A series RLC circuit consists of a $50\Omega$resistor, a localid="1650491262241" $3.3mH$inductor, and a $480nF$capacitor. It is connected to a localid="1650491276532" $5.0kHz$oscillator with a peak voltage of localid="1650491288645" $5.0V$. What is the instantaneous current $i$when

a. $\epsilon ={\epsilon }_0$?

b. $\epsilon =0V$and is decreasing?

We are given that,

Resistance of resistor is, $R=50\Omega$

Inductance of inductor is,localid="1650491308818" $L=3.3mH$

Capacitance of capacitor is, $C=480nF$

Frequency is,localid="1650491328746" $f=5.0kHz$

Peak voltage is, localid="1650491343315" $V=5.0V$

And$\epsilon ={\epsilon }_0$

We know that inductive reactance is,

$$\begin{gathered} X_L=2\mathrm{\pi fL} \\ =2\times 3.14\times 3\times {10}^3\times 3.3\times {10}^{-3} \\ =62\mathrm{\Omega } \end{gathered}$$

And capacitive reactance is,

$$\begin{gathered} X_C=\frac{1}{2\mathrm{\pi fC}} \\ =\frac{1}{2\times 3.14\times 3\times {10}^3\times 480\times {10}^{-9}} \\ =110.5\mathrm{\Omega } \end{gathered}$$

And phase angle is,

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right) \\ ={\tan }^{-1}\left(\frac{62-110.5}{50}\right)=-44° \end{gathered}$$

Peak current is,

$$\begin{gathered} I=\frac{{\epsilon }_0}{\sqrt{R^2+{\left(X_L-X_C\right)}^2}} \\ =\frac{5}{\sqrt{{50}^2+{\left(62-110.5\right)}^2}}=72mA \end{gathered}$$

As instantaneous emf is given as,

$\epsilon ={\epsilon }_0\cos \left(\omega t\right)$

Now, for$\epsilon ={\epsilon }_0$

Equation becomes,

$$\begin{gathered} \epsilon =\epsilon \cos \left(\omega t\right) \\ 1=\cos \left(\omega t\right) \\ {\cos }^{-1}\left(1\right)=\omega t \\ \omega t=0° \end{gathered}$$

Instantaneous current is,

$$\begin{gathered} i=I\cos \left(\omega t-\varphi \right) \\ =72\times {10}^{-3}\cos \left(0°-\left(-44°\right)\right) \\ =52mA \end{gathered}$$

We are given that,$\epsilon =0V$and is decreasing

We have, $\epsilon ={\epsilon }_0\cos \left(\omega t\right)$

At $\epsilon =0V$

Equation becomes,

$$\begin{gathered} 0=\epsilon \cos \left(\omega t\right) \\ \cos \left(\omega t\right)=0 \\ \omega t={\cos }^{-1}\left(0\right)=90° \end{gathered}$$

Therefore instantaneous current is,

$$\begin{gathered} i=I\cos \left(\omega t-\varphi \right) \\ =72\times {10}^{-3}\cos \left(90°-\left(-44°\right)\right) \\ =72\times {10}^{-3}\cos \left(90°+44°\right)=-72\times {10}^{-3}\sin \left(44°\right) \\ =-50mA \end{gathered}$$