# Circuits Problem (Physics)

h t t p ://img266.imageshack.us/my.php?image=circuit2ru6.png

Given the circuit above:
a) Solve for unknown currents.
b) What is the potential difference between points a and b?
c) If both batteries start with 20 A*h of charge, how long does it take the first of the batteries to run out? Which battery is it (left or right)?
d) If both 12V batteries are replaced by batteries of identical voltage, what is the smallest voltage which will result in integer currents throughout the circuit?

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1. Sorry, can not open your image :(

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posted by Damon
2. hi,
i put spaces between the "h t t p :" because it wouldnt let me post it otherwise. if you delete the spaces it should load (i hope!).

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posted by Eileen
3. I call the current coming out the top of the left battery i1
I call the current coming out the top of the right battery i2
Then the current headed down from a to b is (i1+i2) (Kirkhoff)
In the left loop
12 = 4 i1 + 6(i1+i2)
In the right loop
12 = 3 i2 + 6(i1+i2)
so
12 = 10 i1 + 6 i2
12 = 6 i1 + 9 i2
solve those two. I will multiply the first by three and the second by 2
36 = 30 i1 + 18 i2
24 = 12 i1 + 18 i2
subtract
12 = 18 i1
i1 = 2/3
then i2
12 = 10(2/3) + 6 i2
12 = 20/3 + 6 i2
36 = 20 + 18 i2
16 = 18 i2
i2 = 8/9

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posted by Damon
4. b)
i1+i2 = 2/3 + 8/9 = 14/9
times the resistance which is 6 = 28/3 volts a above b

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posted by Damon
5. 2/3 = 6/9 = i1
8/9 = i2
so i2>i1 so the one on the right will run out first
(8/9) t = 20 amp hr
t = 9*20/8 = 9*5/2 = 22.5 hr

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posted by Damon
6. at present i1 = 6/9 and i2 = 8/9
clearly if I multiply the voltages by 9 I will get 6 amps and 8 amps
then if I divide both voltages by two I will get 3 amps and 4 amps
so multiply the voltage by 9/2
(9/2)12 = 54 volts

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posted by Damon
7. omg! thankss!!

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posted by Eileen

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