Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A voltage $V = {V_0}\sin \omega t$ is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is1) No power dissipated even though the current flows through the circuit2) Maximum power dissipated in the circuit?

Last updated date: 20th Jun 2024
Total views: 394.2k
Views today: 3.94k
Verified
394.2k+ views
Hint: The power factor, $\cos \phi$ determines the average power dissipated in a LCR circuit, An LCR circuit, capacitance, resistance and inductance will affect the power dissipated within the circuit.

Step by step solution:Let’s define the data given in the question:

It is given that voltage applied to the LCR circuit, $V = {V_0}\sin \omega t$

From this equation, we get the current flowing in the LCR circuit by the formula:

$I = {I_0}\sin (\omega t - \phi )$

Here $\phi$ is the phase lag of the current with respect to the applied voltage $V = {V_0}\sin \omega t$

And ${I_0}$ is given by equation, ${I_0} = {V_0}Z\phi = {\tan ^{ - 1}}({X_C} - {X_L})R$
Now we are heading to the power dissipated in the LCR circuit,

The power dissipated in the circuit is given by two equations,

$P = {I^2}R$ And $P = VI$

Here we are taking second equation, that is:
The power dissipated, $P = VI$

Applying the values $V$and $I$ in the above equation, we get,

$P = {V_0}\sin \omega t \times {I_0}\sin (\omega t - \phi )$

Applying trigonometric conversions we get,

$\Rightarrow P = \dfrac{{{V_0}{I_0}}}{2}\left[ {\sin \omega t - \sin (\omega t - \phi )} \right]$
$\Rightarrow P = \dfrac{{{V_0}{I_0}}}{2}\left[ {\cos \phi - \cos (2\omega t - \phi )} \right]$

Now we get the power dissipated in the LCR circuit
For the average power dissipated we are taking the average of the above value.

Therefore, average power for one complete cycle= average of $\left\{ {\dfrac{{{V_0}{I_0}}}{2}\left[ {\cos \phi - \cos (2\omega t - \phi )} \right]} \right\}$
In the above expression the average of the second term over a complete cycle will be equal zero.

That is, $\dfrac{{{V_0}{I_0}}}{2}\cos (2\omega t - \phi )$ will be equal to zero

Therefore, the average power dissipated over one complete circle will be equal to

$\dfrac{{{V_0}{I_0}}}{2}\cos \phi$ . (Here $\cos \phi$ is the power factor and is given by$\cos \phi = \dfrac{R}{Z}$.)
Now we are applying the two conditions which are given in the question to the average power dissipated, that is

Conditions:
1. No power dissipated even though the current flows through the circuit:
If the power dissipated is zero, the value of $\cos \phi$ should be equal to zero, that is,
$\phi = \dfrac{\pi }{2}$
This is the condition in which the circuit is purely capacitive or inductive.
2. Maximum power dissipated in the circuit
If the power dissipated is maximum, the value of $\cos \phi$ should be equal to one, that is,
$\phi = 0$
This means the circuit is purely resistive.

Note:There is always loss of power in LCR circuits depending on the quality of materials and electrical conditions where the circuitry operates. The power dissipated in a series resonant circuit can also be expressed in terms of the rms voltage and current.