Current in a Complicated Circuit

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The circuit analysis involves determining the current through a 2-ohm resistor given that the current through an 8-ohm resistor is 0.5 A. Using Ohm's Law and the principles of series and parallel resistors, the current through the 8-ohm resistor leads to the calculation of the current through the 2-ohm resistor. The voltage across the 8-ohm resistor is calculated as 4 volts, and the total voltage from the battery is determined to be 19 volts. This results in a calculated current of 9.5 A through the 2-ohm resistor. The solution effectively illustrates the application of circuit analysis techniques to find unknown currents in complex circuits.
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Homework Statement


In the figure, the current of the circuit in the 8 ohm resistor is 0.5 A. What is the current in the 2 ohm resistor?

Figure http://img181.imageshack.us/img181/5576/helpfk7.jpg

I'm trying to solve for the current in a complicated circuit. I'm given R1 = 16ohms, R2 = 8ohms, R3 = 20ohms, R4 = 6ohms, and R5 = 2ohms. Also, the question for the problem states that the current in R2 = 0.5 A

Homework Equations


Ohms Law: Potential Difference = Current(Resistance)
Formula for Resistors in Series: Rt = R1 + R2 + R3...
Formula for Resistors in Paralell: 1/Rt = 1/R1 + 1/R2 + 1/R3


The Attempt at a Solution



I'm not sure how to go about finding the current through R5 without knowing the voltage of the battery. But I think I may be able to figure this out by trying to compare R2's current with the entire circuit.

Idea 1:
R1 is twice R2, and being in parallel, R1 receives half R2's current.
So R1 has a current of .25 A.
Then from here somehow figure out how to find the current in R5.

Idea 2:
Using Ohm's law I can find the voltage in R2 V=0.5(8), V=4v.
Then applying it to R5, 4v=I(2), I=8 A.
 
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You know the current through R2, it's given. You know the current through R1, as in your Idea 1. You thus know the total current through R3 (it's the sum of the currents through R1 and R2).

You can use Ohm's law to find the voltage across R2, and the voltage across R3. The sum of those two voltage drops is the voltage from the battery. From there, it's pretty straightforward.

Your idea 2 is incorrect, because the same voltage is not applied to R2 as to R5 -- these resistors have different nets at their right hand terminals! Instead, the voltage applied to the entire "subcircuit" of R1-3 is the same as that applied to the entire "subcircuit" of R4-5, because each of these "subcircuits" is wired in parallel to the battery.

- Warren
 
Alright so my reworked solution is as follows:

Current of R1 = 1/2 Current of R2 so, Current of R1 = .25 A
Current Through R3 = Current through R1 + R2 so, Current of R3 = .75 A
Voltage through R2 = .5(8) = 4v
Voltage through R3 = .75(20) = 15v
Battery Voltage = 4v+15v = 19v

V = IA
19v = I(2)
9.5 A = I
 
Cefari said:
Alright so my reworked solution is as follows:

Current of R1 = 1/2 Current of R2 so, Current of R1 = .25 A
Current Through R3 = Current through R1 + R2 so, Current of R3 = .75 A
Voltage through R2 = .5(8) = 4v
Voltage through R3 = .75(20) = 15v
Battery Voltage = 4v+15v = 19v

V = IA
19v = I(2)
9.5 A = I

That looks right. Good job.
 
Thank you both for the help. :approve:
 
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