Determining the equivalent resistence of a circuit

In summary, the conversation discusses a problem of determining the equivalent resistance between points A and D in a circuit. The suggested methods for solving the problem include using the equations for resistors in parallel and series, applying the Δ-Y transformation, or using KVL and KCL with a hypothetical voltage source. The final solution is 7/12 Ω.
  • #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]
 
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  • #2
Eman 5 said:
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
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
NFuller said:
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
collinsmark said:
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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Related to Determining the equivalent resistence of a circuit

1. How do you calculate the equivalent resistance of a circuit?

The equivalent resistance of a circuit can be calculated by using Ohm's law, which states that resistance (R) is equal to voltage (V) divided by current (I). You can also use the formula Req = R1 + R2 + ... + Rn, where R1, R2, etc. are the individual resistances in the circuit.

2. What factors affect the equivalent resistance of a circuit?

The equivalent resistance of a circuit is affected by the individual resistances of the components in the circuit, the arrangement of those components (series or parallel), and the total number of components. Other factors that may affect resistance include temperature and the material the components are made of.

3. Why is it important to determine the equivalent resistance of a circuit?

Determining the equivalent resistance of a circuit is important because it allows us to understand and analyze the behavior of the circuit. It helps us to calculate the current and voltage at different points in the circuit, and to determine the power dissipated by the circuit. This information is crucial for designing and troubleshooting circuits.

4. Can the equivalent resistance of a circuit be lower than the individual resistances?

Yes, the equivalent resistance of a circuit can be lower than the individual resistances. This occurs when the components are arranged in parallel, which allows for multiple paths for current to flow. In this case, the equivalent resistance is calculated differently (Req = 1/(1/R1 + 1/R2 + ... + 1/Rn)) and will be lower than the individual resistances.

5. How does adding or removing components affect the equivalent resistance of a circuit?

Adding or removing components can change the equivalent resistance of a circuit. Adding components in series will increase the equivalent resistance, while adding them in parallel will decrease it. Removing components will have the opposite effect. The equivalent resistance is also affected by the value of the components added or removed.

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