• Power : Power is the rate at which energy is delivered by an AC source. It is given by :-

 \(P = VI\)

where \(V\)= Voltage 

\(I\) = Current 

For a purely resistive AC circuit,

Voltage is \( V_R = V_0 \,sin\) \(\omega\;t\)

Current is \(I_R = I_0\,sin \;\omega\;t\)

Instantaneous Power of a Purely Resistive Circuit

• Consider a purely resistive circuit, as shown in figure.
• The instantaneous power of a purely resistive circuit is given by \(P_R = I_R\;V_R\)

\(\because\;V_R=I_R\;R\)

\(\therefore\;P_R = I_R^2\;R\)

\(P_R = \left[I_0^2\;sin^2\;\omega t\right]R\;\;\;\;\;\;\;\;\left[I_R=I_0\;sin\;\omega t\right]\)

\(P_R = I_0^2R\;sin^2\;\omega t\)

Here, \(P_R,\;I_R\;and\; V_R\) are instantaneous values.

Average Power of a Purely Resistive Circuit

• Average power \((\overline P_R)\) is defined as the total energy dissipated in a given time.
• The instantaneous power can be written as 

\(P_R =\dfrac{I_0^2\;R}{2} (1-cos 2\;\omega t)\)                      

\(\left(\because cos \;2\theta = 1 - 2\; sin^2 \theta\right)\)

\(P_R =\dfrac{I_0^2\;R}{2} - \dfrac{I_0^2R}{2}cos 2 \;\omega t\)

\(cos 2\; \omega t\) is positive for half cycle and negative for another half cycle. Thus, its average is zero for one full cycle.

\(\therefore\;\overline{P}_R\;=\;\dfrac{I_0^2R}{2} - 0\)

\(\;\overline{P}_R\;=\;\dfrac{I_0^2R}{2} \)

Average Power of a Purely Resistive Circuit in terms of RMS Current and Voltage

\(\overline{P}_R = \left(\dfrac{I_0}{\sqrt2}\right)^2\;R\)

⇒   \(\overline{P}_R= I^2_{rms}\;R\)         \(\because\;I_{rms}\;=\;\dfrac{I_0}{\sqrt2}\)

⇒  \(\overline P_R =\dfrac{V^2_{rms}}{R}\)          \(\because V_{rms}= I_{rms} R\)

⇒ \(\overline P_R = I_{rms}\times V_{rms} \)

• Power : Power is the rate at which energy is delivered by an AC source. It is given by 

 \(P = VI\)

where \(V\) = Voltage 

\(I\) = Current 

• For a purely capacitive AC circuit, 

Voltage is \(V_C = V_0\, sin\; \omega t\)

Current is \(I_C = - \omega CV_0\,cos \; \omega t\)

Instantaneous Power of a Purely Capacitive Circuit 

• Consider a purely capacitive AC circuit, as shown in figure.
• The instantaneous power of a purely capacitive circuit is given by 

\(P_C = V_C I_C\)

\(\because V_C = V_0 \,sin \;\omega t \;\;\;and \;\;\; I_C = - \omega C V_0 \,cos \;\omega t\)

\(P_C = - ( V_0 \,sin \;\omega t)(\omega C V_0 \,cos \;\omega t)\)

\(P_C = - \omega C\;V^2_0\, sin \;\omega t \; cos \;\omega t\)

\(P_C = \dfrac{-1}{2} \omega CV_0^2 \; sin \; 2\omega t\)

Here, \(P_C, \; I_C\) & \(V_C\) are instantaneous values. 

Average Power of a Purely Capacitive Circuit 

• Average power \( (\overline P_C)\) is defined as the total energy dissipated in given time.
• The instantaneous power is \(P_C = \dfrac{-1}{2} \omega C V_0^2 sin \;2 \omega t\)
• \(sin \,2 \omega t\) is positive for half cycle and negative for another half cycle. Thus, its average is zero one full cycle.

\(\therefore \overline P_C = 0\)

• Thus, average power loss for capacitor is zero. 

• Power : Power is the rate at which energy is delivered by an AC source. It is given by 

\(P = V I\)

where \(V \)= Voltage 

\(I\) = Current 

• For a purely inductive AC circuit, 

Voltage is \(V_L = V_0\, sin \;\omega t\)

Current is \(I_L = -I_0 \,cos \;\omega t \)

where  \(I_0 = \dfrac{V_0}{\omega L}\) 

Instantaneous Power of a Purely Inductive Circuit 

• Consider a purely inductive AC circuit, as shown in figure.
• The instantaneous power of a purely inductive circuit is given by 

\(P_L = V_L\; I_L\)

\(P_L = - V_0 \; I_0 \,sin\, \omega t\,\,cos\,\omega t\)

\(P_L=-\dfrac{V_0I_0\, sin2\omega t}{2}\)

Here, \(P_L , \; V_L \; and\; I_L\)are instantaneous values.

