- #1
sirfederation
- 20
- 0
ok the question I have is why is the voltage between the 220 olm resistor and the 330 olm resistor 0 V.
All resistors in the image are in series so Is=I1=I2=I3
Is = 10/1020 (v/olm) = .0098 or 9.8 mA
V(subscript220)= 220*9.8 mA = 2.157 V (this is the voltage between the DC source and the 220 olm resistor)
V(subscript330) = 330 * 9.8 mA = 3.234 V (this is the voltage between the 330 olm and the 470 olm resistor)
V(subscript470) = 470 * 9.8 mA = 4.606 ( I have to add 3.234 to 4.606 to find the voltage between the DC source and 470 which is 7.84 V)
I am taking the ground up above the the 330 olm resistor.
Ok now I am not sure but I think the reason why the voltage across the 220 to the 330 olm resistor is zero is because I have to make a open circuit in that area which means that no current is actually passing through the wire since current takes the path of least resistence.
All resistors in the image are in series so Is=I1=I2=I3
Is = 10/1020 (v/olm) = .0098 or 9.8 mA
V(subscript220)= 220*9.8 mA = 2.157 V (this is the voltage between the DC source and the 220 olm resistor)
V(subscript330) = 330 * 9.8 mA = 3.234 V (this is the voltage between the 330 olm and the 470 olm resistor)
V(subscript470) = 470 * 9.8 mA = 4.606 ( I have to add 3.234 to 4.606 to find the voltage between the DC source and 470 which is 7.84 V)
I am taking the ground up above the the 330 olm resistor.
Ok now I am not sure but I think the reason why the voltage across the 220 to the 330 olm resistor is zero is because I have to make a open circuit in that area which means that no current is actually passing through the wire since current takes the path of least resistence.