Question

A parallel-plate capacitor having square plates 4.50 cm on each side and 8.00 mm apart is placed in series with the following: an ac source of angular frequency 650 rad/s and voltage amplitude 22.5 V; a 75.0-Ω resistor; and an ideal solenoid that is 9.00 cm long, has a circular cross section 0.500 cm in diameter, and carries 125 coils per centimeter. What is the resonance angular frequency of this circuit?

Short answer

The resonance angular frequency of this circuit is $3.6\times {10}^7\mathrm{rad}/\mathrm{s}$.

Step-by-step solution

Formula of capacitance, inductance and concept of resonant angular frequency

Capacitance of a parallel plate combination is given by the formula, $C=\frac{{\epsilon }_{0}A}{d}$, where C is the capacitance, ${\epsilon }_0$ is the permittivity of free space,A is the area of the plates, d is the distance between the plates.

Inductance of a solenoid is given by the formula $L=\frac{{\mu }_0{N}^{2}A}{I}$, where, L is the inductance,N is the number of turns in the solenoid, A is the area of cross section of the solenoid, I is the length of the solenoid.

Resonant angular frequency of an A/C circuit is the angular frequency corresponding to which, the current in the circuit is maximum, or in other words the resistance is the circuit is minimum.The formula for resonant frequency is given by,$\omega =\frac{1}{\sqrt{LC}}$

Calculation of resonant angular frequency

We just discussed that, the capacitance is calculated by the formula,$C=\frac{{\epsilon }_{0}A}{d}$ . Given that,

$$\begin{gathered} {\mathrm{\epsilon }}_0=8.85\times {10}^{-12}{\mathrm{m}}^{-3}{\mathrm{kg}}^{-1}{\mathrm{s}}^4{\mathrm{A}}^2 \\ \mathrm{A}={\left(4.5\times {10}^{-2}\right)}^2{\mathrm{m}}^2 \\ \mathrm{d}=8\times {10}^{-3}\mathrm{m} \end{gathered}$$

Therefore that value of the capacitance is-

$$\begin{gathered} \mathrm{C}=\frac{8.85\times {10}^{-12}\times {\left(4.5\times {10}^{-2}\right)}^2}{8\times {10}^{-3}} \\ =22.4\times {10}^{-13}\mathrm{F} \end{gathered}$$

Similarly, for the inductor,$L=\frac{{\mu }_0{N}^{2}A}{I}$ . The given parameters are-

$$\begin{gathered} {\mu }_0=12.57\times {10}^{-7}H/m \\ N=125\times 9 \\ =1125 \\ A=\mathrm{\pi }{\left(\frac{0.5}{\mathrm{A}}\times {10}^{-2}\right)}^2 \\ =1.96\times {10}^{-5}{\mathrm{m}}^2 \\ \mathrm{I}=9\times {10}^{-2}\mathrm{m} \end{gathered}$$

Therefore, the value of inductance is-

$$\begin{gathered} L=\frac{12.57\times {10}^{-7}\times {\left(1125\right)}^2\times 1.96\times {10}^{-5}}{9\times {10}^{-2}} \\ =3.45\times {10}^{-4}H \end{gathered}$$

Hence the resonant frequency can be calculated using the formula as mentioned above-

$$\begin{gathered} \mathrm{\omega }=\frac{1}{\sqrt{\mathrm{LC}}} \\ =\frac{1}{\sqrt{3.45\times {10}^{-4}\times 22.4\times {10}^{-13}}} \\ =3.6\times {10}^7\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the resonant angular frequency of the circuit is$3.6\times {10}^7\mathrm{rad}/\mathrm{s}$