Question

An RLC circuit is connected to a AC power supply. The amplitude of the power supply is 180.0V. To determine the maximum voltage across any element of the circuit one must determine the frequency at which it occurs. For the resistor, it is the resonance frequency. But what about the inductor or capacitor? Determine the maximum amplitude of the voltage across the inductor if R=90.0 ohms, C=5.4F, and L=25.0mH.

Answer 1

The maximum voltage across the inductor is
`V_i = 0.136 V`

Step-by-step explanation:

From the question we are told that

The amplitude of the power is
`E_o = 180.0V`

The resistance is
`R = 90 \Omega`

The capacitance is
`C= 5.4 F`

The inductance is
`L = 25.0mH = 25 *10^(-3) H`

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

`X_c = (1)/(wC)`

Where
`w = (1)/(√(LC) )`

Substituting values

`w = \frac{1}{\sqrt{5.4 * 25 *10^(-2)} }`

`w = \frac{1}{\sqrt{5.4 * 25 *10^(-2)} }`

`w = 2.72\ rad /s`

So
`X_c = (1)/(2.72 * 5.4 )`

`X_c =0.068`

The inductive Reactance is mathematically represented as

`X_L = wL`

Substituting values

`X_L = 2.72 * 25*10^(-3)`

`X_L =0.068`

The impedance of the circuit is mathematically represented as

`z = √((X_L -X_c) ^2 + (R)^2)`

Substituting values

`z = √((0.068 - 0.068) ^2 + (90)^2)`

`z = 90`

The maximum current supplied to the circuit is

`I_(max) = (E_o)/(z)`

So
`I_(max) = (180)/(90)`

=>
`I_(max) = 2A`

Now the maximum voltage across the inductor is

`V_i = I_(max) * X_L`

So
`V_i =2 * 0.068`

=>
`V_i = 0.136 V`