Calculating Current in Parallel DC Circuit - Solving for I3

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The discussion revolves around calculating the current through a 12-ohm resistor in a parallel DC circuit with a total current of 2.0A. The user has derived two equations based on the circuit's configuration but is struggling to find the current through the 12-ohm resistor. A suggestion was made to apply Ohm's Law, noting that the voltage across the 12-ohm resistor is equal to the battery voltage, which has already been calculated as 8/3V. The user acknowledged this oversight and expressed gratitude for the guidance. The conversation emphasizes the importance of understanding voltage and resistance relationships in parallel circuits.
TheMadCapBeta
I'm having trouble solving this problem:

Three resistors(parallel) across a DC voltage source...If the total current through the circuit is I = 2.0A, what is the currect through the 12ohm resistor?

The circuit runs into a junction which splits 3 ways. Along each path there's are resistors 2ohm, 6ohm, and 12ohm, respectively, and all meet again andthen flows back to the battery.


It's parallel so I breaks up into I1, I2,I3
I found two equations:

8/3V = I1(2ohm) + I2(6ohm) + I3(12ohm)
I = I1 + I2 + I3

I found the voltage by first find the equivalent resistor for R1, R2, and R3, (R = 4/3 ohm) So V = IR => V = 2(4/3) => V = 8/3V.

I need to find the current through R3, and I could if I had one more equation, but I'm stuck. I can't see any other equations by loop rule or Kirchoff's law. So, any help will be appreciated, thanks.
 
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Originally posted by TheMadCapBeta
I'm having trouble solving this problem:

Three resistors(parallel) across a DC voltage source...If the total current through the circuit is I = 2.0A, what is the currect through the 12ohm resistor?

The circuit runs into a junction which splits 3 ways. Along each path there's are resistors 2ohm, 6ohm, and 12ohm, respectively, and all meet again andthen flows back to the battery.


It's parallel so I breaks up into I1, I2,I3
I found two equations:

8/3V = I1(2ohm) + I2(6ohm) + I3(12ohm)
I = I1 + I2 + I3

I found the voltage by first find the equivalent resistor for R1, R2, and R3, (R = 4/3 ohm) So V = IR => V = 2(4/3) => V = 8/3V.


Good start!

I need to find the current through R3, and I could if I had one more equation, but I'm stuck. I can't see any other equations by loop rule or Kirchoff's law. So, any help will be appreciated, thanks.

Two words: Ohm's Law

You know the voltage across the 12Ω resistor (it's the same as the voltage of the battery). You also know the resistance, so just use V=IR and you're home free.

edit: typo
 
I knew I was forgetting something. Thanks.
 
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