DC Circuit trouble

[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.

[Tom Mattson]

Originally posted by TheMadCapBeta
[B]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

[TheMadCapBeta]

I knew I was forgetting something. Thanks.