Answer:
The maximum voltage across the inductor is [tex]V_i = 0.136 V[/tex]
Explanation:
From the question we are told that
The amplitude of the power is [tex]E_o = 180.0V[/tex]
The resistance is [tex]R = 90 \Omega[/tex]
The capacitance is [tex]C= 5.4 F[/tex]
The inductance is [tex]L = 25.0mH = 25 *10^{-3} H[/tex]
According to the question the the circuit frequency is resonance frequency
At resonance frequency
capacitive Reactance is equal to the Inductive Reactance
The capacitive Reactance is mathematically represented as
[tex]X_c = \frac{1}{wC}[/tex]
Where [tex]w = \frac{1}{\sqrt{LC} }[/tex]
Substituting values
[tex]w = \frac{1}{\sqrt{5.4 * 25 *10^{-2}} }[/tex]
[tex]w = \frac{1}{\sqrt{5.4 * 25 *10^{-2}} }[/tex]
[tex]w = 2.72\ rad /s[/tex]
So [tex]X_c = \frac{1}{2.72 * 5.4 }[/tex]
[tex]X_c =0.068[/tex]
The inductive Reactance is mathematically represented as
[tex]X_L = wL[/tex]
Substituting values
[tex]X_L = 2.72 * 25*10^{-3}[/tex]
[tex]X_L =0.068[/tex]
The impedance of the circuit is mathematically represented as
[tex]z = \sqrt{(X_L -X_c) ^2 + (R)^2}[/tex]
Substituting values
[tex]z = \sqrt{(0.068 - 0.068) ^2 + (90)^2}[/tex]
[tex]z = 90[/tex]
The maximum current supplied to the circuit is
[tex]I_{max} = \frac{E_o}{z}[/tex]
So [tex]I_{max} = \frac{180}{90}[/tex]
=> [tex]I_{max} = 2A[/tex]
Now the maximum voltage across the inductor is
[tex]V_i = I_{max} * X_L[/tex]
So [tex]V_i =2 * 0.068[/tex]
=> [tex]V_i = 0.136 V[/tex]
