he capacitor can withstand a peak voltage of 550 VV . If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

Suppose, In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is [tex]1.20\times10^{-2}\mu F[/tex]

We need to calculate the resonant frequency

Using formula of resonant frequency

[tex]f=\dfrac{1}{2\pi\sqrt{LC}}[/tex]

Put the value into the formula

[tex]f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}[/tex]

[tex]f=2356.8\ Hz[/tex]

We need to calculate the maximum current

Using formula of current

[tex]I=\dfrac{V_{c}}{X_{c}}[/tex]

[tex]I=2\pi f C\times V_{c}[/tex]

Put the value into the formula

[tex]I=2\pi\times2356.8\times1.20\times10^{-8}\times550[/tex]

[tex]I=0.0977\ A[/tex]

We need to calculate the impedance of the circuit

Using formula of impedance

[tex]Z=\sqrt{R^2+(X_{L}-X_{C})^2}[/tex]

At resonant frequency , [tex]X_{L}=X_{C}[/tex]

So, Z = R

We need to calculate the maximum voltage

Using formula of voltage

[tex]V=IR[/tex]

Put the value into the formula

[tex]V=0.0977\times400[/tex]

[tex]V=39.08\ V[/tex]

Hence, The maximum voltage is 39.08 V.