Mixed Series & Parallel circuit - how to allocate watts ?

Discussion in 'The Projects Forum' started by DrAlloway, Apr 12, 2011.

1. DrAlloway Thread Starter New Member

Apr 7, 2011
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0
This is not really homework ( I am modifying my DIY heated motorcycle jacket ) but the question is simple enough for textbook homework. But this is not my field and I am confused.

I have a long 32 gauge wire (48 feet) that snakes around my jacket to provide heat. It has 6 ohms of resistance so at 12volts, that is 2 amps and 24 watts of heat.

Now I want to add gloves to this wire. Each glove has 2 ohms of resistance.

If I add the gloves in series that gives me 6 + 2 + 2 = 10 ohms or only 1.2 amps and 14.4 watts of heat which is not nearly enough.

So I want to add the gloves in PARALLEL to give me more heat. This circuit looks like one really long wire (20 feet) with a loop sticking out, then the long wire continues another 20 feet; then the second loop for the other glove.

So I think I have to analyze this circuit in two halves: 3 ohms of long wire with a 2 ohm parallel loop sticking out of the middle. Then the same thing for the other half: a parallel circuit with 2 ohms on one path and 3 ohms on the other path.

That gves me 1.2 ohms for each parallel half (2*3)/(2+3) = 6 / 5
The halves are in series so 2.4 ohms for the whole thing
Hence, 5 amps (12/2.4)
and 60 watts (5 * 12 ) for the whole circuit

I think this part is right - if not, stop me now.

Where I get confused is allocating the total 60 watts to each glove.
The series circuit is symmetric so each half has a 6 volt drop ( 12 /2 )
Each half has 30 watts and 5 amps.

Then I allocate a half (6v, 5a, 30w) among the two parallel resistances of 2 for gloves and 3 for long wire

So one glove is:
18 watts = (E * E) / R = 6volts * 6 volts / 2 ohms
or
3 amps in a glove = E / R = 6 volts / 2 ohms
18 watts = (I * I) * R = 3 amps * 3 amps * 2 ohms

I have obviously simplified the arithmetic BUT are my concepts right??
? This is a mixed circuit of series and parallel
? Partition the circuit into two series pieces
? Calculate the total parallel resistance within each piece
? Add the resistances of each piece together as they are in series
? Calculate the whole circuit
? Take the circuit back apart into its two pieces (now its 6v each)
? Calculate watts for a glove in its piece (half) of the circuit

thank you - my hands are really cold and I hope this is right

2. beenthere Retired Moderator

Apr 20, 2004
15,815
283
Since this is not homework, the thread got moved.

Is your heated jacket warm enough as is? If so, why not add the gloves as a separate circuit? The gloves in series are 4 ohms, or 36 watts - 18 per glove.

3. wayneh Expert

Sep 9, 2010
12,394
3,246
A picture (circuit schematic) would go a long way here. I think I understand but it's still confusing. Sounds like you've got a good handle on the tool you need - Ohm's law.

4. DrAlloway Thread Starter New Member

Apr 7, 2011
7
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Without sfw for this purpose:: this is the idea

2Ω : 2Ω
X : X
X X : X X
X X : X X
X X : X X
XXXXXXXXXXXXXXXXXXXXXXX:XXXxXXXXXXXXXXXXXXXXXXXxXXX
X 6Ω for the : whole wire X
12vDC : ground

Note the 6 ohm wire is like 46 feet long and the distance between connections of gloves to long wire (shock faces) is just a couple of inches. I think the theory says that does not matter - that no matter how long the wire segment is between connections - it is one circuit path.

I think the way to analyze this circuit is to partition it in half (at my dotted line) and treat each half like a simple parallel circuit with 2 ohms on one path and 3 ohms on the other path. Solve that half. Then add the two Series halves together.

5. DrAlloway Thread Starter New Member

Apr 7, 2011
7
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Well - that did not work - all my blank spaces went away.
Anybody have FREE, EASY, SIMPLE software for basic circuit diagrams ??

Apr 20, 2004
15,815
283
7. SgtWookie Expert

Jul 17, 2007
22,183
1,728
Just added CODE blocks. If you use the "Go Advanced" button, there is a # sign which will insert the code blocks around the selected text. It preserves the formatting.

Apr 7, 2011
7
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9. wayneh Expert

Sep 9, 2010
12,394
3,246
Ahh, now I see. I think. If the drawing is correct, then your first analysis was correct. You have two loads in series, each with 1.2Ω of resistance, supplied with 12v. Each load consists of a 2Ω and a 3Ω load in parallel. Voltage at the midpoint between the two loads is 6v. Current thru the 3Ω loads is 6v/3Ω=2A each. Current thru the 2Ω loads is 6v/2Ω=3A each. Total current is 5A. Respective power in each is I^2R = 2^23=12w and 3^22=18w. Total power is 24 watts in the jacket and 36w in the gloves, and 60w overall.

10. DrAlloway Thread Starter New Member

Apr 7, 2011
7
0
Thank you !! So, if my schematic is correct, then I did the circuit analysis right. Yippie !

Now I would like to check if my schematic is right - if so, the following two circuits are the same electrically.

Actual:

one long 20 foot wire with a 7 foot loop tapped into the middle - the distance between the loop taps is 1 inch

Theroetical equivalent (I think):

one long 20 foot wire with a 7 foot loop attached - the distance between the loop attachment points is 6 feet

I think the theory says: no matter how far apart the loop connections are ( 1 inch or 6 feet ) that there are only 2 paths: along the long wire and around the loop

This confuses me because textbook schematics assume Zero resistance in the wires connecting all the components. However, in reality - and especially my case where all the resistance is provided by the actual wires (there are no components) - I am afraid the parallel paths are the 7 foot loop and the 1 inch of wire between its connections.

Do you see my concern/confusion ?

Are the parallel resistances really 3 ohms (long wire) and 2 ohms (glove)
or
Rather the parallel resistances are really 2 ohms (glove) and 0.01 (resistance of 1 inch) ?

Theory says All Points on the 20 foot wire are The Same and it does not matter if the connections for the glove loop are 1 inch apart or 6 feet apart.
? right ?
? really ?

thanks again, Bob

11. wayneh Expert

Sep 9, 2010
12,394
3,246
Well now you're losing me again. If all of your conducting wire is really heating element - meaning it has more resistance per foot than "normal" wire, than you do need to consider that. However the resistance of any segment is proportional to its length, so a few inches has essentially zero resistance compared to the multi-foot sections.

I do think you need to give special attention to any segment that is carrying the full 5A. It needs to be able to handle that, and you'll still get I^2R watts dissipated in that section. Hopefully, the section is short, the resistance is therefore low, and the required heat dissipation is not a problem.

12. DrAlloway Thread Starter New Member

Apr 7, 2011
7
0
All the wire in this circuit is 32 gauge, 7 strand, twisted copper with a Teflon insulating cover. It has a Book rated Resistance of 170 ohms per 1,000 feet.