Definition of Power in Alternating Current Circuit:
The rate of power consumption in an alternating current circuit is known as power in the circuit.
Mathematical Analysis:
Let us consider an alternating current circuit in which the voltage and current at any instant are given by
$V=V_{\circ} sin \omega t \qquad(1)$
$i=i_{\circ} sin \left( \omega t - \phi \right) \qquad(2)$
Where $\phi$ $\rightarrow$ Phase difference between voltage and current
So instantaneous Power
$P=Vi$
$P=\left\{V_{\circ} sin \omega t \right\} \left\{ i_{\circ} sin \left( \omega t - \phi \right) \right\}$
$P=V_{\circ} i_{\circ} \: sin \omega t \: sin \left( \omega t - \phi \right)$
$P=\frac{V_{\circ} i_{\circ}}{2} \left[2\: sin \: \omega t \: sin \left( \omega t - \phi \right)\right]$
$P=\frac{V_{\circ} i_{\circ}}{2} \left[ cos \left( \omega t -\omega t + \phi \right) - cos \left( \omega t + \omega t - \phi \right) \right]$
$P=\frac{V_{\circ} i_{\circ}}{2} \left[ cos \left(\phi \right) - cos \left( 2\omega t - \phi \right) \right]$
The average power for one cycle is:
$P_{avg}=The \: average \: value \: of \left\{ \frac{V_{\circ} i_{\circ}}{2} \left[ cos \left(\phi \right) - cos \left( 2\omega t - \phi \right) \right] \right\}$
The term $cos \left( 2\omega t - \phi \right)$ is time-dependent and the average value of term $cos \left( 2\omega t - \phi \right)$ for complete one cycle is zero. So the average power dissipation in one cycle is
$P_{avg}= \frac{V_{\circ} i_{\circ}}{2} cos \phi $
$P_{avg}= \frac{V_{\circ}}{\sqrt{2}} \frac{i_{\circ}}{\sqrt{2}} cos \phi $
$P_{avg}= V_{rms} \: i_{rms} cos \phi $
Case -1 The circuit containing Pure Resistor only:
If the circuit contains a pure resistance only then voltage $V$ and current $i$ are always in the same phase. i.e. $\phi=0^{\circ}$ and $cos \phi = +1$, then average power
$P_{avg}= V_{rms} \: i_{rms} $
Case -2 The circuit containing Pure Inductor only:
If the circuit containing pure inductor only then voltage $V$ leads over current $i$ by $90^{\circ}$ i.e. $\phi=90^{\circ}$ and $cos 90^{\circ} = 0$, then average power
$P_{avg}= 0 $
Case -3 The circuit containing Pure Capacitor only:
If the circuit containing pure capacitor only then voltage $V$ lags behind current $i$ by $90^{\circ}$ i.e. $\phi=90^{\circ}$ and $cos 90^{\circ} = 0$, then average power
$P_{avg}= 0 $
Case -4 The circuit containing inductor and resistor only (L-R Circuit):
We know that the phase in the L-R circuit is
$tan \phi = \frac{X_{L}}{R}$
From the above equation, Draw the triangle as shown in the figure below:
From the above figure, the value of $cos\phi$ is
$cos \phi = \frac{R}{\sqrt{R^{2} + X^{2}_{L}}}$
So average power
$P_{avg}= V_{rms} \: i_{rms} \frac{R}{\sqrt{R^{2} + X^{2}_{L}}} $
$\left(P_{avg}\right)_{L-R} \lt \left(P_{avg}\right)_{R}$
Thus, the average power consumption in an L-R circuit is less than the power consumed in a purely resistive circuit
Case -5 The circuit containing capacitor and resistor only (C-R Circuit):
We know that the phase in the C-R circuit is
$tan \phi = \frac{X_{C}}{R}$
From the above equation, Draw the triangle as shown in the figure below:
From the above figure, the value of $cos\phi$ is
$cos \phi = \frac{R}{\sqrt{R^{2} + X^{2}_{C}}}$
So average power
$P_{avg}= V_{rms} \: i_{rms} \frac{R}{\sqrt{R^{2} + X^{2}_{C}}} $
$\left(P_{avg}\right)_{C-R} \lt \left(P_{avg}\right)_{R}$
Thus, the average power consumption in a C-R circuit is less than the power consumed in a purely resistive circuit
Case -6 The circuit containing capacitor and inductor only (L-C Circuit):
We know that the phase in the L-R circuit is
$\phi = 90^{\circ}$ and $cos 90^{\circ} = 0$
So average power
$P_{avg}= 0 $
Case -7 The circuit containing inductor, capacitor, and resistor only (L-C-R Circuit):
We know that the phase in the L-C-R circuit is
$tan \phi = \frac{X_{L} - X_{C}}{R}$
From the above equation, Draw the triangle as shown in the figure below:
From the above figure, the value of $cos\phi$ is
$cos \phi = \frac{R^{2}}{\sqrt{R^{2} + \left( X^{2}_{L} - X^{2}_{C} \right) }}$
So average power
$P_{avg}= V_{rms} \: i_{rms} \frac{R^{2}}{\sqrt{R^{2} + \left( X^{2}_{L} - X^{2}_{C} \right) }} $
$\left(P_{avg}\right)_{L-C-R} \lt \left(P_{avg}\right)_{R}$
Thus, the average power consumption in an L-C-R circuit is less than the power consumed in a purely resistive circuit
Wattless Current:
If the circuit contains only resistor or only inductor or only inductor and capacitor then power consumption in the circuit will be zero (i.e. there will be no energy dissipation in the circuit) due to the phase difference $90^{\circ}$ between voltage(or impedance) and current. So the current in the circuit is called the "Wattless current".
Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig
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