Finding the current in a resistor

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The discussion focuses on calculating the current in a resistor circuit with both series and parallel components. The total equivalent resistance (Req) is determined to be 6 ohms, leading to a total current of 2A using Ohm's Law (I = V/R). The current splits between the parallel resistors, with 1.33A flowing through one branch and 0.67A through the other, confirming the total current remains 2A. Voltage across the main branch is calculated as 4V, consistent with the overall circuit voltage of 12V. The calculations validate the initial assumptions and provide a clear understanding of current distribution in the circuit.
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Homework Statement


I have the picture attached.




Homework Equations


I = V/R


The Attempt at a Solution



I labelled the 2ohm and the 1ohm resisters in series (R1 and R2). Labelled 5ohm and 1 ohm in series R3, R4. So i found the Req

So R1 and R2 are in series so add them 2+1 = 3ohm. And R3 and R4 are in series so 5+1 = 6ohm. Then I found Req using 1/Req =
1/3ohm + 1/6ohm = 3/6 ohm, then inverse to get Req = 2ohm. And the final resistor is in series so add 4ohm + 2 ohm = 6 ohm in total for the circuit.

Then I did I=V/R

I = 12v/6ohm = 2A.

So that's the current for the entire circuit. How do I Find the current of the 2ohm resistor, when its going up the path.. does the 2A just split into the 2ohm and 5ohm resistors so it's just 1ohm going to the 2ohm resistor?
 

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Workout said:
So that's the current for the entire circuit. How do I Find the current of the 2ohm resistor, when its going up the path.. does the 2A just split into the 2ohm and 5ohm resistors so it's just 1ohm going to the 2ohm resistor?
The 2A is what flows through the 4Ω resistor and into the parallel resistors.
How do you determine the current through parallel resistors?
 
http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.example.parallel.html
this may be of help :)
 
Last edited:
More current will flow to the branch with less resistance in a parallel circuit.

Total Current 2A

simplify the parallel branch as 3ohms vs 6 ohms, as (2+1) in series and (5+1) in series.

voltage across the main branch = 2A x 2ohms = 4V
because total resistance of the parallel branch is 2ohms.

reality check when the current is coming out from the parallel branch = 2A x 4ohms = 8V

yes, 8+4= supply 12V

Now apply ohm's law again,

current for the parallel branch 1= 4V / (2+1ohms) = 1.33A
current for the parallel branch 2= 4V / (5+1ohms) = 0.67A

Reality check = 1.33 + 0.67 = 2A
Reality check = 8V / 4ohms = 2A
yes, 2A is your total current. The answer also proved that my first quote is correct.
 
Last edited:
Thank you!
 

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