Ohhhh oops, I forgot to add the voltage from the battery... This isn't intuitive to me at all. So Δ5 across the 2Ω resistor is 10V.
9V - 10V = -1 V
-1 V / 3 Ω = -1/3 A
Thank you so much!
Okay. I'm still unsure of how to do this.
I split the circuit into the loop on the left. Using the fact that the current through the 2ohm resistor is 5A, I found that the change in potential at this point is 10V. I know that the total change in potential must be 0, so the change in potential...
Homework Statement
The ammeter reads 5.0A. Find I1, I2, and ε
Homework Equations
ΔV=IR
The Attempt at a Solution
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Homework Statement
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Homework Statement
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