Answer:
The maximum voltage is 41.92 V.
Explanation:
Given that,
Peak voltage = 590 volts
Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is [tex]1.20×10^{-2}\ \mu F[/tex].
We need to calculate the resonance frequency
Using formula of 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=2356.88\ 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\times f\times C\times V_{c}[/tex]
[tex]I=2\pi\times2356.88\times1.20\times10^{-8}\times590[/tex]
[tex]I=0.1048\ A[/tex]
Impedance of the circuit is
[tex]z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}[/tex]
At resonance frequency [tex]X_{L}=X_{C}[/tex]
[tex]Z=R[/tex]
We need to calculate the maximum voltage
Using ohm's law
[tex]V=I\times R[/tex]
[tex]V=0.1048\times400[/tex]
[tex]V=41.92\ V[/tex]
Hence, The maximum voltage is 41.92 V.
