Posted by Eileen on Saturday, April 12, 2008 at 5:29pm.
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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?
- Circuits Problem (Physics) - Damon, Saturday, April 12, 2008 at 6:11pm
Sorry, can not open your image :(
- Circuits Problem (Physics) - Damon, Saturday, April 12, 2008 at 7:33pm
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)
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
12 = 18 i1
i1 = 2/3
12 = 10(2/3) + 6 i2
12 = 20/3 + 6 i2
36 = 20 + 18 i2
16 = 18 i2
i2 = 8/9
- Circuits Problem (Physics) - Damon, Saturday, April 12, 2008 at 7:35pm
i1+i2 = 2/3 + 8/9 = 14/9
times the resistance which is 6 = 28/3 volts a above b
- Circuits Problem (Physics) - Damon, Saturday, April 12, 2008 at 7:39pm
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
- Circuits Problem (Physics) - Damon, Saturday, April 12, 2008 at 7:44pm
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
- Circuits Problem (Physics) - Eileen, Saturday, April 12, 2008 at 9:31pm
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