Calculating total resistance of a combination circuit

AI Thread Summary
The discussion focuses on calculating the total resistance of a complex circuit involving a 1470 ohm resistor in parallel with 50 ohm and 1000 ohm resistors. Participants suggest using Kirchhoff's laws and a Delta to Y transformation to simplify the circuit analysis. One method proposed involves drawing loop equations to determine the current in each loop, which can then be used to find the total resistance. There is some confusion about whether the total resistance can be found directly or if it requires calculating individual currents first. Ultimately, using loop analysis and Kirchhoff's laws is emphasized as an effective approach to solve the circuit.
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combination circuit.png

Homework Statement


Find the total resistance of the circuit

Homework Equations


Parallel resistance: 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ..., Series resistance: R1 + R2 + R3...

The Attempt at a Solution


The 1470 ohm resistor is parallel to the 50 ohm resistor and the 1000 ohm resistor, but there is a junction between them, which confuses me on how to add the 1470 ohm resistor to each of them... that probably didn't make sense but it's the best way I could word it. Once that is solved I know how to do the rest.
 
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Ketchup4684 said:

Homework Statement


Find the total resistance of the circuit

Homework Equations


Parallel resistance: 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ..., Series resistance: R1 + R2 + R3...

The Attempt at a Solution


The 1470 ohm resistor is parallel to the 50 ohm resistor and the 1000 ohm resistor, but there is a junction between them, which confuses me on how to add the 1470 ohm resistor to each of them... that probably didn't make sense but it's the best way I could word it. Once that is solved I know how to do the rest.
Try adding a picture.
 
Isaac0427 said:
Try adding a picture.
(facepalm) oops... there it is
 
None of the resistors are in series or in parallel with any other.
 
vela said:
None of the resistors are in series or in parallel with any other.
How would i go about adding up the resistances then?
 
Connect the combination to a voltage source V and calculate the current I it draws. The equivalent resistance is V/I.
 
vela said:
Connect the combination to a voltage source V and calculate the current I it draws. The equivalent resistance is V/I.
But to find the current that is drawn from the battery (6V) i need to know the total resistance of the circuit...
 
You're just going to have to write down equations using Kirchoff's laws and solve the resulting system of equations.
 
I would perform a Delta to Y transform on the 1470, 50, & 1000, which are currently in Delta (a triangle). See these for info on Delta-Y
http://ux.brookdalecc.edu/fac/engtech/andy/elec111/powerpoint/12_deltatowye.pdf and https://en.wikipedia.org/wiki/Y-Δ_transform
 
  • #10
vela said:
Connect the combination to a voltage source V and calculate the current I it draws. The equivalent resistance is V/I.
I would think that there is no point to this. You need the total resistance to calculate the current, then use I=V/R, and then go back to use R=V/I. That's just a lot of unnecessary extra steps that you would take after you find the total resistance.

To solve this you would need to either use kirchhoffs laws or kirchhoffs laws with the delta y transform.
 
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  • #11
Isaac0427 said:
I would think that there is no point to this. You need the total resistance to calculate the current, then use I=V/R, and then go back to use R=V/I. That's just a lot of unnecessary extra steps that you would take after you find the total resistance.

To solve this you would need to either use kirchhoffs laws or kirchhoffs laws with the delta y transform.
At first, I thought @vela was joking (like actually get a battery and some resistors.. Ha). But I think what was meant is to create a circuit on paper, then go through the loop analysis. Once you find the current delivered by the test voltage, then R = Vsource/Isource.
 
  • #12
Isaac0427 said:
I would think that there is no point to this. You need the total resistance to calculate the current, then use I=V/R, and then go back to use R=V/I. That's just a lot of unnecessary extra steps that you would take after you find the total resistance.
Really? There's no other way to find the current I? Like perhaps the way you mention below...

To solve this you would need to either use kirchhoffs laws or kirchhoffs laws with the delta y transform.
 
  • #13
vela said:
Really? There's no other way to find the current I? Like perhaps the way you mention below...
Right, but to find the current that way wouldn't you have to find the total resistance first?
 
  • #14
I can't help but feel you're trolling here.
 
  • #15
Isaac0427 said:
Right, but to find the current that way wouldn't you have to find the total resistance first?
No. Solve the circuit, and you'll find all the currents even if you only really require one of them.
 
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  • #16
gneill said:
No. Solve the circuit, and you'll find all the currents even if you only really require one of them.
Ah ok.
 
  • #17
Ketchup4684 said:
View attachment 203597

Homework Statement


Find the total resistance of the circuit

Homework Equations


Parallel resistance: 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ..., Series resistance: R1 + R2 + R3...

The Attempt at a Solution


The 1470 ohm resistor is parallel to the 50 ohm resistor and the 1000 ohm resistor, but there is a junction between them, which confuses me on how to add the 1470 ohm resistor to each of them... that probably didn't make sense but it's the best way I could word it. Once that is solved I know how to do the rest.

One method is the loop-equation method. Draw out the possible current loops for the circuit, write an equation for each loop, and then solve the simultaneous equations to find each loop current. The sum of these give you the total current for any given source voltage, and from that you can determine the total resistance of the circuit. Here's the circuit redrawn with the loop currents marked:
bridge.gif
 
  • #18
Janus said:
Here's the circuit redrawn with the loop currents marked:

just out of personal curiosity for my learning

can you do that with just the three loops you have done or should it really be for all loops ?
as there is a 4th loop
 
  • #19
davenn said:
just out of personal curiosity for my learning

can you do that with just the three loops you have done or should it really be for all loops ?
as there is a 4th loop
You'd find that only three of the loop equations are independent. The equations from the other loops would be linear combinations of the three.
 
  • #20
davenn said:
just out of personal curiosity for my learning

can you do that with just the three loops you have done or should it really be for all loops ?
as there is a 4th loop
Independent loops are needed, and there are three in this case. Assign current to each loops, so that there is one or two loop currents flowing through each resistor. If it is possible, the loops are separate. @davenn's construction is too complicated. See the next one:
upload_2017-5-18_6-25-8.png


The currents through the resistors are: I1, through 100Ω, I2, through 330Ω, I2-I1, through 220Ω, I2-I3, through 50Ω, and I3-I1, through 1000Ω. The generator current is I1. Anti-clockwise direction is positive. Write Kirchhoff's Loop Rule for each loops, that makes three equations for the three currents, eliminate I2 and I3, solve for I1. The total resistance is R=V/I1.
 

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