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

Options

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

Answer

B

Explanation

Potential difference across `3Omega` and `6 Omega` resistance is same.
Thus, `V=IR=0.8xx3=2.4 V`
Thus, current flowing through parallel combination of `3 Omega` and `6 Omega` resistance is ,
`I=(V)/(R_(P))=(2.4)/(2) = 1.2 A`
Thus, potential difference across `4Omega` resistance is ,
`V_(1)=1.2xx4=4.8 V`