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

 

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.

 

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.

 

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.

 

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

 

One way is to use a pencil and scan the result. Here's a link to some other suggestions - http://forum.allaboutcircuits.com/showthread.php?t=19134

 

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
XXXXXXXXXXXXX:eek:XXXXX:eek:XXXXX:XXXxXX:eek:XXXXX:eek:XXXXXXXXXXXXxXXX
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.

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.

 

Attempt to post schematic:

View attachment Question on Parallel gloves.bmp

 

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.

 

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

 

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.

 

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.