In the circuit shown in figure , `V` must be

Options

A. `50V`
B. `80V`
C. `100V`
D. `1290V`

Answer

B

Explanation

`100 Omega, 25 Omega` and `20 Omega` are in parallel.
Their net resistance is `10 Omega`
`:. R_("net")=4Omega+10Omega+6Omega=20 Omega`
Hence the main current
`i=E/R_("net")=1A`
Current through either of the resistance
`is i/3 or 1/3 A`
`:.V=iR=(1/3)(9)=3V`