Learn power definition and formula in Physics, practice example of Average & Instantaneous power. Find average power if change in kinetic energy is known and instantaneous power when work is given as a function of time.
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 \(I_R = I_0\,sin \;\omega\;t\)
\(\because\;V_R=I_R\;R\)
\(\therefore\;P_R = I_R^2\;R\)
\(P_R = \left[I_0^2\;sin^2\;\omega t\right]R\;\;\;\;\;\;\;\;\left[I_R=I_0\;sin\;\omega t\right]\)
\(P_R = I_0^2R\;sin^2\;\omega t\)
Here, \(P_R,\;I_R\;and\; V_R\) are instantaneous values.
\(P_R =\dfrac{I_0^2\;R}{2} (1-cos 2\;\omega t)\)
\(\left(\because cos \;2\theta = 1 - 2\; sin^2 \theta\right)\)
\(P_R =\dfrac{I_0^2\;R}{2} - \dfrac{I_0^2R}{2}cos 2 \;\omega t\)
\(cos 2\; \omega t\) is positive for half cycle and negative for another half cycle. Thus, its average is zero for one full cycle.
\(\therefore\;\overline{P}_R\;=\;\dfrac{I_0^2R}{2} - 0\)
\(\;\overline{P}_R\;=\;\dfrac{I_0^2R}{2} \)
\(\overline{P}_R = \left(\dfrac{I_0}{\sqrt2}\right)^2\;R\)
⇒ \(\overline{P}_R= I^2_{rms}\;R\) \(\because\;I_{rms}\;=\;\dfrac{I_0}{\sqrt2}\)
⇒ \(\overline P_R =\dfrac{V^2_{rms}}{R}\) \(\because V_{rms}= I_{rms} R\)
⇒ \(\overline P_R = I_{rms}\times V_{rms} \)
\(P = VI\)
where \(V\) = Voltage
\(I\) = Current
Voltage is \(V_C = V_0\, sin\; \omega t\)
Current is \(I_C = - \omega CV_0\,cos \; \omega t\)
\(P_C = V_C I_C\)
\(\because V_C = V_0 \,sin \;\omega t \;\;\;and \;\;\; I_C = - \omega C V_0 \,cos \;\omega t\)
\(P_C = - ( V_0 \,sin \;\omega t)(\omega C V_0 \,cos \;\omega t)\)
\(P_C = - \omega C\;V^2_0\, sin \;\omega t \; cos \;\omega t\)
\(P_C = \dfrac{-1}{2} \omega CV_0^2 \; sin \; 2\omega t\)
Here, \(P_C, \; I_C\) & \(V_C\) are instantaneous values.
\(\therefore \overline P_C = 0\)
\(P = V I\)
where \(V \)= Voltage
\(I\) = Current
Voltage is \(V_L = V_0\, sin \;\omega t\)
Current is \(I_L = -I_0 \,cos \;\omega t \)
where \(I_0 = \dfrac{V_0}{\omega L}\)
\(P_L = V_L\; I_L\)
\(P_L = - V_0 \; I_0 \,sin\, \omega t\,\,cos\,\omega t\)
\(P_L=-\dfrac{V_0I_0\, sin2\omega t}{2}\)
Here, \(P_L , \; V_L \; and\; I_L\)are instantaneous values.
\(P_L = \dfrac{-1}{2} V_0 \; I_0 \,sin \; 2 \omega t \)
\(\therefore \overline P_L = 0\)
\(Z = \sqrt{ R^2 + (X_L)^2}\)
\(\phi = tan^{-1} \left(\dfrac{X_L}{R}\right)\)
\(P_{avg} = V_{rms }\;I_{rms }\; cos \phi \)
\(\because \; I _{rms} = \dfrac{V_{rms}}{Z}\)
\(\therefore\; P_{avg} = \dfrac{V^2_{rms}}{R^2 + (X_L)^2}\;cos\left[tan^{-1}\left(\dfrac{X_L}{R}\right)\right]\)
\(Z = \sqrt{R^2 + (X_C)^2}\)
\(\phi = tan^{-1} \left(\dfrac{-X_C}{R}\right)\)
\(\phi = -tan^{-1} \left(\dfrac{X_C}{R}\right)\)
\(P_{avg} = V_{rms}\;I_{rms} \; cos \phi\)
\(\because\; I_{rms } = \;\dfrac{V_{rms}}{Z}\)
\(\therefore\; P_{avg} = \dfrac{V^2_{rms}}{\sqrt{R^2 + (X_C)^2}}\;\;\;cos \left[-tan^{-1}\left(\dfrac{X_C}{R}\right)\right]\)
\( P_{avg} = \dfrac{V^2_{rms}}{\sqrt{R^2 + (X_C)^2}}\;\;\;cos \left[tan^{-1}\left(\dfrac{X_C}{R}\right)\right]\)
\(Z = \sqrt{(X_L + X_C)^2}\)
\(Z = \pm(X_L + X_C)\)
\(\phi = tan ^{-1} \left(\dfrac{X_L - X_C}{R}\right)\)
\(\phi = tan ^{-1} \left(\dfrac{X_L - X_C}{0}\right)\)
