Anonymous

# Physics Problem?

Update:

Thank you

### 1 Answer

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- billrussell42Lv 72 months agoFavorite Answer
replace the two batteries with one of 66 volts with Ri = 4Ω, oriented same as the 80 volt one

the 3 r's in parallel equal

1/R = 1/6 + 1/12 + 1/4

1/R = 2/12 + 1/12 + 3/12

R = 12/6 = 2 Ω

that is in series with 5+4, so total R is 11 Ω

total I = 66/11 = 6 amps

power in 5Ω = I²R = 36x5 = 180 watts

voltage across the parallel combo is

V = IR = 6x2 = 12 volts

current thru the 6 Ω is

I = E/R = 12/6 = 2 amps

drop across the two Ri's is 6x3 and 6x1 or 18 and 6 volts

battery voltages are therefore 80–18 = 62 and 14+6 = 20 volts

note that the 14 volt battery is actually being charged by the larger one

as a check, voltages around the loop are

62 – 5•6 – 20 – 12

62 –30 – 32

0

ok

note that I made a change to your drawing to avoid ambiguities.

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