \end{aligned}$

• An equivalent resistance $\small R$ to those array of resistors is such, that it conducts the same total current $\small I$ having the same potential drop $\small V$ across it.
• Then,

$\qquad\qquad \begin{aligned} \small I&=\small\frac{V}{R} \end{aligned}$

• Then replacing the total current $\small I$ from both the equations we get

$\qquad\qquad \begin{aligned} \small \frac{V}{R}&=\small \frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}+\frac{V}{R_4}+....+\frac{V}{R_n}\\ \small \frac{1}{R}&=\small \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+.......+\frac{1}{R_n}\cdots(\,\cancel{\small V}\,) \end{aligned}$