Solving Circuit Problems: R, E, and Current

[daimlerpogi]

Homework Statement

In the circuit shown in the figure, find (a) the current in resistor R; (b) the resistance R; (c) the unknown emf E. (d) If the circuit is broken at point x, what is the current in the 28v battery?

Homework Equations

KVL and KCL

The Attempt at a Solution

I don't know if these are correct :S
KVL:
3(6) + 6(4) - E = 0
E = 42v
So my E will be 42 volts?

And in question (a) since 6A will pass in node a and the current in I2 is 4A, then logically, the remaining 2A will flow through R. So the current in R will be 2 Amperes.

So now I can solve for the Resistance in R (question B)
4(6) - 48 - 28 + 2(R) = 0

24 - 48 - 28 = -2(R)

R = 26 ohms

Are my solutions correct? In question D, the current will be 2A if the circuit is broken at point x. Correct me if I'm wrong. Thanks in advance!

 

[Defennder]

3(6) + 6(4) - E = 0
E = 42v
So my E will be 42 volts?

And in question (a) since 6A will pass in node a and the current in I2 is 4A, then logically, the remaining 2A will flow through R. So the current in R will be 2 Amperes.

Correct so far.

So now I can solve for the Resistance in R (question B)
4(6) - 48 - 28 + 2(R) = 0

24 - 48 - 28 = -2(R)

R = 26 ohms

Where did -48 come from? Also check your sign convention for the 28V voltage source. If you're going in a clockwise loop, and you enter the 28V voltage source through the +ve terminal, then you add, not subtract the potential drop.

Are my solutions correct? In question D, the current will be 2A if the circuit is broken at point x. Correct me if I'm wrong. Thanks in advance!

Why would it be 2A? Remember that once x is cut, E doesn't factor into the calculations at all. You would then have a single loop consisting of currents through top and bottom wires.

 

[daimlerpogi]

Okay I already get it now. Thanks Defennder!