Question

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?

Answer 1

Answer:

The maximum voltage is 39.08 V.

Explanation:

Given that,

Voltage = 550 V

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

We need to calculate the resonant frequency

Using formula of resonant frequency

$f=\dfrac{1}{2\pi\sqrt{LC}}$

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.8\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi f C\times V_{c}$

Put the value into the formula

$I=2\pi\times2356.8\times1.20\times10^{-8}\times550$

$I=0.0977\ A$

We need to calculate the impedance of the circuit

Using formula of impedance

$Z=\sqrt{R^2+(X_{L}-X_{C})^2}$

At resonant frequency , $X_{L}=X_{C}$

So, Z = R

We need to calculate the maximum voltage

Using formula of voltage

$V=IR$

Put the value into the formula

$V=0.0977\times400$

$V=39.08\ V$

Hence, The maximum voltage is 39.08 V.