Question

prove that for 2 resistances in parallel , the effective resistance is given by Rp = R1R2/ R1 + R2​

Answer 1

Solution:-

For two restiance R₁ and R₂ in parallel

$: \implies \rm \: \dfrac{1}{R_e} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

$\rm : \implies \dfrac{1}{R_e} = \dfrac{R_1 + R_2}{R_1R_2}$

$\rm : \implies \: R_e = \frac{R_1R_2}{R_1 + R_2}$

Hence proved

More information

=> If 'n' resistor of equal resistance 'R' are connected in parallel

$\rm \: \implies \dfrac{1}{R_e} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + .........n \: times$

$\rm\dfrac{1}{R_e} = \dfrac{n}{R}$

$\boxed{ \rm \: R_e = \frac{R}{n} }$

Some noteworthy point about parallel combination

i) Equivalent resistance is always smaller than any of the individual resistance

ii) Potential V in each resistance is same current where i is divided