a. `I_(R ) = 6A - 4A = 2A` b. Using a Kirchhoff's loop around the exterior of the circuit, we acquire `28V - (6A)(3Omega)-(2A) R = 0` or `R = 5 Omega` c. Utilizing a conuterclockwise loop in the bottom fifty percent of the circuit, we acquire `epsilon - (6A) (3 Omega) - (4A)(6Omega) = 0 or epsilon = 42V` d. If the circuit is broken at point x then the current in the battery is `I= (Sigma epsilon)/(SigmaR) = (28V)/(3Sigma + 5 Sigma) = 3.5A`


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