Question

The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude Vmax can the source have if the maximum capacitor voltage is not exceeded?

Answer 1

The maximum voltage is 41.92 V.

Step-by-step 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
`1.20×10^(-2)\ \mu F`.

We need to calculate the resonance frequency

Using formula of frequency

`f=(1)/(2\pi√(LC))`

Put the value into the formula

`f=\frac{1}{2\pi\sqrt{0.380*1.20*10^(-8)}}`

`f=2356.88\ Hz`

We need to calculate the maximum current

Using formula of current

`I=(V_(c))/(X_(c))`

`I=2\pi* f* C* V_(c)`

`I=2\pi*2356.88*1.20*10^(-8)*590`

`I=0.1048\ A`

Impedance of the circuit is

`z=\sqrt{R^2+(X_(L)^2-X_(C)^2)}`

At resonance frequency
`X_(L)=X_(C)`

`Z=R`

We need to calculate the maximum voltage

Using ohm's law

`V=I* R`

`V=0.1048*400`

`V=41.92\ V`

Hence, The maximum voltage is 41.92 V.