Question

10 resistors of 2 ohm connected in parallel.calculate the effective resistance

Answer 1

Effective Resistance = 1/R₁ + 1/R₂ + . . . . + 1/R₁₀

$\small \tt \: \frac{1}{Effective \: \: Resistance }= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \\ \\ \tt \: \frac{1} {Effective \: \: Resistance} = \frac{10}{2} \\ \\ \tt \: Effective \: \: Resistance = \frac{1}{5}$

Answer 2

Explanation:

As we know that in parallel connection

Req1=R11+R21 + R31 + R41+..+ R n1

If R 1=R2=R3=..=R n =2Ω and n=10

Then,Req1 = 21 + 21 + 21 + 21 +.+ 21 Required= 102Ω

=0.2 Ω is the answer

Hope its help..