# Series Parallel Help

Discussion in 'Homework Help' started by foleystefan, Nov 6, 2010.

1. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
0
I've been doing quite well with studying my electronics book on series, parallel, and series-parallel but there were a few of the series-parallel that get me stuck. I'm not sure how to go about getting different values for items with just a small amount of information.

I have included such a problem that has stumped me.

http://img248.imageshack.us/img2

I need to find all the unknowns which I assume include:

V1-V5
VT
I1-I6
IT
PT

If someone could solve these out for me cause I'm quite perplexed as to how to derive anything from these (other than I6). I want to see how each value is derived and what equations were used. I understand all the base equations but honestly can't seem to come up with a solution unless the only unknown to be gotten is just I6 (25mA). Please include if you use REQ too please.

Thanks.

Stefan

2. ### JoeJester AAC Fanatic!

Apr 26, 2005
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use the advanced features and post your message in this thread and not at imageshack. .png format is preferred.

Jan 28, 2005
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hgmjr

4. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
0

sorry hopefully this will work

5. ### Fraser_Integration Member

Nov 28, 2009
142
5
In a series circuit, the same current flows through every component. So if you know I6, you know I5 and I4 because they are in the same branch and must have the same current through them.

Knowing this you can work out I4 and I5 voltage drops because V = IR.

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6. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
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thanks for that, I don't know why that seemed so hard. I'm going to post my answers so that if anyone if anyone has the time to check them and see if I'm doing it right. Here they are:

REQ=500 ohms
RT=1.19 k-ohms
I6=25mA
I6=I5=I4=25mA
V5=11.25V
V4=6.25
VTr456=25V
V2=VTr456=25V
VEQ=50V
I2=25mA
VT=V1=V3=VEQ=50V
IT=42mA
V1=16.38V
V3=12.6V
PT=2.1W

If someone could check these for me that would be great. Let me know where I went wrong if there is anywhere I went wrong.

Thanks again.

Stefan

7. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
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just wanted to bump

8. ### hgmjr Moderator

Jan 28, 2005
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214
You answer to the voltage across R2 is correct at 25V. Your calculation of the voltage Vt of 50V is not correct.

hgmjr

9. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
0
ahh thanks, 119V is what I got after, is that correct?

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
That just so happens to be wrong.

hgmjr

11. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
0
ok, can you give me the values for the unknowns so i can have something to go by?

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The one you seem to be snagged on is the value of Vt. You have the value of the voltage across R2 correct at 25V. That means that you are well on your way. With the value of V2 in hand you can determine the current that is flowing through R2. You can also find the current through the series resistors R4, R5, and R6. Then add the two currents to get the current that is flowing through R3 and R1. This current can be used to calculate the voltage Vt since you know the voltage across V2.

hgmjr

13. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,402
1,227
You need to revisited how you arrived at the following:

VEQ=50V
VT=V1=V3=VEQ=50V
V1=16.38V
V3=12.6V
IT=42mA
PT=2.1W

All the rest are OK.

14. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
0
ok, redone it again, is it 59.5V?

15. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Do you know where you made the mistake?

16. ### foleystefan Thread Starter New Member

Oct 4, 2010
11
0
so 59.5V is correct then? I figured that the voltage total from R4,5,6 is equal to voltage 2 which is equal to voltage equivalency. also current equivalency is equal to current total from R4,5,6 added to current 2. therefore IEQ = 25mA+25mA=50mA. which then gives me the values needed to make the Vt and so on.

17. ### hgmjr Moderator

Jan 28, 2005
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You got it. Congratulations.

hgmjr

foleystefan likes this.

Apr 26, 2005
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Nice work.