# 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.

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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