Question

A series RLC circuit is driven by an alternating source at a frequency of400Hz and an emf amplitude of 90.0V. The resistance is$\mathbf{20}\mathbf{.}\mathbf{0}\mathit{\Omega }$, the capacitance is$\mathbf{12}\mathbf{.}\mathbf{1}\mathit{\mu }\mathit{F}$, and the inductance is$\mathbf{24}\mathbf{.}\mathbf{2}\mathit{m}\mathit{H}$. What is the RMS potential difference across (a) the resistor, (b) the capacitor, and (c) the inductor? (d) What is the average rate at which energy is dissipated?

Short answer

a. The RMS potential difference across the resistor is 37.0V.

b. The RMS potential difference across the capacitor is 60.9V.

c. The RMS potential difference across the inductor is 113V

d. The average rate at which energy is dissipated is 68.6W.

Step-by-step solution

Step 1: Given Information

i. Frequency $f=400Hz$

ii. Maximum voltage $n=90.0V$

iii. Resistance $R=20.0\Omega$

iv. Inductance localid="1663169536458" $L=24.2mH=24.2\times {10}^{-3}H$

v. Capacitance$C=12.1\mu F=12.1\times {10}^{-6}F$

Determining the concept

Use the formula of capacitive reactance and capacitive inductance to find the impendence of the circuit. Using this, find the RMS current,RMS potential difference across the resistor,RMS potential difference across the capacitor, the rms potential difference across the inductor, and the average rate at which energy is dissipated.

Capacitive reactance-$X_c=1/2\mathrm{\pi fc}$

.

Capacitive inductance-

$X_L=2\mathrm{\pi fL}$

Impedance of the circuit-

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

Current in the circuit-

$I=n/Z$

Rms value of current-

$I_{rms}=I/\sqrt{\text{2}}$

Where, V is volt, I is current, R is resistance, Z is impedance.

(a) Determining the rms potential difference across the resistor  

The capacitive reactance is,

$X_c=\frac{1}{2\mathrm{\pi fc}}$

$$\begin{gathered} X_c=\frac{1}{2\mathrm{\pi }\left(400\mathrm{Hz}\right)\left(12.1\times {10}^{-6}\mathrm{F}\right)} \\ {X}_c=32.88\Omega \end{gathered}$$

Now, the inductive reactance is,

$$\begin{gathered} X_L=2\mathrm{\pi fL} \\ {\mathrm{X}}_{\mathrm{L}}=2\mathrm{\pi }\left(400\mathrm{Hz}\right)\left(24.2\times {10}^{-3}\mathrm{H}\right) \\ {\mathrm{X}}_{\mathrm{L}}=60.82\mathrm{\Omega } \end{gathered}$$

Thus, the impedance is,

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

$$\begin{gathered} Z=\sqrt{{\left(20.0\Omega \right)}^2+\left[\left(60.82\Omega \right)-\left(32.88\Omega \right)\right]} \\ Z=34.36\Omega \end{gathered}$$.

The current is given by,

$I=\frac{n}{Z}=\frac{90.0V}{34.36\Omega }$

$I=2.62A$

Therefore,

$$\begin{gathered} I_{rms}=\frac{1}{\sqrt{2}}=\frac{2.62A}{\sqrt{2}} \\ {I}_{rms}=1.8526A \end{gathered}$$

Thus, the rms potential difference across the resistor is,

$$\begin{gathered} V_{rms}=\left(1.85A\right)\left(20.0\Omega \right) \\ {V}_{rms}=37.0V \end{gathered}$$.

Therefore, the rms value of potential difference across the resistor is 37.0V.

(b) Determining the rms potential difference across the capacitor

The RMS potential difference across the capacitor is given by,

$V_{Crms}=I_{rms}{X}_C$

$$\begin{gathered} V_{crms}=\left(1.85A\right)\left(32.88\Omega \right) \\ {V}_{crms}=60.9V \end{gathered}$$

Therefore, the rms potential difference across the capacitor is .

(c) Determining the rms potential difference across the inductor

TheRMS potential differenceacross the inductor is given by,

$V_{Lrms}=I_{rms}{X}_L$

$$\begin{gathered} V_{crms}=\left(1.85A\right)\left(60.82\Omega \right) \\ {V}_{crms}=113V \end{gathered}$$

Therefore, the rms potential difference across the inductor is .

(d) Determining the average rate at which energy is dissipated

Theaverage rate of energy dissipation is given by,

$P_{avg}={\left(I_{rms}\right)}^{2}R$

$P_{avg}={\left(1.8526A\right)}^2\left(20.0\Omega \right)$

$P_{avg}=68.6W$

Therefore, the average rate of energy dissipation is $68.6W$.