In the circuit shown in fig
a. find the current in resistor R
b. find the reistor R
c. find the the unkown emf `epsilon`
d. if the circuit is broken at point x,
what is the current in resistor R ?

Explanation

a. `I_(R ) = 6A - 4A = 2A`
b. using a Kirchhoff's loop around the outside of the circuit,
we get
`28V - (6A)(3Omega)-(2A) R = 0`
or `R = 5 Omega`
c. Using a conuterclockwise loop in the bottom half of the circuit, we get
`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`