A voltage $V = V_o$ sin $\omega\text{t}$ is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit?
CBSE OUTSIDE DELHI - SET 2 2014
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Applied voltage $= V_o$ sin $\omega\text{t}$
Current in the circuit $=I_o$ sin $(\omega\text{t} - \phi)$
where $\phi$is the phase lag of the current with respect to the voltage applied, Hence instantaneous power dissipation
$ = \text{V}_{0} \sin \omega\text{t}\times\text{I}_{o}\sin(\omega\text{t} - \phi) $
$ = \frac{\text{V}_{0}\text{I}_{0}}{2}[2\sin\omega\text{t}.\sin(\omega\text{t} - \phi]$
$ = \frac{\text{V}_{0}\text{I}_{0}}{2}[\cos\phi - \cos(2\omega\text{t} - \phi]$
Therefore, average power for one complete cycle
= average of $ [ \frac{\text{V}_{0}\text{I}_{0}}{2}[\cos\phi - \cos(2\omega\text{t} - \phi]]$
The average of the second term over a complete cycle is zero.
Hence, average power dissipated over one complete cycle $ = \frac{\text{V}_{0}\text{I}_{0}}{2}\cos\phi$
Conditions:
$\omega\text{L} = \frac{1}{\text{wc}}(\text{or } \phi = 0).$
Current in the circuit $=I_o$ sin $(\omega\text{t} - \phi)$
where $\phi$is the phase lag of the current with respect to the voltage applied, Hence instantaneous power dissipation
$ = \text{V}_{0} \sin \omega\text{t}\times\text{I}_{o}\sin(\omega\text{t} - \phi) $
$ = \frac{\text{V}_{0}\text{I}_{0}}{2}[2\sin\omega\text{t}.\sin(\omega\text{t} - \phi]$
$ = \frac{\text{V}_{0}\text{I}_{0}}{2}[\cos\phi - \cos(2\omega\text{t} - \phi]$
Therefore, average power for one complete cycle
= average of $ [ \frac{\text{V}_{0}\text{I}_{0}}{2}[\cos\phi - \cos(2\omega\text{t} - \phi]]$
The average of the second term over a complete cycle is zero.
Hence, average power dissipated over one complete cycle $ = \frac{\text{V}_{0}\text{I}_{0}}{2}\cos\phi$
Conditions:
- No power is dissipated when R = 0 $(\text{or }\phi = 90^{o} ) $
- Maximum power is dissipated when $X_L= X_C$
$\omega\text{L} = \frac{1}{\text{wc}}(\text{or } \phi = 0).$
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