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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

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