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Determining the equivalent resistence of a circuit

  • Thread starter Eman 5
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  • #1
Eman 5

Homework Statement


I want to determine the equivalent resistance between A and D in the drawing below:
[/B]
yuVKS.jpg


Homework Equations


1/Req=1/R1+1/R2+1/R3+........
And
Req=R1+R2+R3+....

The Attempt at a Solution


I don't really know how to simplify this drawing, I'm not sure if one resistor isn't calculated in this case. I determined the equivalent resistance of a similar circuit but the resistors in it were equal so, I don't calculate one of the resistor inside the triangle as the voltages at its two ends are equal.[/B]
 

Answers and Replies

  • #2
I don't really know how to simplify this drawing
Start by labeling the two end points, A and D, in your new drawing. Then, build up the circuit by adding in each resistor one at a time. So to start, notice that the bottom resistor of ##1\Omega## connects straight from A to D. Next, notice the top two resistors of ##2\Omega## and ##1\Omega## are in series between A and D. Continue drawing in the other resistors and it should become more clear which resistors are in parallel and which are in series. It will just be a matter of applying the formulas for equivalent resistance thereafter.

EDIT: It looks like using a ##\Delta##-Y transformation for the Y branch will allow you to draw an equivalent circuit where you will be able to apply your known formulas for resistors in series and in parallel. Without this transformation, the problem is still easily solvable but you must set up a system of equations for the voltage drop across each resistor and solve the linear system.
 
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  • #3
collinsmark
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While equations such as [itex] \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} [/itex] for parallel and [itex] R_{eq} = R_1 + R_2 + R_3 [/itex] for series configurations are often useful, they won't help you much here. You'll find that some resistors are neither completely parallel nor completely series. You need a different approach.

Attach a hypothetical voltage source between A and D. Give it some specified voltage, whatever you want. You might as well make it 1 V for convenience. Then solve for the currents in the circuit using KVL and/or KCL. You'll need 4 simultaneous equations and 4 unkowns.

With that you can calculate the current passing through the hypothetical voltage source. And at that point, since you know the voltage of the voltage source and the current passing through it, you can calculate the equivalent resistance of the circuit.

Edit: Oh, and @Eman 5, Welcome to Physics Forums (PF)! :welcome:
 
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  • #4
collinsmark
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EDIT: It looks like using a ##\Delta##-Y transformation for the Y branch will allow you to draw an equivalent circuit where you will be able to apply your known formulas for resistors in series and in parallel. Without this transformation, the problem is still easily solvable but you must set up a system of equations for the voltage drop across each resistor and solve the linear system.
Right, that's another approach that will work too, if you happen to know/remember the Y to [itex] \Delta [/itex] (or [itex] \Delta [/itex] to Y) transformation equations. :smile:
 
  • #5
Eman 5
While equations such as [itex] \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} [/itex] for parallel and [itex] R_{eq} = R_1 + R_2 + R_3 [/itex] for series configurations are often useful, they won't help you much here. You'll find that some resistors are neither completely parallel nor completely series. You need a different approach.

Attach a hypothetical voltage source between A and D. Give it some specified voltage, whatever you want. You might as well make it 1 V for convenience. Then solve for the currents in the circuit using KVL and/or KCL. You'll need 4 simultaneous equations and 4 unkowns.

With that you can calculate the current passing through the hypothetical voltage source. And at that point, since you know the voltage of the voltage source and the current passing through it, you can calculate the equivalent resistance of the circuit.
Thank you:smile:. I followed your steps and I could solve it. The equivalent resistance=7/12 Ω which is the correct answer shown in the book.
 
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