The average power dissipated in an A.C. circuit per cycle is given as:
P=12I0V0
But, I0=2(Irms)−−−−−−√ and
V0=2(Vrms)−−−−−−√∴P=12(2(Irms)−−−−−−√×2(Vrms)−−−−−−√)∴P=2IrmsVrms2∴P=IrmsVrms
According to Ohm’s law, V = IR
∴P=I2R=V2R
Power is always dissipated in a resistor. This because the current and the voltage are in phase. In an inductor, power dissipation is zero due to the fact that power is positive when energy is stored in the magnetic field of the coil and negative when the energy is given back to the voltage supply on the next part of the cycle.
In a capacitor, power dissipation is also zero due to the fact that power is positive when energy is stored in the electric field between the capacitor’s plates and negative when the energy is given back to the voltage supply.
In essence, power dissipation is zero both in an inductor and a capacitor because the voltage and current are out of phase by90°.
Therefore, in an a.c. circuit containing inductor, capacitor and a resistor in series, power is only dissipated in the resistor.
Hence, P=IV=I2R=V2R
Example 5:
A circuit consists of a resistor 500ohms and a capacitor of 5μF connected in series. If an alternating voltage of 10v and frequency 50 Hz is applied across the series circuit, calculate
(a) the reactance of the capacitor
(b) the current flowing in the circuit
(c) the voltage across the capacitor
(d) If the capacitor is replaced with an inductor of 150mH, calculate the impedance and voltage across the inductor
Solution:
(a) XC=12πfL=12π×50×5.0×10−6=636.62ohms
(b) Z=R2+X2C−−−−−−−√ since there is no L
Z=5002+636.622−−−−−−−−−−−−√=809.5ohmsI=VZ=10809.5A=12.35×10−3A
(c) VC=IXC=12.35×10−3×636.62=7.86V
(d) XL=ωL=2πfL=2×π×50×10−3ohms=47.12ohmsZ=R2+X2L−−−−−−−√=5002+47.122−−−−−−−−−−−√=502.2ohmsI=VZ=10502.2=19.9×10−3AV=IXL=19.9×10−3×47.12=938×10−3V
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