Question

Three resistors of 5, 10 and 15 Ωare connected in a parallel and the combination is

connected to a battery of 30V.


Find:
(i) the current flowing through it
(ii) The effective resistance​

Answer 1

Step-by-step explanation:

Given= three resistors R1,R2and R3 are connected in parallel

potential difference,V = 30

(ii) we know that,

1/R1+1R2+1/R3 = 1/5+1/10+1/15

= 1/R= 6+3+2/30 = 11/30

R = 30/11

(i)using ohm's law,

V = IR

30 = I x 30/11

I = 30 x 11/30 = 330/30 = 11 Ampere