\(\phi = tan ^{-1} (\infty)\)
\(\phi = \dfrac{\pi}{2}\)
\(P_{avg} = V_{rms}\;I_{rms} \; cos \phi\)
\(\because\; I_{rms } = \;\dfrac{V_{rms}}{Z}\)
\(\therefore\; P_{avg} = \dfrac{V^2_{rms}}{{\pm(X_L-X_C)}}\;cos \dfrac{\pi}{2}\)
\(P_{avg} = \;0\)
\(P=V\)\(I\)
where \(V \)= Voltage
\(I\) = Current
Voltage is \(V\) = \(V_0 sin \; \omega t\)
Current is \(I = I_0 sin (\omega t- \phi)\)
\(P_{LCR} = VI\)
\(P_{LCR} = V_0 \; I_0 sin \; \omega t \; sin (\omega t -\phi )\)
\([\therefore sin (\omega t – \;\phi ) = sin \;\omega t \;cos \phi \;- \;cos \omega t \;sin \phi]\)
\(P_{LCR} = V_0\;I_0 \; sin \omega t [sin \omega t \; cos \phi - cos \omega t \; sin \phi]\)
\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t \,cos \phi - V_0\; I_0 \; cos \omega t \; sin \phi \; sin \omega t\)
\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t \; cos \phi \;-\dfrac{V_0\;I_0}{2} sin 2 \;\omega t \; sin \phi \)
\(P_{LCR} = V_0\;I_0\; sin^2 \omega t \; cos \phi\; - \dfrac{V_0I_0}{2}sin 2\omega t \; sin \phi\)
\(P_{LCR} = \dfrac{V_0I_0}{2} \;cos \phi \;(1 – cos 2 \;\omega t) - \dfrac{V_0I_0}{2}\; sin 2 \;\omega t sin \phi \)
\(P_{LCR} = \dfrac{V_0I_0}{2} cos \phi - \dfrac{V_0I_0}{2} \; cos \phi \; cos 2\; \omega t - \dfrac{V_0I_0}{2} sin 2 \; \omega t \; sin \phi \)
\(\therefore \overline{P_{LCR}} = \dfrac{V_0I_0}{2} cos \phi\)
\( \overline{P_{LCR}} = \dfrac{V_0}{\sqrt2} × \dfrac{I_0}{\sqrt2} cos \phi\)
\( \overline{P_{LCR}} = V_{rms} × I_{rms}\, cos \phi \)
\(P = VI\)
where \(V \)= Voltage
\(I\) = Current
Voltage is \(V \)= \(V_0 sin \; \omega t\)
Current is \(I = I_0 sin (\omega t- \phi)\)
\(P_{LCR} = VI\)
\(P_{LCR} = V_0 \; I_0 sin \; \omega t \; sin (\omega t -\phi )\)
[\(\therefore sin (\omega t – \;\phi ) = sin \;\omega t \;cos \phi \;- \;cos \omega t \;sin \phi\)]
\(P_{LCR} = V_0\;I_0 \; sin \omega t [sin \omega t \; cos \phi - cos \omega t \; sin \phi]\)
\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t cos \phi - V_0\; I_0 \; cos \omega t \; sin \phi \; sin \omega t\)
\(P_{LCR} = V_0\;I_0\; sin ^2 \; \omega t \; cos \phi \;-\dfrac{V_0\;I_0}{2} sin 2 \;\omega t \; sin \phi \)
\(P_{LCR} = V_0\;I_0\; sin^2 \omega t \; cos \phi\; - \dfrac{V_0I_0}{2}sin 2\omega t \; sin \phi\)
\(P_{LCR} = \dfrac{V_0I_0}{2} \;cos \phi \;(1 – cos 2 \;\omega t) - \dfrac{V_0I_0}{2}\; sin 2 \;\omega t sin \phi \)
\(P_{LCR} = \dfrac{V_0I_0}{2} cos \phi - \dfrac{V_0I_0}{2} \; cos \phi \; cos 2\; \omega t - \dfrac{V_0I_0}{2} sin 2 \; \omega t \; sin \phi \)
\(\therefore \overline{P_{LCR}} = \dfrac{V_0I_0}{2} cos \phi\)
\( \overline{P_{LCR}} = \dfrac{V_0}{\sqrt2} × \dfrac{I_0}{\sqrt2} cos \phi\)
\( \overline{P_{LCR}} = V_{rms} × I_{rms}\, cos \phi \)
\(P_{avg} = \dfrac{1}{2} V_0\; I_0 \; cos \phi\)
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.
cos \(\phi\) = 1
⇒ \(\phi\) = 0°
\(\phi = tan^{-1} \left(\dfrac{X_L-X_C}{R}\right)\)
\(0° = tan^{-1} \left(\dfrac{X_L-X_C}{R}\right)\)
\(tan\;0° = \dfrac{X_L-X_C}{R}\)
0 = X_{L} – X_{C}
\(\Rightarrow\)X_{L} = X_{C}
A \(\dfrac{4}{5}\)
B \(\dfrac{3}{5}\)
C \(\dfrac{2}{5}\)
D \(\dfrac{5}{4}\)