In the circuit shown, the currents `i_(1)` and `i_(2)` are

Options

A. `i_(1)=1.5A,i_(2)=0.5A`
B. `i_(1)=0.5A, i_(2)=1.5 A`
C. `i_(1)=1A, i_(2)=3A`
D. `i_(1)3A, i_(2)=1A`

Answer

B

Explanation

Resistance `R=(12xx4)/(12+4)+2=(48)/(16)+2=5Omega`
Electric current `i=(E )/(R+r)=(12)/(6)=2A`
`therefore " " i_(1)+i_(2)=2 A` …..(i)
`i prop (1)/(R )rArr l_(1)R_(1)=l_(2)R_(2)`
`l_(1)=(R_(2))/(R_(1))l_(2) rArr l_(1)=(1)/(3)l_(2)` ltbr This value in the eq. (i) and we get
`therefore " " i_(2)=1.5A, i_(1)=0.5 A`