Capacitive Reactance Capacitors in AC Circuits Capacitance l

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Capacitive Reactance

Capacitive Reactance

Capacitors in AC Circuits

Capacitors in AC Circuits

Capacitance l Capacitance also causes opposition to current flow. l Blocks DC.

Capacitance l Capacitance also causes opposition to current flow. l Blocks DC.

Capacitance l The farad (symbol: F) is the SI unit of capacitance. l It

Capacitance l The farad (symbol: F) is the SI unit of capacitance. l It is named after the British physicist Michael Faraday.

Voltage Current Capacitor in DC

Voltage Current Capacitor in DC

ICE

ICE

I C E Current Capacitance Voltage Current leads Voltage in a Capacitive Circuit

I C E Current Capacitance Voltage Current leads Voltage in a Capacitive Circuit

Voltage Current Capacitor in AC

Voltage Current Capacitor in AC

Voltage V = Vpsin(2πft) Current I = C dv/dt Vp-Peak Voltage 2π-Cycle I =

Voltage V = Vpsin(2πft) Current I = C dv/dt Vp-Peak Voltage 2π-Cycle I = C Vp(2πf)cos(2πft) f-frequency(Hz) t-time(seconds) Ip = Vp(2πf)C

Ip = Vp(2πf)C Ip = Vp/R 1/R = 2πf. C R = 1/(2πf. C)

Ip = Vp(2πf)C Ip = Vp/R 1/R = 2πf. C R = 1/(2πf. C) Capacitive Reactance Xc = 1/(2πf. C) C is an Active Component

Capacitance in AC Circuits

Capacitance in AC Circuits

Capacitive Reactance l The opposition to current flow caused by a capacitor is known

Capacitive Reactance l The opposition to current flow caused by a capacitor is known as capacitive reactance. l Capacitive reactance, as it is opposition to current flow, is measured in Ohms Ω. l The symbol for capacitive reactance is XC. l The formula is:

Capacitor Values 1 m. F = 1 X 10 -3 F 1μF = 1

Capacitor Values 1 m. F = 1 X 10 -3 F 1μF = 1 X 10 -6 F 1 n. F = 1 X 10 -9 F 1 p. F = 1 X 10 -12 F

Find the current flowing in a circuit when a 4μF capacitor is connected across

Find the current flowing in a circuit when a 4μF capacitor is connected across a 880 v, 2000 Hz power supply. C = 4μF = 4 X 10 -6 F Vp = 880 V f = 2000 Hz Find: Ip

Xc = 1/(2πf. C) Xc = 1/(2π(2000 Hz)(4 x 10 -6 F) Xc =

Xc = 1/(2πf. C) Xc = 1/(2π(2000 Hz)(4 x 10 -6 F) Xc = 19. 9Ω Ip = Vp/Xc Ip = 880 v/19. 9Ω Ip = 44. 2 A