Average Power of a Purely Inductive Circuit 

• Average power \( (\overline P_L)\) is defined as the total energy dissipated in given time. 
• The instantaneous power is

\(P_L = \dfrac{-1}{2} V_0 \; I_0 \,sin \; 2 \omega t \)

• \(sin \,2 \omega t\) is positive for half cycle and negative for another half cycle. Thus, its average is zero one full cycle.

\(\therefore \overline P_L = 0\)

• An inductor stores energy in the form of magnetic field for increasing current and it transfers the energy back for decreasing current. 
• Thus, average power for inductor is zero.

• Consider a series R-L circuit, as shown in figure

• The impedance of a series R-L circuit is given by

 \(Z = \sqrt{ R^2 + (X_L)^2}\)

• The phase angle of a series R-L circuit is given by

 \(\phi = tan^{-1} \left(\dfrac{X_L}{R}\right)\)

• The average power of a series R-L circuit is given by 

\(P_{avg} = V_{rms }\;I_{rms }\; cos \phi \)

\(\because \; I _{rms} = \dfrac{V_{rms}}{Z}\)

\(\therefore\; P_{avg} = \dfrac{V^2_{rms}}{R^2 + (X_L)^2}\;cos\left[tan^{-1}\left(\dfrac{X_L}{R}\right)\right]\)

• Consider a series R-C circuit, as shown in figure.

• The impedance of a series R-C circuit is given by

 \(Z = \sqrt{R^2 + (X_C)^2}\)

• The phase angle of a series R-C circuit is given by 

\(\phi = tan^{-1} \left(\dfrac{-X_C}{R}\right)\)

\(\phi = -tan^{-1} \left(\dfrac{X_C}{R}\right)\)

• The average power of a series R-C circuit is given by 

\(P_{avg} = V_{rms}\;I_{rms} \; cos \phi\)

\(\because\; I_{rms } = \;\dfrac{V_{rms}}{Z}\)

\(\therefore\; P_{avg} = \dfrac{V^2_{rms}}{\sqrt{R^2 + (X_C)^2}}\;\;\;cos \left[-tan^{-1}\left(\dfrac{X_C}{R}\right)\right]\)

\( P_{avg} = \dfrac{V^2_{rms}}{\sqrt{R^2 + (X_C)^2}}\;\;\;cos \left[tan^{-1}\left(\dfrac{X_C}{R}\right)\right]\)

• Consider a series L-C circuit, as shown in figure.

• The impedance of a series L-C circuit is given by 

\(Z = \sqrt{(X_L + X_C)^2}\)

\(Z = \pm(X_L + X_C)\)

• The phase angle of a series L-C circuit is given by 

\(\phi = tan ^{-1} \left(\dfrac{X_L - X_C}{R}\right)\)

\(\phi = tan ^{-1} \left(\dfrac{X_L - X_C}{0}\right)\)

\(\phi = tan ^{-1} (\infty)\)

\(\phi = \dfrac{\pi}{2}\)

• Average power of a series L-C circuit is given by 

\(P_{avg} = V_{rms}\;I_{rms} \; cos \phi\)

\(\because\; I_{rms } = \;\dfrac{V_{rms}}{Z}\)

\(\therefore\; P_{avg} = \dfrac{V^2_{rms}}{{\pm(X_L-X_C)}}\;cos \dfrac{\pi}{2}\)

\(P_{avg} = \;0\)

• Power : Power is the rate at which energy is delivered by an AC source. It is given by 

\(P=V\)\(I\)

where \(V \)= Voltage 

\(I\) = Current 

• For series LCR circuit, 

Voltage is \(V\) = \(V_0 sin \; \omega t\)

Current is \(I = I_0 sin (\omega t- \phi)\)

Instantaneous Power of a Series LCR Circuit

• Consider a series LCR circuit as shown in figure.
• The instantaneous power of a series LCR circuit is given by 

\(P_{LCR} = VI\)

\(P_{LCR} = V_0 \; I_0 sin \; \omega t \; sin (\omega t -\phi )\)

\([\therefore sin (\omega t – \;\phi ) = sin \;\omega t \;cos \phi \;- \;cos \omega t \;sin \phi]\)

\(P_{LCR} = V_0\;I_0 \; sin \omega t [sin \omega t \; cos \phi - cos \omega t \; sin \phi]\)

\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t \,cos \phi - V_0\; I_0 \; cos \omega t \; sin \phi \; sin \omega t\)

\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t \; cos \phi \;-\dfrac{V_0\;I_0}{2} sin 2 \;\omega t \; sin \phi \)

Average Power of a Series LCR Circuit 

• Average power \(\left(\overline{P_{LCR}}\right)\)is defined as the total energy dissipated per second.
• The instantaneous power can be written as

