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Diagram: http://gyazo.com/0b719e6d399ad894d2c48ce06d994672
In the figure the resistances are R1 = 0.77 Ω and R2 = 1.8 Ω, and the ideal batteries have emfs ε1 = 2.1 V, and ε2 = ε3 = 3.5 V. What are the (a) current (in A) in battery 1, (b) the current (in A) in battery 2 and (c) the current (in A) in battery 3? (d) What is the potential difference Va - Vb (in V)?
V=IR
Kirchoff's Loop Rules
Okay, for some reason I solved a part of this or two before and hoped to get back to it.
For part (a), I set up the equation of I being the current and R being the resistance; B being battery:
I= B2-B1/(4R1+R2)
And I got the answer .287A and then multiplied that by 2 and got the first part which is .587A. I understand the logic of its the potential difference divided by the equivalent resistance but why is that multiplied by 2, any reason for that? I didn't understand.
For parts (b)+ (c), I tried to set it up in a similar way but they are both wrong:
For part (b): B2-B1/(4R1+R2) and I got .287A
For part (c): I did the same equation as above putting in B3 instead of B2 and got .287APart (d) I was able to do by finding the potential difference between points (a) and (b) which I took the .287A value that I calculated in (b) and (c) and set it up like this:
ΔV=-I2R2^2+(3.5V) which comes out to 2.98V which was correct. I could use some help with this problem, I understand how the equation to solve B is found out, but I don't understand why you would have the EQ resistance to find the first battery, rather than the resistance the battery goes through. Could use a nice explanation on that. Thanks!
If someone could help with doing it the Kirchhoff rule way, I tried to assign currents with directions but for some reason I get something completely different from the correct answer I got in part (a).
I have a test on this material Tuesday and I REALLY need to do well, I blanked out on the first exam and forgot everything and did really, really bad .
I really appreciate it! Thank you!
Homework Statement
In the figure the resistances are R1 = 0.77 Ω and R2 = 1.8 Ω, and the ideal batteries have emfs ε1 = 2.1 V, and ε2 = ε3 = 3.5 V. What are the (a) current (in A) in battery 1, (b) the current (in A) in battery 2 and (c) the current (in A) in battery 3? (d) What is the potential difference Va - Vb (in V)?
Homework Equations
V=IR
Kirchoff's Loop Rules
The Attempt at a Solution
Okay, for some reason I solved a part of this or two before and hoped to get back to it.
For part (a), I set up the equation of I being the current and R being the resistance; B being battery:
I= B2-B1/(4R1+R2)
And I got the answer .287A and then multiplied that by 2 and got the first part which is .587A. I understand the logic of its the potential difference divided by the equivalent resistance but why is that multiplied by 2, any reason for that? I didn't understand.
For parts (b)+ (c), I tried to set it up in a similar way but they are both wrong:
For part (b): B2-B1/(4R1+R2) and I got .287A
For part (c): I did the same equation as above putting in B3 instead of B2 and got .287APart (d) I was able to do by finding the potential difference between points (a) and (b) which I took the .287A value that I calculated in (b) and (c) and set it up like this:
ΔV=-I2R2^2+(3.5V) which comes out to 2.98V which was correct. I could use some help with this problem, I understand how the equation to solve B is found out, but I don't understand why you would have the EQ resistance to find the first battery, rather than the resistance the battery goes through. Could use a nice explanation on that. Thanks!
If someone could help with doing it the Kirchhoff rule way, I tried to assign currents with directions but for some reason I get something completely different from the correct answer I got in part (a).
I have a test on this material Tuesday and I REALLY need to do well, I blanked out on the first exam and forgot everything and did really, really bad .
I really appreciate it! Thank you!
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