How to Calculate Equivalent Resistance and Current in a Simple Circuit Problem?

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SUMMARY

The forum discussion focuses on calculating the equivalent resistance and current in a simple electrical circuit involving resistors of 1500Ω, 900Ω, and 1200Ω, with a source voltage of 30V. The correct equivalent resistance was determined to be 750Ω after applying series and parallel resistor formulas. The current through the circuit was calculated using Ohm's Law, yielding a total current of 40mA. Kirchhoff's Laws were discussed as a method for analyzing current flow through the circuit, emphasizing the importance of understanding series and parallel configurations.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of series and parallel resistor combinations
  • Familiarity with Kirchhoff's Laws for circuit analysis
  • Basic concepts of electrical potential and current flow
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  • Learn how to apply Kirchhoff's Voltage and Current Laws in circuit analysis
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Callix
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Homework Statement


Assume the electric potential is zero at the negative terminal of the battery. Calculate:

a). the equivalent resistance of the circuit
b). the electric potential at position a
c). the magnitude and direction of the current through each resistor

Circuit.png


Homework Equations

The Attempt at a Solution


a). I said that the whole equivalent resistance of the circuit is 3000\Omega. I combined the two on the right in series, then the 1200 with that in parallel. Then I added 1500 and 900 to that and got 3000.

b). Since the source voltage is 30V and the equivalent resistance is 3000\Omega, then the current is 30/3000=0.01 A. So then I said that the voltage at point a is 0.01A*900\Omega= 9V

c). I wasn't sure how to go about this one.. Does this use Kirchoff's law?
 
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Callix said:
I said that the whole equivalent resistance of the circuit is 3000
03A9.png?rev=2.5.png
That can't be correct, because there is a path between the battery terminals with resistance of only 1500, so the equivalent resistance must be less than that.

The 1500 is in parallel with the other branches not containing the cell, not in series.

It may help to redraw the diagram, rearranging it to put parallel wires next to one another and series wires above and below one another.
 
andrewkirk said:
That can't be correct, because there is a path between the battery terminals with resistance of only 1500, so the equivalent resistance must be less than that.

The 1500 is in parallel with the other branches not containing the cell, not in series.

It may help to redraw the diagram, rearranging it to put parallel wires next to one another and series wires above and below one another.

Would it just be 1500 then? I'm a bit confused then about how to calculate it.. The two on the far right are in series so you add them and then next is the two that are on the right in parallel, so you would add the reciprocals. Then..? =/
 
Callix said:
The two on the far right are in series so you add them and then next is the two that are on the right in parallel, so you would add the reciprocals. Then..? =/

You add those two together and that combination is in parallel with the 1200 ##\Omega## resistor. So figure out the value and redraw the circuit with that equivalent resistor. You will then see that it is not in parallel with the 900 ##\Omega## resistor.

b). Since the source voltage is 30V and the equivalent resistance is 3000\Omega, then the current is 30/3000=0.01 A. So then I said that the voltage at point a is 0.01A*900\Omega= 9V

Even if 0.01 A were the current through the equivalent resistance, the current through the 900 ##\Omega## resistor would not be 0.01 A.
 
Mister T said:
You add those two together and that combination is in parallel with the 1200 ##\Omega## resistor. So figure out the value and redraw the circuit with that equivalent resistor. You will then see that it is not in parallel with the 900 ##\Omega## resistor.
Even if 0.01 A were the current through the equivalent resistance, the current through the 900 ##\Omega## resistor would not be 0.01 A.

Circuit2.jpg

Is this correct so far?
 
Callix said:
Circuit2.jpg

Is this correct so far?
Yes.
 
SammyS said:
Yes.

Okay, so now aren't these in series then? So you would just add them normally?
 
Callix said:
Okay, so now aren't these in series then? So you would just add them normally?
Two resistors are in series with each other, and in parallel with another...
 
krebs said:
Two resistors are in series with each other, and in parallel with another...

Okay, so the two that are in series on the two on the right, correct? And then in parallel would be the two ends?
 
  • #10
Callix said:
Okay, so the two that are in series on the two on the right, correct? And then in parallel would be the two ends?
Yep, exactly. Use the same formulas you used to condense the right side to condense the entire circuit. ##\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_n...}## is one of them
 
  • #11
krebs said:
Yep, exactly. Use the same formulas you used to condense the right side to condense the entire circuit. ##\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_n...}## is one of them

Ok! So I got that the equivalence resistance is 750 ohms. So now would I take 30V/750ohms to find the current?
 
  • #12
Callix said:
Ok! So I got that the equivalence resistance is 750 ohms. So now would I take 30V/750ohms to find the current?
Correct. Then you need Kirchhoff. Start with your 40 mA total, and reconstruct your circuit piece by piece to calculate the current and voltage through each resistor.
 
  • #13
krebs said:
Correct. Then you need Kirchhoff.

Okay, so if I look at the joint above the voltage source.. there is a current entering from the bottom and current leaving on both the left and right sides. So
I1-I2-I3=0

At join 2 (to the right of the resistor of the top), there is current entering from the left, and leaving out the right and bottom,so
I2-I3-I4=0

Is this correct?
 
Last edited:
  • #14
Callix said:
Okay, so if I look at the joint above the voltage source.. there is a current entering from the bottom and current leaving on both the left and right sides. So
I1-I2-I3=0

At join 2 (to the right of the resistor of the top), there is current entering from the left, and leaving out the right and bottom,so
I2-I3-I4=0

Is this correct?
Yes.
 
  • #15
krebs said:
Yes.

Will I have to each of the 4 joins since so far I only have 2 equations and 4 unknowns? Won't I end up with some more unknowns?
 
  • #16
Callix said:
Will I have to each of the 4 joins since so far I only have 2 equations and 4 unknowns?
Yes, there are 4 junctions. You will end up seeing that several of the unknowns are equal to each other. Also the ratio of ##\frac{l_2}{l_1} = \frac{\text{Resistance of the left side of the circuit}}{\text{Total resistance of the circuit}}##
 
  • #17
Callix said:
Okay, so if I look at the joint above the voltage source.. there is a current entering from the bottom and current leaving on both the left and right sides. So
I1-I2-I3=0

At join 2 (to the right of the resistor of the top),

If you look at the circuit you drew in Post #5 there is no "Joint 2".

There is no need to use Kirchhoff's Laws to analyze this circuit. Just look at the equivalent circuits you constructed on your way towards the one you drew in Post #5 and use the facts that the current through resistors in series is the same, and the voltage across resistors in parallel is the same.

For example, if you look at the 1500 ##\Omega## resistor in the original circuit you can see that the voltage across it is 30 V (because it's in parallel with the battery). You can therefore immediately find the current through it.

Alternatively, you could use Kirchhoff's Laws to analyze the original circuit. You would have four unknowns and four equations.

Or use Kirchhoff's Laws to analyze any of the equivalent circuits you drew, such as the one in Post #5.
 

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