Answer
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Hint: In an alternating current circuit, values of both electromotive force and current change continuously with respect to time. So, we cannot calculate the power directly. the average power for the sinusoidal function will be obtained by averaging over one period. The average power of an alternating current circuit is also known as the true power of the electrical circuit.
First we calculate root mean square (rms) values of current and voltage. Then we find the phase difference between them. Phase difference is denoted by symbol. In question phase difference is \[\dfrac{\pi }{3}\]in radian or\[{60^0}\].
Formula used: We know that the power is equal to the product of voltage and time. We calculate average power by the formula \[{P_{av}} = {V_{rms}} \times {i_{rms}}\cos \phi \].
Complete step by step solution:
Given: \[i = \dfrac{1}{{\sqrt 2 }}\sin \left( {100\pi t} \right)Ampere\] and \[e = \dfrac{1}{{\sqrt 2 }}\sin \left( {100\pi t + \dfrac{\pi }{3}} \right)Volt\]
i.e., \[{i_0} = \dfrac{1}{{\sqrt 2 }}\] and \[{V_0} = \dfrac{1}{{\sqrt 2 }}\].
Therefore, average power consumed in the circuit is given by
\[{P_{av}} = {V_{rms}} \times {i_{rms}}\cos \phi \].
\[{P_{av}} = \dfrac{1}{2} \times \dfrac{1}{2} \times \cos {60^0}\]
We know that \[{V_{rms}} = \dfrac{1}{2}\], \[{i_{rms}} = \dfrac{1}{2}\]and \[\phi = {60^0}\]
\[ \Rightarrow {P_{av}} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}W\]
\[\therefore {P_{av}} = \dfrac{1}{8}W\]
Hence, Average power consumed in AC circuit is \[\therefore {P_{av}} = \dfrac{1}{8}W\].
Additional information: For AC circuits, the instantaneous power is constantly changing with varying amounts of time. The curve of power vs time is sinusoidal. As we get this type of curve i.e., it makes it difficult to measure power. Calculating average or mean value of the power, It is therefore more convenient to calculate. The average value of the instantaneous power over a fixed number of cycles, the average power of sinusoidal curve is given simply as: \[{P_{av}} = {V_{rms}} \times {i_{rms}}\cos \phi \].
Note: Students must be careful to calculate root mean square values of current and voltage. They must be careful to find phase difference between voltage and current. In question rms values are given. Don’t put \[\dfrac{1}{{\sqrt 2 }}\]as \[{V_{rms}}\] and \[{i_{rms}}\]in formula. Here \[\dfrac{1}{{\sqrt 2 }}\]is \[{i_0}\] for current and \[\dfrac{1}{{\sqrt 2 }}\]is \[{V_0}\].
First we calculate root mean square (rms) values of current and voltage. Then we find the phase difference between them. Phase difference is denoted by symbol. In question phase difference is \[\dfrac{\pi }{3}\]in radian or\[{60^0}\].
Formula used: We know that the power is equal to the product of voltage and time. We calculate average power by the formula \[{P_{av}} = {V_{rms}} \times {i_{rms}}\cos \phi \].
Complete step by step solution:
Given: \[i = \dfrac{1}{{\sqrt 2 }}\sin \left( {100\pi t} \right)Ampere\] and \[e = \dfrac{1}{{\sqrt 2 }}\sin \left( {100\pi t + \dfrac{\pi }{3}} \right)Volt\]
i.e., \[{i_0} = \dfrac{1}{{\sqrt 2 }}\] and \[{V_0} = \dfrac{1}{{\sqrt 2 }}\].
Therefore, average power consumed in the circuit is given by
\[{P_{av}} = {V_{rms}} \times {i_{rms}}\cos \phi \].
\[{P_{av}} = \dfrac{1}{2} \times \dfrac{1}{2} \times \cos {60^0}\]
We know that \[{V_{rms}} = \dfrac{1}{2}\], \[{i_{rms}} = \dfrac{1}{2}\]and \[\phi = {60^0}\]
\[ \Rightarrow {P_{av}} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}W\]
\[\therefore {P_{av}} = \dfrac{1}{8}W\]
Hence, Average power consumed in AC circuit is \[\therefore {P_{av}} = \dfrac{1}{8}W\].
Additional information: For AC circuits, the instantaneous power is constantly changing with varying amounts of time. The curve of power vs time is sinusoidal. As we get this type of curve i.e., it makes it difficult to measure power. Calculating average or mean value of the power, It is therefore more convenient to calculate. The average value of the instantaneous power over a fixed number of cycles, the average power of sinusoidal curve is given simply as: \[{P_{av}} = {V_{rms}} \times {i_{rms}}\cos \phi \].
Note: Students must be careful to calculate root mean square values of current and voltage. They must be careful to find phase difference between voltage and current. In question rms values are given. Don’t put \[\dfrac{1}{{\sqrt 2 }}\]as \[{V_{rms}}\] and \[{i_{rms}}\]in formula. Here \[\dfrac{1}{{\sqrt 2 }}\]is \[{i_0}\] for current and \[\dfrac{1}{{\sqrt 2 }}\]is \[{V_0}\].
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