Answer
346.2k+ views
Hint: Power dissipation is the process by which an electronics device produces heat. Here we need to calculate the power dissipated in a circuit where voltage and current is known. Apply ac power dissipated formula we can calculate the power i.e. $P = VI\cos \phi $ .
Complete step by step solution:
According to the problem,
Ac voltage across the circuit is given by
$V = {V_o}\sin \left( {\omega t + \phi } \right)$
As per the question
$V = 230\sin \left( {\omega t + \dfrac{\pi }{2}} \right) \cdot \cdot \cdot \cdot \left( 1 \right)$
Where ${V_o} = 230V$
And $\phi = \dfrac{\pi }{2}$
Now Ac current across the circuit is given by
$I = {I_o}\sin \left( {\omega t + \phi } \right)$
As per the question
$I = 10\sin \left( {\omega t} \right) \cdot \cdot \cdot \cdot \left( 2 \right)$
Where ${I_o} = 10A$
And $\phi = 0^\circ $
We know that,
Power dissipated in an ac circuit
$P = {V_o}{I_o}\cos \phi \cdot \cdot \cdot \cdot \left( 3 \right)$
Where, $\cos \phi $ is to power factor and $\phi $ be the phase difference
From equation $\left( 1 \right)\,\,and\,\,\left( 2 \right)$, we get
$\phi = \dfrac{\pi }{2}$
Where $\phi $ is the phase difference between V and I
Now putting the value ${V_o}\,\,and\,\,{I_o}$ respectively in equation $\left( 3 \right)$,we get
$P = 230 \times 10 \times \cos \dfrac{\pi }{2}$
$ \Rightarrow P = 0W$
The SI unit of power is watt.
Therefore, the power dissipated in an ac circuit will be $zero\,W$.
Additional Information:
$V$=Instantaneous value of the voltage
$I$=Instantaneous value of the current
${V_o}$= Amplitude of voltage
${I_o}$= Amplitude of current
Note: Generally there will be mistakes while putting the value of V and I in the power dissipated formula. Many of us put the instantaneous value of voltage and current instead of putting the amplitude of voltage and current respectively.
Complete step by step solution:
According to the problem,
Ac voltage across the circuit is given by
$V = {V_o}\sin \left( {\omega t + \phi } \right)$
As per the question
$V = 230\sin \left( {\omega t + \dfrac{\pi }{2}} \right) \cdot \cdot \cdot \cdot \left( 1 \right)$
Where ${V_o} = 230V$
And $\phi = \dfrac{\pi }{2}$
Now Ac current across the circuit is given by
$I = {I_o}\sin \left( {\omega t + \phi } \right)$
As per the question
$I = 10\sin \left( {\omega t} \right) \cdot \cdot \cdot \cdot \left( 2 \right)$
Where ${I_o} = 10A$
And $\phi = 0^\circ $
We know that,
Power dissipated in an ac circuit
$P = {V_o}{I_o}\cos \phi \cdot \cdot \cdot \cdot \left( 3 \right)$
Where, $\cos \phi $ is to power factor and $\phi $ be the phase difference
From equation $\left( 1 \right)\,\,and\,\,\left( 2 \right)$, we get
$\phi = \dfrac{\pi }{2}$
Where $\phi $ is the phase difference between V and I
Now putting the value ${V_o}\,\,and\,\,{I_o}$ respectively in equation $\left( 3 \right)$,we get
$P = 230 \times 10 \times \cos \dfrac{\pi }{2}$
$ \Rightarrow P = 0W$
The SI unit of power is watt.
Therefore, the power dissipated in an ac circuit will be $zero\,W$.
Additional Information:
$V$=Instantaneous value of the voltage
$I$=Instantaneous value of the current
${V_o}$= Amplitude of voltage
${I_o}$= Amplitude of current
Note: Generally there will be mistakes while putting the value of V and I in the power dissipated formula. Many of us put the instantaneous value of voltage and current instead of putting the amplitude of voltage and current respectively.
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