1.

Three resistances of 2Ω each are connected to a 2V battery as shown in the front figure. Then ........

Answer»

Given circuit resistance are parallel combination then 

\(\frac1R=\frac1R_1-\frac1R_2+\frac1R_2\) 

\(\frac1R=\frac12+\frac12+\frac12\) 

\(\frac1R=\frac32\) 

R = 2/3

We use V = iR

i = V/R

i = \(\frac{2\times3}2\) = 3

i = 3 ampeare

Then I1 = 2A

it divided two branches then

I1 = 1 + 1

Then I2 = 2A



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