how much power is consumed in (1)purely inductive and (2)purely capacitive a.c. circuit?
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No real power is consumed on an average during a cycle in a pure inductive or pure capacitive AC circuit.
During one part of the cycle, inductance stores energy coming in through the current, and during another part of the cycle inductance releases energy back to the circuit.
During one part of cycle the capacitor charges (as current comes in) to maximum charge and maximum potential, then during another part of the cycle (AC), the charge is returned to the circuit in the form of current.
Both these elements store and release power during 1 cycle of AC current.
Real power is consumed only in circuits with resistance, dissipated only through the resistance.
During one part of the cycle, inductance stores energy coming in through the current, and during another part of the cycle inductance releases energy back to the circuit.
During one part of cycle the capacitor charges (as current comes in) to maximum charge and maximum potential, then during another part of the cycle (AC), the charge is returned to the circuit in the form of current.
Both these elements store and release power during 1 cycle of AC current.
Real power is consumed only in circuits with resistance, dissipated only through the resistance.
kvnmurty:
sorry I did not answer completely. with wave form equations
Inductance: X_L = w L
V = Vm Sin wt ; ; ; I = Vm/(wL) Sin (wt - pi/2) ;;;; Im = Vm / X_L
Power (t) = V (t) * I(t) = (Vm * Im/2) Sin 2wt = Vrms * I_rms Sin 2wt
Its Time period = pi/w. Power average over a cycle = 0 , as it is a sine wave.
But average Power absorbed during T/2 = pi/2w is Vm Im / pi
The same amount is released during the next T/2.
V = Vm Sin wt ; ; ; I = Vm/(wL) Sin (wt - pi/2) ;;;; Im = Vm / X_L
Power (t) = V (t) * I(t) = (Vm * Im/2) Sin 2wt = Vrms * I_rms Sin 2wt
Its Time period = pi/w. Power average over a cycle = 0 , as it is a sine wave.
But average Power absorbed during T/2 = pi/2w is Vm Im / pi
The same amount is released during the next T/2.
Capacitance: X_c = 1/wc
V = Vm Sin wt ;;; I = Im Cos wt ;;; Im = Vm *wc ;; let T = pi/w
Power(t) = (Vm Im / 2) Sin 2wt => Average over T= pi/w is 0
Average over T/2 , = pi/2w is Vm im / pi
This amount is absorbed and released alternately during T/2.
V = Vm Sin wt ;;; I = Im Cos wt ;;; Im = Vm *wc ;; let T = pi/w
Power(t) = (Vm Im / 2) Sin 2wt => Average over T= pi/w is 0
Average over T/2 , = pi/2w is Vm im / pi
This amount is absorbed and released alternately during T/2.
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