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October 21, 2014

October 21, 2014

Posted by **Eileen** on Saturday, April 12, 2008 at 5:29pm.

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 6:11pmSorry, can not open your image :(

- Circuits Problem (Physics) -
**Eileen**, Saturday, April 12, 2008 at 6:17pmhi,

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!).

- Circuits Problem (Physics) -
- Circuits Problem (Physics) -
**Damon**, Saturday, April 12, 2008 at 7:33pmI 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

- Circuits Problem (Physics) -
**Damon**, Saturday, April 12, 2008 at 7:35pmb)

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 7:39pm2/3 = 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 7:44pmat 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 9:31pmomg! thankss!!

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