Electricity/Alternating current

Engineering

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AC (Alternating Current)

Electricity provides a sinusoidal time varying voltage over time

$v(t)=VSin\omega t$

Symbol

o---[~]---o

Alternating Current and conductor

Resistors

Voltage

$v(t)=i(t)Z_{R}$

Current

$i(t)={\frac {v(t)}{Z_{R}}}$

Power

$p(t)=i(t)v(t)$

Impedance

$Z_{R}={\frac {v(t)}{i(t)}}=R+X_{R}=R$

Reactance

$X_{R}=0$

Capacitors

Voltage

$v(t)={\frac {1}{C}}\int i(t)$

Current

$i(t)=C{\frac {d}{dt}}v(t)$

Power

$p(t)={\frac {1}{2}}Cv^{2}(t)$

Impedance

$Z_{C}={\frac {v_{C}(t)}{i_{C}(t)}}=R_{C}+X_{C}$

$Z_{C}=R+{\frac {1}{\omega C}}\angle -90^{o}=R+{\frac {1}{j\omega C}}=R+{\frac {1}{sC}}$

Reactance

$X_{C}={\frac {1}{\omega C}}\angle -90^{o}={\frac {1}{j\omega C}}={\frac {1}{sC}}$

Phase angle difference

$Tan\theta =\omega T$

Time constant

$T=RC$

$X_{R}=0$

Frequency respond

Low frequency . $\omega =0$ , $X_{C}={\frac {1}{\omega C}}=00$ . Capacitor open circuit

High frequency. $\omega =00$ , $X_{C}={\frac {1}{\omega C}}=0$ . Capacitor short circuit

Inductors

Voltage

$v(t)=L{\frac {d}{dt}}i(t)$

Current

$i(t)={\frac {1}{L}}\int v(t)dt$

Power

$p(t)={\frac {1}{2}}Li^{2}(t)$

Impedance

$Z_{L}={\frac {v_{L}(t)}{i_{L}(t)}}=R_{L}+X_{L}$

$Z_{L}=R+\omega L\angle 90^{o}=R+j\omega L=R+sL$

Reactance

$X_{L}=\omega L\angle 90^{o}=j\omega L=sL$

Phase angle difference

$Tan\theta =\omega T$

Time constant

$T={\frac {L}{R}}$

Frequency respond

Low frequency . $\omega =0$ , $X_{L}=\omega L=0$ . Inductor shorts circuit

High frequency. $\omega =00$ , $X_{L}=\omega L=00$ . Inductor opens circuit

See Also

• Capacitor
• Maxwell's Equations