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Power

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 for Resistors

  • 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

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

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

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

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

Illustration Questions

In a purely resistive circuit, the peak current is \(I_0\) = 2A and resistance is R = 10 \(\Omega\). Find the average power of resistor.

A 5 W

B 10 W

C 20 W

D 15 W

×

Peak current, \(I_0\) = 2 A

Resistance of resistor, \(R\) = 10 \(\Omega\)

Average power of resistor is given by,

\(P_{avg} = \dfrac{I_0^2R}{2}\)

\(P_{avg} = \dfrac{(2)^2 × 10}{2}\)

\(P_{avg} = 20 \;W\)

In a purely resistive circuit, the peak current is \(I_0\) = 2A and resistance is R = 10 \(\Omega\). Find the average power of resistor.

A

5 W

.

B

10 W

C

20 W

D

15 W

Option C is Correct

Power of Capacitor

  • 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.

Illustration Questions

Find the average power for a purely inductive circuit for full cycle, if the inductance of inductor is L = 20 mH, and the source voltage V = 2 sin 120 \(\pi\;t\).

A 3 W

B 5 W

C Zero 

D 10 W

×

Inductance of inductor, \( L = 20 \,mH \)

angular frequency, \(\omega\) \(= 120 \pi\)

peak voltage,  \(V_o= 2\,V\)

The average power for a purely inductive circuit is zero.

\(P_L =\;\;^- \dfrac{1}{2} V_0\; I_0 sin \; 2 \;(120 \pi )t\)

\(sin \,2 \omega t\) is positive for half cycle and negative for another half cycle. Thus it's average is zero one full cycle.

\(\therefore \overline P_L = 0\)

 

 

Find the average power for a purely inductive circuit for full cycle, if the inductance of inductor is L = 20 mH, and the source voltage V = 2 sin 120 \(\pi\;t\).

A

3 W

.

B

5 W

C

Zero 

D

10 W

Option C is Correct

Average Power for Series LCR Circuit 

  • 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. 

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

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

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

Illustration Questions

For a series LCR circuit, the inductive reactance is XL = 5 \(\Omega\), the capacitive reactance is XC = 2 \(\Omega\) and the resistance is R = 4 \(\Omega\). Calculate the average power of series LCR circuit, if the peak voltage is \(V_o\)= 20 V. (\(cos 37 ° =\dfrac{4}{5}\))

A 30 W

B 32 W

C 28 W

D 20 W

×

Inductive reactance, XL = 5 \(\Omega\)

Capacitive reactance, XC = 2 \(\Omega\)

Resistance, R = 4 \(\Omega\)

Peak voltage, \(V_0 \)= 20 V

Impedance of series LCR circuit is 

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

\(Z = \sqrt {(4)^2 + (5 - 2)^2} \)

\(Z = \sqrt {16 + 9} \)

\(Z = 5\; \Omega \)

 

Peak current is given by 

\(I_0 = \dfrac{V_0}{Z}\)

\(I_0 = \dfrac{20}{5}\;= 4 \; A\)

Phase angle is given by 

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

\(\phi = tan^{-1}\left(\dfrac{3}{4}\right)\)

\(\phi = 37 ° \)

Average power is given by 

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

\(P_{avg} = \dfrac{1}{2} (4)(20)\; cos 37° \)

\(P_{avg} = \dfrac{1}{2}(4)(20)\left(\dfrac{4}{5}\right)\)

\(P_{avg} = \;32 \;W\)

For a series LCR circuit, the inductive reactance is XL = 5 \(\Omega\), the capacitive reactance is XC = 2 \(\Omega\) and the resistance is R = 4 \(\Omega\). Calculate the average power of series LCR circuit, if the peak voltage is \(V_o\)= 20 V. (\(cos 37 ° =\dfrac{4}{5}\))

A

30 W

.

