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1. Homework Statement
I have the picture attached.
2. Homework Equations
I = V/R
3. 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?
I have the picture attached.
2. Homework Equations
I = V/R
3. 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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