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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?
7 answers
Sorry, can not open your image :(
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!).
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!).
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
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
b)
i1+i2 = 2/3 + 8/9 = 14/9
times the resistance which is 6 = 28/3 volts a above b
i1+i2 = 2/3 + 8/9 = 14/9
times the resistance which is 6 = 28/3 volts a above b
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
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
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
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
omg! thankss!!
3 answers