# Simple Circuit Help.

Discussion in 'Homework Help' started by Kaiser, Sep 18, 2007.

1. ### Kaiser Thread Starter New Member

Sep 18, 2007
6
0
Hello everyone. I have some brief questions about a simple circuits problem in my homework.

First off, here is the image of the circuit (sorry..drawn in MSPaint.) http://img211.imageshack.us/img211/3271/circuitgh9.jpg

The Current Source = 0.5A
R1 = 5 Ohms
R2 = 25 Ohms
R3 = 10 Ohms
R4 = 50 Ohms
R5 = 5 Ohms

It asks to find the total resistance of resistors 2, 3, 4, and 5. I've done this by counting R2 as part of a series circuit, and figuring R3,R4, and R5 as a parallel.

So, 25 + (1/((1/10)+(1/50)+(1/5))) = 28.13 Ohms (assuming this is correct..)

Now it asks to find the Voltage drop across the current source..I'm not sure how to figure this since I don't have distance..

Next it asks to find voltage drop across resistor 1, How much power is dissipated by this resistor (V=I^2*R)...But once again...voltage drop?

And finally it asks for the voltage drop across R2,3,4,5...and once again, voltage drop?

I've read and re-read all of the notes taken in class, and have no idea how to figure voltage drop for this type of circuit, I checked out an Ugly book and the NEC book, but everything there is to do with voltage drop requires distance.

Thanks in advance,

-Kaiser

2. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,227
Voltage drop is the voltage "dropped" across the resistances.

The voltage drop you're reading about is long line runs and the difference between the source and terminal ends.

You need to reconsider your calculations again.

3. ### Kaiser Thread Starter New Member

Sep 18, 2007
6
0
Well I'm lost on how to find the voltage drop, would it be the voltages in each resistor (after dissipation) added together??

4. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,227
Kaiser,

First you can find the total resistance. Attached is the beginining of my response.

Once you find that, then post your attempts at finding the voltage drops.

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5. ### Eduard Munteanu Active Member

Sep 1, 2007
86
0
The supply is a current source, so it outputs whatever voltage is needed to maintain that 0.5A current through the circuit. Replace all those resistors with an equivalent one (that is, find the Norton equivalent circuit) and apply Ohm's law: voltage_drop = current x equivalent_resistance.

6. ### niftydog Active Member

Jun 13, 2007
95
0
Look again. R2 and R5 are in series with each other, and the two are in parallel with R3 and R4. Add R2 and R5 and draw the combination as one resistor. Add R3 and R4 and do the same. Find the parallel resistance of those two combinations and you're half way there.

Ignore the word "drop", forget about dissipation - just find the voltage across each resistor.

7. ### Distort10n Active Member

Dec 25, 2006
429
1
Your calculation is incorrect. R2 is in series with R5, and their resistive sum is in parellel with...what?

Once you find the total resistance in the circuit, then you can simply use Ohm's Law.

The total resistance of R2, R3, R4, and R5, is in series with R1. Current is equal in all elements in a series circuit.

KVL would apply here.

Your homework is asking you about the votlage drop across an element. If the books are talking about distance, then it may be talking about communications; e.g., sending a signal down a cable, etc.

8. ### Distort10n Active Member

Dec 25, 2006
429
1
Man...I type too slow.

9. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,227
Yeah, voltage drop is too new age for me. They still are IR drops to me.

10. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,227
Mavis Beacon teaches typing ... and she has since Win 3.1

11. ### Kaiser Thread Starter New Member

Sep 18, 2007
6
0
First of all..thank you all VERY much..

So now I get the total resistance is 8.571 Ohms.

Now part 2 asks, "What is the voltage drop across the current source?"

So I'll assume that it's the voltage in each resistor added together? Actually, it can't be due to the next questions..

Part 3 asks, "What is the voltage drop across resistor 1? How much power is dissipated by this resistor?" So using V=R(total) * I ; V=8.571*.5, V=4.288.
The power dissipated can be defined as I^2*R(1) So.. .5^2* 5=1.25V

Part 4 asks, "What is the voltage drop across resistors 2,3,4, and 5.

So I've got to find the various currents..The currents through each side of the parallel circuit are the same all the way through right..

so basically..I'm seeing that I have three currents(the initial, and the two running down each side of the parallel.)

Unfortunately I have no example of how to find these..let me google for a while..The only example I've got is the same circuit, but without R2 (and different Ohm amounts as well.) Anyway, the professor used Kirchhoff's laws to determine them..and I can't use them here.

12. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
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How did you get 8.571 ohms for Rt?

Why can't you use Kirchoff's Law?

13. ### niftydog Active Member

Jun 13, 2007
95
0
Just to make sure we're all on the same page, I've redrawn the schematic. Kaiser, is this correct?
Code ( (Unknown Language)):
1.
2.        R1         R2
3.        5R         25R
4.        ___        ___
5.     o-|___|--o---|___|--o
6.     |        |          |
7.     |        |          |
8.     |        |          |
9.     |        |          |
10.     |       .-.        .-.
11.     |       | | R3     | | R5
12.   ,---.     | | 10R    | | 5R
13.   | ^ |0.5A '-'        '-'
14.   '---'      |          |
15.     |        |          |
16.     |        |          |
17.     |       .-.         |
18.     |       | | R4      |
19.     |       | | 50R     |
20.     |       '-'         |
21.     |        |          |
22.     o--------o----------o
It seems correct to me because all the maths works out very neatly.

First, redraw it, it will make more sense drawn this way;
Code ( (Unknown Language)):
1.
2.        R1
3.        5R
4.        ___
5.     o-|___|--o----------o
6.     |        |          |
7.     |        |          |
8.     |        |          |
9.     |        |          |
10.     |       .-.        .-.
11.     |       | | R3     | | R2
12.   ,---.     | | 10R    | | 25R
13.   | ^ |0.5A '-'        '-'
14.   '---'      |          |
15.     |        |          |
16.     |        |          |
17.     |       .-.        .-.
18.     |       | | R4     | | R5
19.     |       | | 50R    | | 5R
20.     |       '-'        '-'
21.     |        |          |
22.     o--------o----------o
Add the series resistors and redraw it again to simplify it further.
Code ( (Unknown Language)):
1.
2.        R1
3.        5R
4.        ___
5.     o-|___|--o----------o
6.     |        |          |
7.     |        |          |
8.     |        |          |
9.     |        |          |
10.     |       .-.        .-.
11.     |       | | R3+R4  | | R2+R5
12.   ,---.     | | 60R    | | 30R
13.   | ^ |0.5A '-'        '-'
14.   '---'      |          |
15.     |        |          |
16.     |        |          |
17.     |        |          |
18.     o--------o----------o
Find the parallel combination and redraw;
Code ( (Unknown Language)):
1.
2.        R1
3.        5R
4.        ___
5.     o-|___|--o
6.     |        |
7.     |        |
8.     |        |
9.     |        |
10.     |       .-.
11.     |       | | R3+R4||R2+R5
12.   ,---.     | | 20R
13.   | ^ |0.5A '-'
14.   '---'      |
15.     |        |
16.     |        |
17.     |        |
18.     o--------o
The voltage across the current source is 0.5A x 25 ohms = 12.5V.

Now use the voltage divider equations to find the voltage across R1 and the rest should hopefully fall into place for you.