T6.8 Power in LCR Circuit:

In the series LCR circuit, the power delivered by the AC circuit is defined as the product of current and emf.
Mathematically,

Power ($P$) = $I * E$  .......... (i)                  (∵ P = I * IR = I$^2$R)

The instantaneous value of current and emf are,

$\left.\begin{matrix}
I = I_0 Sin \omega t
\\
E = E_0 Sin (\omega t + \theta)
\end{matrix}\right\}$ .......... (ii)

Then,

Power (P) = $I_0$ Sin $\omega t . E_0 $ Sin $(\omega t + \theta)$

                 = $I_0 . E_0$ Sin $\omega t$ (Sin $\omega t$ Cos $\theta$ + Cos $\omega t$ Sin $\theta)$

                 = $I_0 . E_0$ (Cos $\theta$ . Sin$^2 \omega t$ + Sin$\theta$ . Cos $\omega t$ . Sin $\omega t)$

Now the small amount of work done ($dW$) is given by,

       $dW = P.dt$

or,   $dW = I_0 . E_0$ (Cos$\theta$ . Sin$^2 \omega t$ + Sin$\theta$ . Cos $\omega t$ . Sin $\omega t)$ .......... (iii)

Hence, the total amount of work done $(W)$ can be obtained by integrating the above equation (iii) from $0$ to $T$, then

or, $\int dW = \int_{0}^{T} I_0 . E_0$ (Cos$\theta$ . Sin$^2 \omega t$ + Sin$\theta$ . Cos $\omega t$ . Sin $\omega t) $

By Integrating,

$W = \frac{I_0 . E_0}{2} T. Cos\theta$

Therefore, the average Power $(P_{av})$ in the AC circuit is,

$P_{av} = \frac{W}{T} = \frac{I_0 . E_0}{2\;*\;T} T. Cos \theta $

             $  = \frac{I_0 . E_0}{2} Cos\theta$

             $  = \frac{I_0}{\sqrt{2}} * \frac{E_0}{\sqrt{2}} * Cos\theta$

$\Rightarrow$  $P_{av} = I_{rms} * E_{rms} * Cos \theta$ .......... (iv)

Here I$_{rms}$ and E$_{rms}$ are called apparent power and Cos$\theta$ is called power factor.

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Special Cases:

Case I: Power Consumption across R, If $\theta = 0$; Then, 

$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} $

Case II: Power Consumption across L, If $\theta = \frac{\pi}{2}$; Then, 

$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} * Cos(\frac{\pi}{2}) = 0 $

Case III: Power Consumption across C, If $\theta = \frac{-\pi}{2}$; Then, 

$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} Cos(\frac{-\pi}{2}) = 0 $

Case IV: Power Consumption across RL, If $\theta = \frac{R}{Z}$; Then, 

$\Rightarrow$  $P_{av} = I_{rms} * E_{rms} (\frac{R}{\sqrt{X_{L}{^2} + R^2}})$

Case V: Power Consumption across RC, If $\theta = \frac{R}{Z}$; Then, 

$\Rightarrow$  $P_{av} = I_{rms} * E_{rms} (\frac{R}{\sqrt{X_{C}{^2} + R^2}})$

Case VI: Power Consumption across LCR, If $\theta = \frac{R}{Z}$; Then, 

$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} (\frac{R}{\sqrt{{X_{L} - X_{C}){^2} + R^2}}}) $

E1.5 Work - Energy Theorem:

Net work on an object causes a change  in the kinetic energy. The work done in a body due to a force is equal to the change in its kinetic energy. 

i.e. Work done = Change in Kinetic Energy

Consider a body of mass $M$, moving with a initial velocity $u$  when a constant force $F$ is acts on it. Let final velocity of the body becomes $v$ after covering a distance $d$ in a direction of force. Then, work done by the force is given by

$W = F.d$ .......... (i)

From the Newton's law of Motion:

$F = ma$ .......... (ii)

From equation (i) and (ii)

$F = ma.d$ .......... (iii)

And from the Kinematics, we know the formula:

$v^2 = u^2 + 2ad$

$d = \frac{v^2 - u^2}{2a}$ .......... (iv)

From the equation (i), (iii) and (iv), we get

$W = ma\;(\frac{v^2 - u^2}{2a})$

      $= m\;(\frac{v^2}{2} - \frac{u^2}{2})$

     $= \frac{1}{2}\,mv^2 - \frac{1}{2}\,mu^2 $

      = Final K.E - Initial K.E.

∴ Work done = Change in Kinetic Energy

So, the work done on moving the body by the force is equal to the increase in its kinetic energy.

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