91Ó°ÊÓ

i. Frequency $f=400Hz$

ii. Maximum voltage $n=90.0V$

iii. Resistance $R=20.0\mathrm{Î\copyright }$

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

v. Capacitance$C=12.1\mathrm{μ}F=12.1×{10}^{-6}F$

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-$X_c=1/2\mathrm{Ï€´Ú³¦}$

.

Capacitive inductance-

$X_L=2\mathrm{Ï€´Ú³¢}$

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.

The capacitive reactance is,

$X_c=\frac{1}{2\mathrm{Ï€´Ú³¦}}$

$$\begin{gathered} X_c=\frac{1}{2\mathrm{Ï€}\left(400\mathrm{Hz}\right)\left(12.1×{10}^{-6}\mathrm{F}\right)} \\ {X}_c=32.88\mathrm{Î\copyright } \end{gathered}$$

Now, the inductive reactance is,

$$\begin{gathered} X_L=2\mathrm{Ï€´Ú³¢} \\ {\mathrm{X}}_{\mathrm{L}}=2\mathrm{Ï€}\left(400\mathrm{Hz}\right)\left(24.2×{10}^{-3}\mathrm{H}\right) \\ {\mathrm{X}}_{\mathrm{L}}=60.82\mathrm{Î\copyright } \end{gathered}$$

Thus, the impedance is,

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

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

The current is given by,

$I=\frac{n}{Z}=\frac{90.0V}{34.36\mathrm{Î\copyright }}$

$I=2.62A$

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,

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

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

The RMS potential difference across the capacitor is given by,

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

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

Therefore, the rms potential difference across the capacitor is .

TheRMS potential differenceacross the inductor is given by,

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

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

Therefore, the rms potential difference across the inductor is .

Theaverage rate of energy dissipation is given by,

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

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

$P_{avg}=68.6W$

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