Asked by Me and I
A 10.0 Ω resistor is in parallel with a 20.0 Ω resistor and supplied by a 9.0 V battery. What is the current passing through the 20.0 Ω resistor?
(Points : 1)
0.45 A
0.90 A
1.35 A <---
6.00 A
(Points : 1)
0.45 A
0.90 A
1.35 A <---
6.00 A
Answers
Answered by
Damon
R = 10*20 /30 = 20/3
i = 9 /(20/3) = 27/20 = 1.35 Amps
Yes !
i = 9 /(20/3) = 27/20 = 1.35 Amps
Yes !
Answered by
Me and I
thank you, i actually understood how to do this from your last post on my other question.
Answered by
Damon
You are very welcome :)
Answered by
Damon
If you have a whole bunch of resistors in parallel you can do
1/R = 1/R1+ 1/R2 + 1/R3 + 1/R4 ....
Since the current has a choice which resistor to go through, you are adding CONDUCTIVITIES
(inverse of resistance)
C = C1+C2+C3 ...
then in the end
R = 1/C
or you can do them consecutively
R12 = R1 R2 /(R1+R2)
R123 = R12 R3 /(R3 + R12)
R1234 = R123 R4 / (R4 +R123)
etc
1/R = 1/R1+ 1/R2 + 1/R3 + 1/R4 ....
Since the current has a choice which resistor to go through, you are adding CONDUCTIVITIES
(inverse of resistance)
C = C1+C2+C3 ...
then in the end
R = 1/C
or you can do them consecutively
R12 = R1 R2 /(R1+R2)
R123 = R12 R3 /(R3 + R12)
R1234 = R123 R4 / (R4 +R123)
etc
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