10. Reactance and Angular Velocity - Alternating Currents
by M. Bourne
Alternating Currents
An alternating current is created by rotating a coil of wire through a magnetic field. If the angular velocity of the wire is ω, the
capacitive reactance is given by:
`X_C=1/(omegaC)`
inductive reactance is given by:
XL = ωL
Example
If `R = 10\ Ω`, `L = 0.6\ "H"`, `C = 200\ mu "F"` and ` ω = 50\ "rad/s"`, find the magnitude of the impedance and the phase difference between the current and the voltage.
Answer
Recall: `μ` (micro) means `10^-6`.
Capacitive Reactance:
`X_C=1/(omegaC)`
`=1/((50)(200xx10^-6))`
`=100\ Omega`
Inductive Reactance `X_L= ωL = 50 × 0.6 = 30\ Ω`
`X_L − X_C= 30 - 100 = -70\ Ω`
`Z = 10 - 70j\ Ω` in rectangular form.
Using calculator, `|Z| = 70.71\ Ω` and ` θ = -81.87^@`.
So `Z = 70.71\ ∠\ -81.87^@\ Ω`
So the magnitude of the impedance is `70.71\ Ω` and the voltage lags the current by `81.87^@` (this is the phase difference).
Angular Velocity and Frequency
In this section, recall that angular velocity and frequency are related by the expression:
ω = 2πf
