Answer:
Explanation:
Given: i. in the AC series circuit, V = √2 [tex]V_{0}[/tex]cos(ωt)], L and R.
ii. in the DC circuit, R, L and [tex]V_{0}[/tex].
1. To show that average power through resistor in the AC circuit is the same in the DC circuit, when L = 0.
For the AC circuit,
Average power = [tex]I^{2}[/tex]Z
where: I is the current flowing and Z is the impedance.
Impedance, Z = [tex]\sqrt{R^{2} + X_{L} ^{2} }[/tex]
But, [tex]X_{L}[/tex] = 2[tex]\pi[/tex]fL
⇒ [tex]X_{L}[/tex] = 0 (∵ L = 0)
So that, Impedance, Z = R
Thus, average power through the resistor in the AC circuit = [tex]I^{2}[/tex]R
For the DC circuit,
Average power = [tex]I^{2}[/tex]R
Therefore,
the average power through the resistor in the AC circuit = the average power through the resistor in the DC circuit = [tex]I^{2}[/tex]R, when L = 0.
2. In the AC circuit, the expression for the peak current, [tex]I_{0}[/tex] can be determined by:
[tex]V_{0}[/tex] = [tex]I_{0}[/tex] × Z
[tex]I_{0}[/tex] = [tex]\frac{V_{0} }{Z}[/tex]
[tex]I_{0}[/tex] = [tex]\frac{V_{0} }{\sqrt{R^{2} + (wL)^{2} } }[/tex]
3. The frequency that would produce average power in the AC circuit is;
Average power = [tex]I^{2}[/tex]Z
Average power = [tex]I^{2}[/tex]([tex]\sqrt{R^{2} + (wL)^{2} }[/tex]
Square both sides to have;
[tex](Average power)^{2}[/tex] = [tex]I^{4}[/tex] ([tex]R^{2}[/tex] + [tex]w^{2}[/tex][tex]L^{2}[/tex])
[tex](Average power)^{2}[/tex] - [tex]I^{4}[/tex][tex]R^{2}[/tex] = [tex]w^{2}[/tex][tex]L^{2}[/tex]
⇒ [tex]w^{2}[/tex] = [ [tex](Average power)^{2}[/tex] - [tex]I^{4}[/tex][tex]R^{2}[/tex]] ÷ [tex]L^{2}[/tex]
ω = [tex]\sqrt{}[/tex]{[ [tex](Average power)^{2}[/tex] - [tex]I^{4}[/tex][tex]R^{2}[/tex]] ÷ [tex]L^{2}[/tex]}
Therefore, frequency, ω = [tex]\sqrt{}[/tex]{[ [tex](Average power)^{2}[/tex] - [tex]I^{4}[/tex][tex]R^{2}[/tex]] ÷ [tex]L^{2}[/tex]} is required to produce average power in the resistor in the AC circuit compared to the DC circuit.
