Answer
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Hint: In contrast to direct current (DC), which flows only in one direction, alternating current (AC) is an electric current that regularly reverses direction and changes its magnitude continuously over time.
Complete step by step answer:
In an AC circuit, resonance is described as the state in which the circuit's inductive reactance equals its capacitive reactance. Resonance frequency is the value of angular frequency of alternating emf (or potential difference across the power source) for which resonance is formed in the circuit.
\[P = IV\], where I is the current through the element and \[V\] is the voltage across it, determines whether a circuit element dissipates or produces electricity. Since the current and voltage in an ac circuit are both time dependent, the instantaneous power \[p\left( t \right) = i\left( t \right)v\left( t \right)\] is also time dependent.
Now, coming to the question, Power dissipated in an AC circuit,
$P = IV\cos \phi $ [$\cos \phi $ is power factor; $\phi $ is phase difference]
\[V = 230\sin \left( {\omega t + \dfrac{\pi }{2}} \right)\]____(i)
$I = 20\sin \left( {\omega t} \right)$_____(ii)
From equation (i) we get peak value ${V_0} = 230\,V$ and phase angle $\theta = \dfrac{\pi }{2}$
From equation (ii) we get peak current ${I_0} = 20\,A$
Power dissipated $P = IV\cos \phi $
$\Rightarrow P = \dfrac{{{V_0}}}{{\sqrt 2 }} \times \dfrac{{{I_0}}}{{\sqrt 2 }}\cos \left( {\dfrac{\pi }{2}} \right)$
$\Rightarrow P = \dfrac{{230}}{{\sqrt 2 }} \times \dfrac{{20}}{{\sqrt 2 }} \times \left( 0 \right)$ $[\because \cos \dfrac{\pi }{2} = 0]$
$\therefore P = 0$
Therefore, power dissipated through the AC circuit is zero.
Note:Peak power dissipation is also normally only relevant in specific applications since peak power is an instantaneous value that occurs only when the voltage and current are at their highest levels. The \[rms\] power is the most important power specification in an AC circuit.
Complete step by step answer:
In an AC circuit, resonance is described as the state in which the circuit's inductive reactance equals its capacitive reactance. Resonance frequency is the value of angular frequency of alternating emf (or potential difference across the power source) for which resonance is formed in the circuit.
\[P = IV\], where I is the current through the element and \[V\] is the voltage across it, determines whether a circuit element dissipates or produces electricity. Since the current and voltage in an ac circuit are both time dependent, the instantaneous power \[p\left( t \right) = i\left( t \right)v\left( t \right)\] is also time dependent.
Now, coming to the question, Power dissipated in an AC circuit,
$P = IV\cos \phi $ [$\cos \phi $ is power factor; $\phi $ is phase difference]
\[V = 230\sin \left( {\omega t + \dfrac{\pi }{2}} \right)\]____(i)
$I = 20\sin \left( {\omega t} \right)$_____(ii)
From equation (i) we get peak value ${V_0} = 230\,V$ and phase angle $\theta = \dfrac{\pi }{2}$
From equation (ii) we get peak current ${I_0} = 20\,A$
Power dissipated $P = IV\cos \phi $
$\Rightarrow P = \dfrac{{{V_0}}}{{\sqrt 2 }} \times \dfrac{{{I_0}}}{{\sqrt 2 }}\cos \left( {\dfrac{\pi }{2}} \right)$
$\Rightarrow P = \dfrac{{230}}{{\sqrt 2 }} \times \dfrac{{20}}{{\sqrt 2 }} \times \left( 0 \right)$ $[\because \cos \dfrac{\pi }{2} = 0]$
$\therefore P = 0$
Therefore, power dissipated through the AC circuit is zero.
Note:Peak power dissipation is also normally only relevant in specific applications since peak power is an instantaneous value that occurs only when the voltage and current are at their highest levels. The \[rms\] power is the most important power specification in an AC circuit.
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