Answer
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Hint: Learn about the power dissipated through an AC circuit. The power dissipated from a circuit is given by the product of the voltage and current through the circuit. For an AC circuit the power dissipated is the average taken over a period of time
Formula used:
The average power dissipated from an AC circuit is given by,
\[P = \dfrac{1}{2}{V_0}{I_0}\cos \varphi \]
where, \[{V_0}\] is the amplitude of the voltage applied \[{I_0}\] is the amplitude of the current through the circuit \[\varphi \] is the phase difference between the current and the voltage.
Complete step by step answer:
We have given here the voltage of the circuit is \[v = 230\sin (\omega t + \dfrac{\pi }{2})\] where, amplitude of the voltage is \[{V_0} = 230V\] , \[\omega \] is the frequency of the source applied \[\dfrac{\pi }{2}\] is the initial phase of the voltage. The current through the circuit is given by, \[I = 20\sin \omega t\] where, \[{I_0} = 20A\] is the amplitude of the current through the circuit\[\omega \] is the frequency of the source applied. So, the phase difference between the current and voltage is \[\varphi = \dfrac{\pi }{2}\].
Now, we know that the average power dissipated from an AC circuit is given by,
\[P = \dfrac{1}{2}{V_0}{I_0}\cos \varphi \]
where, \[{V_0}\] is the amplitude of the voltage applied \[{I_0}\] is the amplitude of the current through the circuit \[\varphi \] is the phase difference between the current and the voltage.
So, here we have, amplitude of the voltage is \[{V_0} = 230V\] , amplitude of the current through the circuit \[{I_0} = 20A\]and phase difference between the current and the voltage is \[\varphi = \dfrac{\pi }{2}\]. Hence putting the values we have,
\[P = \dfrac{1}{2}200 \times 20 \times \cos \dfrac{\pi }{2}\]
\[\Rightarrow P = \dfrac{1}{2}200 \times 20 \times 0\]
\[\therefore P = 0\]
Hence, the average power dissipated through the circuit is \[0\,watt\].
Note: In the given circuit the maxima of voltage meets the minima of the current so the average value of the power becomes zero for a full circle. The average power dissipated through the circuit is zero does not mean that no energy is being exerted by the source in the circuit.
Formula used:
The average power dissipated from an AC circuit is given by,
\[P = \dfrac{1}{2}{V_0}{I_0}\cos \varphi \]
where, \[{V_0}\] is the amplitude of the voltage applied \[{I_0}\] is the amplitude of the current through the circuit \[\varphi \] is the phase difference between the current and the voltage.
Complete step by step answer:
We have given here the voltage of the circuit is \[v = 230\sin (\omega t + \dfrac{\pi }{2})\] where, amplitude of the voltage is \[{V_0} = 230V\] , \[\omega \] is the frequency of the source applied \[\dfrac{\pi }{2}\] is the initial phase of the voltage. The current through the circuit is given by, \[I = 20\sin \omega t\] where, \[{I_0} = 20A\] is the amplitude of the current through the circuit\[\omega \] is the frequency of the source applied. So, the phase difference between the current and voltage is \[\varphi = \dfrac{\pi }{2}\].
Now, we know that the average power dissipated from an AC circuit is given by,
\[P = \dfrac{1}{2}{V_0}{I_0}\cos \varphi \]
where, \[{V_0}\] is the amplitude of the voltage applied \[{I_0}\] is the amplitude of the current through the circuit \[\varphi \] is the phase difference between the current and the voltage.
So, here we have, amplitude of the voltage is \[{V_0} = 230V\] , amplitude of the current through the circuit \[{I_0} = 20A\]and phase difference between the current and the voltage is \[\varphi = \dfrac{\pi }{2}\]. Hence putting the values we have,
\[P = \dfrac{1}{2}200 \times 20 \times \cos \dfrac{\pi }{2}\]
\[\Rightarrow P = \dfrac{1}{2}200 \times 20 \times 0\]
\[\therefore P = 0\]
Hence, the average power dissipated through the circuit is \[0\,watt\].
Note: In the given circuit the maxima of voltage meets the minima of the current so the average value of the power becomes zero for a full circle. The average power dissipated through the circuit is zero does not mean that no energy is being exerted by the source in the circuit.
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