Informative line

### 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

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

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

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

⇒ $$\overline P_R = I_{rms}\times V_{rms}$$

#### 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.

#### 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

# 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

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

$$\because\; I_{rms } = \;\dfrac{V_{rms}}{Z}$$

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

$$P_{avg} = \;0$$

#### 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

#### 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

# 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.

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

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

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

#### 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.

#### 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