\(P_{LCR} = V_0\;I_0\; sin^2 \omega t \; cos \phi\; - \dfrac{V_0I_0}{2}sin 2\omega t \; sin \phi\)

\(P_{LCR} = \dfrac{V_0I_0}{2} \;cos \phi \;(1 – cos 2 \;\omega t) - \dfrac{V_0I_0}{2}\; sin 2 \;\omega t sin \phi \)

\(P_{LCR} = \dfrac{V_0I_0}{2} cos \phi - \dfrac{V_0I_0}{2} \; cos \phi \; cos 2\; \omega t - \dfrac{V_0I_0}{2} sin 2 \; \omega t \; sin \phi \)

• Since \(cos 2 \omega t\) and \(sin 2 \omega t \) are positive for half cycle and negative for another half cycle. Thus, separately their average are zero for one full cycle. 

\(\therefore \overline{P_{LCR}} = \dfrac{V_0I_0}{2} cos \phi\)

\( \overline{P_{LCR}} = \dfrac{V_0}{\sqrt2} × \dfrac{I_0}{\sqrt2} cos \phi\)

\( \overline{P_{LCR}} = V_{rms} × I_{rms}\, cos \phi \)   

• Power : Power is the rate at which energy is delivered by an AC source. It is given by 

 \(P = VI\)

where \(V \)= Voltage 

\(I\) = Current 

• For series LCR circuit, 

Voltage is \(V \)= \(V_0 sin \; \omega t\)

Current is \(I = I_0 sin (\omega t- \phi)\)

Instantaneous Power of a Series LCR Circuit

• Consider a series LCR circuit as shown in figure.
• The instantaneous power of a series LCR circuit is given by 

\(P_{LCR} = VI\)

\(P_{LCR} = V_0 \; I_0 sin \; \omega t \; sin (\omega t -\phi )\)

[\(\therefore sin (\omega t – \;\phi ) = sin \;\omega t \;cos \phi \;- \;cos \omega t \;sin \phi\)]

\(P_{LCR} = V_0\;I_0 \; sin \omega t [sin \omega t \; cos \phi - cos \omega t \; sin \phi]\)

\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t cos \phi - V_0\; I_0 \; cos \omega t \; sin \phi \; sin \omega t\)

\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t \; cos \phi \;-\dfrac{V_0\;I_0}{2} sin 2 \;\omega t \; sin \phi \)

Average Power of a Series LCR Circuit 

• Average power \(\left(\overline{P_{LCR}}\right)\)is defined as the total energy dissipated per second.
• The instantaneous power can be written as

\(P_{LCR} = V_0\;I_0\; sin^2 \omega t \; cos \phi\; - \dfrac{V_0I_0}{2}sin 2\omega t \; sin \phi\)

\(P_{LCR} = \dfrac{V_0I_0}{2} \;cos \phi \;(1 – cos 2 \;\omega t) - \dfrac{V_0I_0}{2}\; sin 2 \;\omega t sin \phi \)

\(P_{LCR} = \dfrac{V_0I_0}{2} cos \phi - \dfrac{V_0I_0}{2} \; cos \phi \; cos 2\; \omega t - \dfrac{V_0I_0}{2} sin 2 \; \omega t \; sin \phi \)

• Since  \(cos 2\omega t\) and \(sin 2 \omega t\) are positive for half cycle and negative for another half cycle. Thus, separately their average are zero for one full cycle. 

\(\therefore \overline{P_{LCR}} = \dfrac{V_0I_0}{2} cos \phi\)

\( \overline{P_{LCR}} = \dfrac{V_0}{\sqrt2} × \dfrac{I_0}{\sqrt2} cos \phi\)

\( \overline{P_{LCR}} = V_{rms} × I_{rms}\, cos \phi \)   

Power factor 

• Average power is given by 

\(P_{avg} = \dfrac{1}{2} V_0\; I_0 \; cos \phi\)

Here, the term cos \(\phi\) is known as power factor.

The phasor diagram of \(V\) and \(I\) for series LCR circuit is shown in figure

Taking components  of \(I\)  along  \(V\) shown in figure.

• \(I sin \,\phi\) is wattless current as it does not contribute any power to circuit.
• The value of cos \(\phi\) oscillates between 1 and – 1. Thus, for the maximum value of average power, the value of cos \(\phi\) will be 1.

cos \(\phi\) = 1

⇒ \(\phi\) = 0° 

• The phase angle is given by 

\(\phi = tan^{-1} \left(\dfrac{X_L-X_C}{R}\right)\)

• For maximum power, \(\phi\) = 0°

\(0° = tan^{-1} \left(\dfrac{X_L-X_C}{R}\right)\)

\(tan\;0° = \dfrac{X_L-X_C}{R}\)

0 = X_L – X_C

\(\Rightarrow\)X_L = X_C

• Therefore, it can be concluded that the AC source delivers maximum power when X_L = X_C i.e, at resonance frequency.