# Electric circuits, finding current with terminal voltage

1. May 19, 2013

### physics604

1. What current flows through the 11.0 Ω resistor?

a) 0.21 A
b) 0.27 A
c) 0.93 A
d) 1.2 A

To summarize:

V1 = 8.0 V
V2 = 5.0 V
R1 = 2.0 Ω
R2 = 11.0 Ω
R2 = 1.0 Ω

I know we shouldn't attach pictures, but the diagram does make the question a lot easier to understand.

2. Relevant equations

V=IR
VT = ε-Ir

3. The attempt at a solution

RT = 11.0 + 2.0 + 1.0 = 14.0 Ω
VT = 8.0 + 5.0 = 13.0 V
IT = V/R = 13.0/14.0 = 0.93 A.

And since current should be the same throughout a series circuit, the current through the 11.0 Ω resistor should be 0.93 too. My answer is wrong though.

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Last edited: May 19, 2013
2. May 19, 2013

### barryj

You are close but notice that the battery's oppose each other

3. May 19, 2013

### physics604

Yes, but what affect does that have? Does it mean the current is different in different parts of the circuit?

Am I doing the VT and the RT correctly?

4. May 19, 2013

### barryj

If you write the kirchoff voltage law around the loop, you will find that the battery voltages will subtract rather than add as you have done.

5. May 19, 2013

### physics604

So are you saying VT= 8.0-5.0 = 3.0 V?

But why would that happen? Kirchoff's Loops Rule just states that potential differences around a complete loop is equal to zero. Wouldn't both batteries create a positive potential change though?

6. May 19, 2013

### barryj

Check the schematic diagram, the batteries do not add, they will subtract.