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
R = 10*20 /30 = 20/3
i = 9 /(20/3) = 27/20 = 1.35 Amps
Yes !
thank you, i actually understood how to do this from your last post on my other question.
You are very welcome :)
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
To find the current passing through the 20.0 Ω resistor, we can use Ohm's Law (I = V/R) and the formulas for resistors in parallel.
First, let's calculate the equivalent resistance of the two resistors in parallel. The formula for resistors in parallel is given by:
1/Req = 1/R1 + 1/R2
Where Req is the equivalent resistance and R1 and R2 are the individual resistances.
Plugging in the values, we get:
1/Req = 1/10 + 1/20
1/Req = (2+1)/20
1/Req = 3/20
Now, let's calculate Req:
Req = 20/3 Ω
Next, we can use Ohm's Law to find the current passing through the 20.0 Ω resistor. The formula for Ohm's Law is given by:
I = V/R
Where I is the current, V is the voltage, and R is the resistance.
Plugging in the values, we get:
I = 9/20/3 A
I = 9/20 * 3/1 A
I = 27/20 A
I ≈ 1.35 A
Therefore, the current passing through the 20.0 Ω resistor is approximately 1.35 A.