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Consider a circuit where 2 resistors are connected in parallel

  • Thread starter Oerg
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This is a conceptual problem

Consider a circuit where 2 resistors are connected in parallel, which are in turn connected to another resistor in series. Let the resistance of the two resistors in parallel be [tex]R_1[/tex] and [tex]R_2[/tex]. The resistance of the resistor in series is [tex]R_3[/tex]

Normally we would add the resistance like this to find the total resistance

[tex] R_{total}=R_3+\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}} [/tex]

however, this works out different when i derive the resistance of the circuit anew from kirchoffs law which would give

[tex] R_{total}=\frac{1}{\frac{1}{R_1+R_3}+\frac{1}{R_2+R_3}} [/tex]

am i missing something? or is it wrong to apply the resistance formula in the case where there is another resistance connected in series?
 

Answers and Replies

ehild
Homework Helper
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Re: Resistor

Show your work. How did you get your formula? Actually, the formula for parallel and series resistors are derived from Kirchhoff's laws, you should get the same result.

ehild
 
345
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Re: Resistor

well if you simplify the expressions you would not get the same result, it is immediately apparent that there will be a [tex]{{R_3}^2}[/tex] term in the second expression but not in the first.

The second expression was obtained by considering 2 different loops and combining the resulting current, same as you would for the derivation of the formula for the combination of resistors in parallel.
 
Last edited:
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219
Re: Resistor

Your 2 loops are: +side of voltage source -> R3 -> R1 -> - side of voltage source
and
+side voltage source -> R3 -> R2 -> - side of voltage source ?

both loops contain R3. The voltage across R3 depends on the currents of both loops.
The voltage across R3 is not equal to R3*(current in one of the loops)

If 2 loops share a voltage source, it isn't a problem, because the voltage across it is
always the same.
 
ehild
Homework Helper
15,361
1,778
Re: Resistor

The picture shows your circuit.
According to Kirchhoff''s Current Law, I3=I1+I2.
According to the loop Law,
UBC+UCA=UBA, that is
E-I3R3-I1R1=0
and E-I3R3-I2R2=0.
The total current is I3, the total voltage is E, and the resultant resistance between A and B is
RAB=E/I3, and it is the same as the first formula.

ehild
 
Last edited:
345
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Re: Resistor

Ahhh, you even have a diagram, I am so touched. Thanks for your effort on the forums.

I understand now, in my loop, I had assumed that the same current run throughs both resistor and that I could get the resultant current by adding up the current from the other loop. This is of course not equivalent.
 

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