Consider a resistance R, an inductor of self inductance L and a capacitor of capacitance C connected in series across an AC source. The applied voltage is given by,
v = v0 sin ωt ......(1)
where, v is the instantaneous value, v0 is the peak value and ω = 2πf, f being the frequency of AC.
If i be the instantaneous current at time t, the instantaneous voltages across R, L and C are respectively iR, iXL and iXC. The vector sum of the voltage amplitudes across R, L, C equals the amplitude v0 of the voltage applied.
Let VR, VL and VC be the voltage amplitudes across R, L and C respectively and y I0 the current amplitude. Then VR = i0R is (VL – VC) in phase with i0.
VL = i0XL= i0(ωL) leads i0 by 90°
The current in a pure resistor is phase, with the voltage, whereas the current in a pure
VC = i0XC = i0(\(\frac{1}{\omega C}\)) lags behind i0 by 90°
The current in a pure resistor is phase with the voltage, whereas the current in a pure inductor lags the voltage by \(\frac{\pi}{2}\) rad. The current in a pure capacitor leads the voltage by \(\frac{\pi}{2}\) rad.
For vL > vC, phase angle Φ between the voltage and the current is positive.
From the right angled triangle OAP,
OP2 = OA2 + AP2
= OA2 + OB2(QAP = OB)
Where Z is the impedance of the circuit. Phase angle between v & i.