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A parallel plate capacitor has a capacitance C=200 pF.it is connected to 230 V AC supply with an angular frequency 300 rad/s.the rms value of conduction current in the circuit and displacement current in the capacitor respectively are :
Asked by dileepkumardasari665 | 08 Feb, 2024, 08:30: AM
answered-by-expert Expert Answer

Displacement current id is given by

img ............................(1)

where A is area of plates of capcitor , εo is permittivity of free space , d is distance between plates of capacitor

and V is rms voltage of alternating current .

if angular frquency ω of alternating current is 300 rad/s, then  rms value (dV/dt) is calculated as follows

img

RMS vaule of (dV)/dt = 230 × 300  V/s = 69000 V/s

Capacitance C = img  = 200 pF

Hence by substituting capacitance and rms value of dV/dt in eqn.(1) , we get

img

---------------------------------------------------

RMS value of onduction current img is

img

where z is impedence of the circuit

img

Conduction current img = 230 / ( 1.667 × 107 )  A = 1.38 × 10-5 A

--------------------------------------

Conduction current = displacement current

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