In the circuit shown in figure, `E_1 and E_2` are two ideal sources of unknown emfs. Some currents are shown. Potential difference appearing across `6Omega` resistance is `V_(A) - V_(B) = 10V. `

Options

A. The current in the `4.00 Omega` resistance between C & B is 5A.
B. The unknown emf `E_(1)` is 36 V.
C. The unknown emf `E_(2)` is 54 V.
D. the resistance R is equal to `9Omega`.

Explanation

after redrawing the circuit
(a) `I_(4)=5A`
(b) From loop (1) to (1)
`-8(3)+E_(1)-4(3)=0impliesE_(1)=36`volt from loop `(2)` to `(2)`
`+4(5)+5(2)-E_(2)+8(3)=0`
`E_(2)=54` volt
(c) from loop (3) to (3)
`-2R-E_(1)+E_(2)=0`
`R=(E_(2)-E_(1)),/(2)=(54)/(2)-36=9Omega`