# Kirchoff's Laws

1. Jan 14, 2010

### physicsfun_12

1. The problem statement, all variables and given/known data
Hello there, hope you are well.

I am having trouble solving this problem involving Kirchoff's laws.

I have attached a copy of the circuit and problem to this post in jpeg format.

Thanks again in advance for any help.

Mike

2. Relevant equations

3. The attempt at a solution
Well I merged to the two 20 ohms into one 10ohm and said that the total voltage in any loop equals zero.

so 60=20Ia + 10Ib

and 20=20Ib + 10Ia

I have tried solving this many times and keep getting a negative current for Ib. My original equation must be wrong.

Thanks alot for any help or input,

Mike

#### Attached Files:

• ###### Kirchhoff's Laws_question.jpg
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2. Jan 14, 2010

### tiny-tim

Hi Mike!
Which loop has Ia, and which has Ib, and in which direction?

(and where do those 20s on the RHS come from?)

3. Jan 14, 2010

### physicsfun_12

I think Ive got it now.

I was using loop analysis so thats why there is only two currents.

The 20s came from 60=10Ia + 10(Ia+Ib) which expands to give 20Ia + 10Ib

You can work out the currents using simultaneous equations, giving a Ia=3.333A and Ib=-0.6666A

I of the load resistor, Il=Ia+Ib = 2.6667A

So Vl=2.6667*10 = 26.667V

Mike

Last edited: Jan 14, 2010
4. Jan 14, 2010

### tiny-tim

ohhh! you should always state an intermediate step like that

(apart from anything else, you stand a good chance of making a mistake with ±s if you don't write this stuff out clearly)

Yes, the answer looks ok now.

5. Jan 14, 2010

### physicsfun_12

Sorry about that, you are absolutely right... that's most probably why I was getting it wrong in the first place!

Thanks for your help and take care,

Mike