Question

In an L-R-C series circuit, R = 300 Ω, L = 0.400 H, and. When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 A. (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

Short answer

The voltage amplitude of the source is 150V, The voltage amplitude across resistor is 150V and that across inductor and capacitor is 322.75 volt each. Also average power supplied by the source is 37.5W.

Step-by-step solution

Step-1: Formulas used  

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

$\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C})}\right)}^2}$, where Z is the impedance

The equivalent Ohm’s law relation to get the amplitude voltage V in the circuit.

V=IZ

Similarly, the amplitude voltage across the resistor, capacitor and inductor is found by the relation

V=IX

Where X is the reactance which is equal to${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{\omega L}$ for inductor and${\mathit{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega }\mathbf{C}}$ for capacitor andRfor resistor.

Also ${\mathbf{P}}_{\mathbf{av}}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{Z}$, where Z is the impedance.

The process of peaking the current at a particular frequency represents the resonance and that frequency is known as resonant frequency.

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}$,whereis the resonant angular frequency.

At resonance,${\mathbf{X}}_{\mathbf{L}}\mathbf{,}{\mathbf{X}}_{\mathbf{C}}$.

At resonance, the impedance is minimum and this minimum value is same as that of resistor

Step-2: Calculations for Voltage amplitude across the source

The current I in the circuit is 0.5A.

$$\begin{gathered} Z=R \\ =300\Omega \end{gathered}$$

Now plug the values for I and Z to get V

$$\begin{gathered} V=\left(0.500A\right)\left(300\Omega \right) \\ =150V \end{gathered}$$

Therefore, the voltage amplitude of the source is 150V.

Step-3: Calculations for voltage amplitude across resistor, capacitor and inductor

The voltage across the inductor and the capacitor have the same magnitude even they have different phase angles. So, they cancel each other and the voltage across the resistor could be calculated using Z=R.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{R}}=\left(0.500\mathrm{A}\right)\left(300\mathrm{\Omega }\right) \\ =150\mathrm{V} \\ {\mathrm{V}}_{\mathrm{L}}=\left(6454.97\right)\left(0.400\right) \\ =322.75\mathrm{V} \\ {\mathrm{V}}_{\mathrm{C}}=\frac{1}{\left(6454.97\right)\left(6\times {10}^{-8}\right)} \\ =322.75\mathrm{V} \end{gathered}$$

Step-4: Calculations for average power supplied by the source

The average rate here represents the average power supplied to the resistor.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{av}}=\left(0.5{\mathrm{A}}^2\right)\left(300\mathrm{\Omega }\right) \\ =37.5\mathrm{W} \end{gathered}$$

Therefore, (a)The voltage across the source is 150V,(b)The voltage amplitude across resistor, capacitor and inductor are 150V, 322.75V and 322.75V respectively and (c)The average power supplied by the source is 37.5W.