What is the answer key doing in this circuit

[flyingpig]

Homework Statement

http://img546.imageshack.us/img546/8831/circuitproblem.th.png

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Solutions

http://img855.imageshack.us/img855/9171/solutioncircuit.th.png

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The Attempt at a Solution

Now I want to tackle this problem in two ways

(1) Add the net resistance and do what I normally do.

(2) Make my life difficult and apply Kirchhoff's rule

I will begin with (1)

Now the answer key seem to have just ignored one of the resistors because I don't understand what happened to the other 5.0Ω resistor

So I added the two 10Ω and one 5Ω resistor in parallel and I got 2.5Ω

Then I added up the resistors in series and I got 27.5Ω

Then I simply did 25V/27.5Ω = 0.909A

Then the current through the 20Ω must be 0.909A since we are in series and we only got one battery

For part (2), I will continue in the next post, but I have class right now so I'll type it out later because it looks ridiculously difficult right now lol so I apologize

 

[flyingpig]

OKay I am back and I realize I cannot apply Kirchhoff's Method unless I really do break it up into equivalent resistors...

 

[gneill]

The three resistors in the inset part of the circuit (between a and b) are not all in parallel; One of them has a series voltage source.

What you could do is convert the voltage source and its series resistor into its Norton equivalent, then all three resistances would be in parallel with the resulting current source and can then be combined into a single resistance. You would then have the option of converting back to a voltage source (Thevenin) yielding a single series loop, or just addressing the resulting current divider directly.

 

[flyingpig]

You better elaborate on those fancy words...LOl

 

[SammyS]

Homework Statement

http://img546.imageshack.us/img546/8831/circuitproblem.th.png

Uploaded with ImageShack.us

Solutions

http://img855.imageshack.us/img855/9171/solutioncircuit.th.png

Uploaded with ImageShack.us

The Attempt at a Solution

Now I want to tackle this problem in two ways

(1) Add the net resistance and do what I normally do.

(2) Make my life difficult and apply Kirchhoff's rule

I will begin with (1)

Now the answer key seem to have just ignored one of the resistors because I don't understand what happened to the other 5.0Ω resistor

No, they included it.

20Ω & 5Ω are in series → 25Ω.

That 25Ω is in parallel with 5Ω & 10Ω → (50/17)Ω ≈ 2.94Ω.

So I added the two 10Ω and one 5Ω resistor in parallel and I got 2.5Ω

Then I added up the resistors in series and I got 27.5Ω

Then I simply did 25V/27.5Ω = 0.909A

Then the current through the 20Ω must be 0.909A since we are in series and we only got one battery

For part (2), I will continue in the next post, but I have class right now so I'll type it out later because it looks ridiculously difficult right now lol so I apologize