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
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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
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