Current through 6 resistors in series and parallel

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Homework Help Overview

The discussion revolves around a circuit problem involving six resistors arranged in series and parallel configurations, with all resistors having the same resistance of 2.00 kilo-Ohm and a voltage supply of 12V. The original poster seeks to calculate the current through specific resistors, R5 and R6, but encounters difficulties in applying Ohm's law and understanding voltage drops across the resistors.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to find equivalent resistance before applying Ohm's law. There are attempts to calculate total resistance and voltage drops, with some questioning the assumptions about voltage across R5 and R6.

Discussion Status

The discussion is ongoing, with participants providing guidance on calculating equivalent resistance and clarifying the relationship between voltage and current in the circuit. There is recognition of misunderstandings regarding voltage drops across specific resistors, particularly R5.

Contextual Notes

Participants note the complexity of the circuit and the importance of correctly identifying voltage drops, as well as the need for accurate calculations of equivalent resistance. There is an acknowledgment of potential errors in previous calculations of resistance values.

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Homework Statement


See attached figure.
All resistors have the same resistance of 2.00kilo-Ohm. Assume V = 12V. Calculate the current through resistor R6 and resistor R5.

Homework Equations


V = IR
I = V/R
R total (series) = R1 + R2...etc
1/R total (parallel) = 1/R1 + 1/R2...etc


The Attempt at a Solution


Since the current would travel towards A, it would first go through R5 (in parallel with R1 and R3), and I = V/R so for R5, I = 12/2000 = .006 A
This is wrong and I don't understand why.
Then for R6, the total resistance for R1 - R5 is 10,000 ohm, so the voltage would drop 10,000*the current, but I don't know how to find the original current.
I'm completely lost. I redrew the figure, but still can't solve.
 

Attachments

  • resistor.gif
    resistor.gif
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You can't just plug in any voltage that happens to be in the problem into Ohm's law. The current through R5 is equal to the voltage across R5 divided by R5, so you need to find that voltage first. The only voltage you know right now is the voltage across the battery.

Start by combining all the resistors into one equivalent resistance.
 
Using series and parallel combination find the resistance across AC. Then find the total resistance of the circuit.
 
I found the resistance across AC to be 1999 ohm, and the total resistance to be 3999 ohm.

There is no voltage drop before the current hits R5, correct? So wouldn't the voltage through R5 still be 12 V? This would make the current .006 A, which is wrong, so I'm not sure what I'm doing wrong.

Also, I don't know how to calculate the voltage drop over AC because I don't know the current?
 
I found the resistance across AC to be 1999 ohm
Check your calculation for this part.
 
R1 and R2 in series= R1 + R2 = 4000 ohm
1/ (R1&2) and R3 in parallel = 1/4000 + 1/2000 = 1333 ohm
(R1,2,3) in series with R4 = 1333 + 2000 = 3333 ohm
1/ (R1,2,3,4) in parallel with R5 = 1/3333 + 1/2000 = 1250 ohm

So total resistance (R1,2,3,4,5) + R6 in series = 1350 ohm
 
sorry I meant total resistance is 3250 ohm
 
The battery is connected across this equivalent resistance, so you know the voltage drop across the resistance. Use Ohm's Law to calculate the current flowing out of the battery and through the circuit.
 
Ok, I understand how to get the current for R6 by calculating the voltage drop from A to C and using the leftover voltage to find the current for R6.

But, I still don't understand why I'm wrong on the current for R5.
R5 gets all 12 volts, and R5 has a resistance of 2000 ohm, so the current should be .006, but this is wrong and I don't understand why?
 
  • #10
R5 doesn't have a 12-volt drop across it. The voltage drop across it is the difference in potential between points A and C.
 
  • #11
Ah, I see. Thank you so much for your help, I really appreciate it!
 

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