ABCDFPA is a network of three batteries of the e.m.f's E, 12 V and 4 V respectively and three resistance `2Omega,4Omega` and `6Omega` connected as shown in the diagram. An ideal ammeter connected between F and P shows a current reading of 0.5 A. Then the value of the e.m.f. E is

Options

A. `6V`
B. `6.6V`
C. `8V`
D. `5.5V`

Answer

B

Explanation

Refer to circuit diagram of te problem. Let the currents in the resistors of `2Omega,4Omega` and `6Omega` be `i_(1),i_(2)` and `i_(3)` in the directions shown in the figure. By junction rule `i_(1)=i_(1)+i_(3)`
Taking kirchhoff's loop law around loop `CDFPC`,
`-E-2i_(2)-4i_(2)+12=0implies2i_(1)+4i_(2)=12-E`
Taking the loop law around the loop CBAPC,
`-4-6i_(3)-4i_(2)+12=0implies4i_(2)+6i_(3)=8`
Solving (1), (2) and (3) and putting `i_(1)=0.5A` we obtain `i_(2)=1.1A,i_(3)=0.6A,E=6.6V`