In the circuit shown in figure find:

a. the current in the `3.00 Omega` resistor,b. the unknown emfs`E_1` and `E_2` and c the resistance `R`.

Explanation

a. The sum of the currents that enter the junction below the
resistor equals
`3A + 5A = 8A`
b. Using the lower left loop, we get
`epsilon _(1) - (4Omega)(3A)- (3Omega)(8A) = 0 or epsion_(1) = 36V`
Using the lower right loop , we get
`epsilon - (6Omega)(5A) - (3Omega)(8A) = 0` or `epsilon_2 =54V`
c. Using the top loop, we get
`54 V-R(2A) - 36 V = 0` or `R(18V)/(2A) = 9Omega`