Current through `3Omega` resistor is 0.8A, then potential drop through `4Omega` resistor is

Options

A. 9.6 V
B. 2.6 V
C. 4.8 V
D. 1.2 V

Answer

C

Explanation

Given : `I_(1)=0.8A`
As the resistances `3Omega and 6Omega` are connected in parallel. Therefore potential drop across these resistances is same.
`therefore 3I_(1)=6I_(2)or(0.8)xx3=I_(2)xx6orI_(2)=(2.4)/(6)=0.4A`
The current flowing through `4Omega` resistor is
`I=I_(1)+I_(2)=0.8A+0.4A=1.2A`
Potential drop across `4Omega` resistance `=Ixx4=(1.2)xx4=4.8V`