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In the given circuit, E=8 V,R1=1Omega, R...

In the given circuit, `E=8 V,R_1=1Omega, R_2=R_3=R_4=5Omega`.The current through the resistance `R_4` is (in A):

A

`1//3`

B

`2//3`

C

1

D

2

Text Solution

Verified by Experts

`R_1,R_2 and R_3` are in parallel. So, equivalent resistance of the circuit, `R_(eq)=8/3 Omega`
So, current through the battery `j_0=8/((8/3))=3A`
Current through `R_4, i=j_0/3=1A`
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