Circuit problem Kirchkoff's law

• johnknee
In summary, current passes through the closed switch and the power dissipated in the 47 ohm resistor is 0.383A.f

Homework Statement

a) With the switch open, what is the current I1 ? (A positive sign means that current
flows in the direction of the arrow.)

b) Once the switch is closed in the circuit, what is the power dissipated in the 47 Ω
resistor?

c) How much current passes through the closed switch?

d) With the switch closed, what is the voltage difference, VA-VB ?

Homework Equations

junction rule: I3 = I2 + I1
Loop rule(s)

The Attempt at a Solution

I have attempted all 4 parts, please check my solutions? (parts a to c should be correct, but d not too sure.)

For a) Uses the same picture as part b

I believe that because there is a open switch, then the whole circuit became a series.
1st step: I got the Req value = 5ohms + 22ohms + 47ohms + 15ohms = 89ohms.
2nd step: I got the Veq value = 10.0V + (-6.0V) = 4.0v
3rd step: I did I = V/R = 4.0V/89ohms = 0.0449 A = I1.

For b) I have very badly drawn out all of the current directions(the switch is also closed) the top loop and the bottom loop would be separate with a junction at the middle piece on either end.

First, I got that I2 + I1 = I3.

Then, I solved for the top loop, starting at the negative terminal side of the battery with 10V moving clockwise.
Top loop(clockwise): +10V1 - 15*I1 - 47*I1 + (0V since I'm assuming no Voltage drop across switch) = 0
Bottom loop(counter clockwise): -5*I2 + 6V2 - 22*I2 + 0V = 0

For the top loop: I isolated I1 and got 10/62 = 0.161 = I1
For the bottom loop: I got 6/27 = 0.222A = I2

Power dissipated in 47 ohm resistor I am assuming would be I1^2 * R = (0.161)^2 * 47 = 1.22 Watts?

Part c)
I'm not totally sure how I do this one, but I think that the current passing through the closed switch is just I3? which is I3 = I1 + I2 = 0.161 + 0.222 = 0.383A.

Part d)
For this part.. I am not quite sure.
I was thinking of like, VA - VB means we have to get from VB to VA in the circuit.
So... like VA - VB = +6V2 - 22*I2 + 47*I1? (going counter clockwise up the loop from the 6V battery.)
which would give me VA - VB = 6 - 22(0.222)+47(0.161) = 8.68 V.

Last edited:
That all looks right.