In an AC circuit, the amplitudes of the voltages across the components and the current in the circuit are treated as vectors. The phase difference between the voltage and current is the angle between these vectors.
Let the amplitude of the current flowing in the circuit be i.
Then the magnitude of the voltage vector of the resistance is VR = iR ……. (i).
Since the voltage across the resistance and the current are always in phase, the vectors of voltage of resistance (VR) and current (i) are parallel.
In an AC circuit, capacitance acts as resistance to the flow of circuit called the reactance of the circuit (Xc).
The value of the capacitive reactance is XC=1/ωC ….. (ii).
where, ω is the angular frequency of the source and C is the capacitance.
The magnitude of the voltage vector of the capacitance is VC=iXC ……(iii).
There is a phase difference of −π/2 between the capacitor voltage and the current.
That is the capacitor voltage lags behind the current by a phase of π/2. Therefore, the vector of capacitor voltage (Vc) is at an angle of −π/2 from the current vector.
Therefore, the resultant or the net voltage in the circuit is equal to Vnet = VR+VC.
(Note that here, VR and VC are vectors and thus Vnet is also a vector.)
Therefore, magnitude of net voltage is Vnet = √V2R+V2C−−−−−−−
Substitute the value of VR and VC from equations (i) and (iii).
⇒ Vnet = √(iR)2+(iXC)2
⇒ Vnet=i√R2+1/ω2C2
Here Vnet= V0.
⇒ I = V0/√R2+1ω2C2
This means that if the value of ω is increased, then the value of the current (i) will increase. Power dissipated through the resistance is given as P=i2R. Therefore, the power in the circuit will also increase. If the power of the bubs increases, it will glow brighter.