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Hi all, I'm new here, been havin' a good ol' tussle with this one. Has me frustrated because it seems pretty elementary.
The circuit diagram has been drawn out in the attachment
The problem asks what is the current through the 20 ohm resistor.
Parallel: 1/Req=1/R1+1/R2+...1/RN
Series: Req=R1+R2+...RN
I=V/R
I've added the 10 and 5 ohm resistors to make 10/3 ohm to simplify that area of the diagram.
I've also added the 20 and 5 ohm resistors on the bottom end of the diagram to make 25 ohm.
Thus, adding these two equivalent resistances gives me 85/3. This leaves the resistor to the left of the battery.
Adding that one to this value as per the series rule of equivalent resistances gives me 115/3.
Using this equivalent resistance, I can find the total current, which comes out to be
I=25/(115/3)=75/115=.652 A.
From here, I am at a loss as to how to use this value to calculate the specific current at the 20 ohm resistor.
Am I even using the proper voltage to calculate the total current?
The correct answer to this problem is 227 mA.
Thanks, any help would be appreciated.
Homework Statement
The circuit diagram has been drawn out in the attachment
The problem asks what is the current through the 20 ohm resistor.
Homework Equations
Parallel: 1/Req=1/R1+1/R2+...1/RN
Series: Req=R1+R2+...RN
I=V/R
The Attempt at a Solution
I've added the 10 and 5 ohm resistors to make 10/3 ohm to simplify that area of the diagram.
I've also added the 20 and 5 ohm resistors on the bottom end of the diagram to make 25 ohm.
Thus, adding these two equivalent resistances gives me 85/3. This leaves the resistor to the left of the battery.
Adding that one to this value as per the series rule of equivalent resistances gives me 115/3.
Using this equivalent resistance, I can find the total current, which comes out to be
I=25/(115/3)=75/115=.652 A.
From here, I am at a loss as to how to use this value to calculate the specific current at the 20 ohm resistor.
Am I even using the proper voltage to calculate the total current?
The correct answer to this problem is 227 mA.
Thanks, any help would be appreciated.
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