B

32 W

C

28 W

D

20 W

Option B is Correct

Power Factor of Series LCR Circuit 

  • 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 = XL – XC

\(\Rightarrow\)XL = XC

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

Illustration Questions

For a series LCR circuit, the inductive reactance is XL = 5 \(\Omega\) , the capacitive reactance is XC = 2 \(\Omega\) and the resistance is R = 4 \(\Omega\). Calculate the power factor of series LCR circuit. \((cos \;37° =\dfrac{4}{5})\)

A \(\dfrac{4}{5}\)

B \(\dfrac{3}{5}\)

C \(\dfrac{2}{5}\)

D \(\dfrac{5}{4}\)

×

Inductive reactance, XL = 5 \(\Omega\)

Capacitive reactance, XC = 2\(\Omega\) 

Resistance, R = 4\(\Omega\) 

 

Phase angle is given by

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

\(\phi = tan^{-1}\left(\dfrac{5-2}{4}\right)\)

\(\phi = tan^{-1}\left(\dfrac{3}{4}\right)\)

\(\phi=37°\)

Power factor is

\(cos 37°\) = \(\dfrac{4}{5}\)

For a series LCR circuit, the inductive reactance is XL = 5 \(\Omega\) , the capacitive reactance is XC = 2 \(\Omega\) and the resistance is R = 4 \(\Omega\). Calculate the power factor of series LCR circuit. \((cos \;37° =\dfrac{4}{5})\)

A

\(\dfrac{4}{5}\)

.

B

\(\dfrac{3}{5}\)

C

\(\dfrac{2}{5}\)

D

\(\dfrac{5}{4}\)

Option A is Correct

Power for a Series R-L Circuit 

  • 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 

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

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

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

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

 

Illustration Questions

Find the average power of a series R-L circuit, if the resistance of resistor is R = 50 \(\Omega\), the inductive reactance is XL = 50 \(\Omega\) and rms voltage is Vrms = 50 V.

A 30 W

B 35 W 

C 25 W

D 80 W

×

Resistance of resistor, R = 50 \(\Omega\)

Inductive reactance, XL = 50 \(\Omega\)

rms voltage, Vrms = 50  V

Impedance of series R-L circuit 

\(Z = \sqrt{R^2 + X^2_L}\)

\(Z = \sqrt{(50)^2 + (50)^2}\)

= 50 \(\sqrt2\) \(\Omega\)

Phase angle of series R-L circuit 

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

\(\phi = tan^{-1} \left(\dfrac{50}{50}\right)\)

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

\(\phi = 45 ° \)

Average power of series R-L circuit

\(P_{avg}\; = \; \dfrac{V^2_{rms}}{Z} \;cos 45 ° \)

\(P_{avg} = \dfrac{50^2}{50 \sqrt2}\;\left(\dfrac{1}{\sqrt2}\right)\)

\(P_{avg}\;=\;25 \;W\)

Find the average power of a series R-L circuit, if the resistance of resistor is R = 50 \(\Omega\), the inductive reactance is XL = 50 \(\Omega\) and rms voltage is Vrms = 50 V.

A

30 W

.

B

35 W 

C

25 W

D

80 W

Option C is Correct

Illustration Questions

For a series LCR circuit, the resistance of resistor is \( R = 15\,\Omega\) the inductance of inductor is \(L = 2\, mH\) and voltage is \( V = 10\,sin 2\) \(\pi t\). Determine the value of capacitance for which the current is maximum. 

A 20.13 F

B 12.67 F

C 120 F

D 30.13 F

×

Resistance of resistor, \( R = 15\) \(\Omega\)

Inductance of inductor,\( L = 2 \,mH\)

Voltage, \( V = 10\, sin 2\pi t\)

angular frequency, \(\omega\) \(= 2\) \(\pi \) \( rad/sec\)

Since, current is given by 

\(I_0 = \dfrac{V_0}{Z}\)

Therefore, for the current to be maximum, the impedance i.e.Z should be minimum.

Since, Impedance in series LCR circuit is given by 

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

Therefore, for the impedance to be minimum, XL should be equal to XC.

Thus, for the current to be maximum, 

\(X_L = X_C\)

\(\omega L = \dfrac{1}{\omega C}\)

\(\omega^2 = \dfrac{1}{LC}\)

\(\omega = \dfrac{1}{\sqrt{LC}}\)

\(\omega = \dfrac{1}{\sqrt{LC}}\)

\(C = \dfrac{1}{\omega ^2L}\)

\(C = \dfrac{1}{4 \pi ^2 × 2 × 10^{-3}}\)

 

\(C = 0.01267 × 10^3\)

\(C = 12.67 \;\;F\)

For the current to be maximum, the capacitance should be \(C = 12.67 F\).

For a series LCR circuit, the resistance of resistor is \( R = 15\,\Omega\) the inductance of inductor is \(L = 2\, mH\) and voltage is \( V = 10\,sin 2\) \(\pi t\). Determine the value of capacitance for which the current is maximum. 

A

20.13 F

.

B

12.67 F

C

120 F

D

30.13 F

Option B is Correct

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