91影视

Resistance is a measure of opposition to the flow of current in a closed electrical circuit. It is measured in Ohm (鈩�).

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. When an inductor is attached to an AC supply, the resistance produced by it is called inductive reactance (X_L).

A capacitor is a device consisting of two conductors in close proximity that are used to store electrical energy. These conductors are insulated from each other. When a capacitor is attached to an AC supply, the resistance produced by it is called capacitive reactance (X_C).

Resistance of the resistor, R = 200 鈩�

The inductance of the coil, L = 0.900 H

Capacitance of capacitor, C = 6.00 碌F

The amplitude of voltage source, V = 30.0 V

Angular frequency of the source, 蝇 = 250 rad/s

Since the capacitor, inductor, and resistor are connected in series, Impedance of the circuit will be given by

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

Here, X_C is the inductive reactance that is given by

$\begin{array}{r}{X}_c=\frac{1}{蝇C}\\ =\frac{1}{(250\mathrm{rad}/\mathrm{s})\left(6.00鈭�{10}^{鈭�6}\mathrm{F}\right)}\\ =666.67\mathrm{惟}\end{array}$

And, X_L is the inductive reactance that is given by

role="math" localid="1668251678662" $\begin{array}{r}{X}_L=蝇L\\ =(250\mathrm{rad}/\mathrm{s})\left(0.9\mathrm{H}\right)\\ =225\mathrm{惟}\end{array}$

Putting the values in Z

$\begin{array}{r}Z=\sqrt{R^2+{\left(X_L鈭�X_C\right)}^2}\\ =\sqrt{{200}^2+(225鈭�666.67{)}^2}\\ =484.84\mathrm{惟}\end{array}$

Therefore, the impedance of the circuit is 484.84 鈩�

By using Ohm鈥檚 Law

$\begin{array}{r}V=I.Z\\ I=\frac{V}{Z}\end{array}$

It is given that voltage amplitude of the source is 30.0 V, so

$I=\frac{V}{Z}=\frac{30.0V}{484.84惟}=0.0619A$

Therefore, the amplitude of the current in circuit is 0.0619 A

The angle between voltage and current phasors in L-C-R circuit is given by

$\begin{array}{r}\tan 鈦�蠒=\frac{\left(X_L鈭�X_C\right)}{R}\\ \tan 鈦�蠒=\frac{(225鈭�666.67)\mathrm{惟}}{200\mathrm{惟}}\\ \tan 鈦�蠒=鈭�2.2084\\ 蠒=鈭�{65.64}^{鈭�}\end{array}$

Therefore, the phase angle of the voltage source with respect to current is -65.64掳 which means voltage lags the current as the phase angle is negative.

The instantaneous voltage across resistor:

$\begin{array}{r}{V}_R=IR\\ =\left(0.0619A\right)\left(200\mathrm{惟}\right)\\ =12.38\mathrm{V}\end{array}$

As the voltage across resistor is in phase with current, it can be represented as:

$v_R=V_{R}cos鈦�蝇t=12.38cos鈦�\left[\right(250rad/s\left)t\right]$

The instantaneous voltage across inductor:

$\begin{array}{r}{V}_L=I{X}_L\\ =\left(0.0619\mathrm{A}\right)\left(225\mathrm{惟}\right)\\ =13.928\mathrm{V}\end{array}$

As the voltage across inductor leads the current, it can be represented as:

$v_L=-V_{L}sin鈦�蝇t=-13.928sin鈦�\left[\right(250rad/s\left)t\right]$

Voltage drop across capacitor is given by

$\begin{array}{r}{V}_C=L{X}_{\mathrm{C}}\\ =\left(0.0619A\right)\left(666.67\mathrm{惟}\right)\\ =41.267\mathrm{V}\end{array}$

As the voltage across capacitor lags the current, it can be represented as:

$v_C=V_{C}sin鈦�蝇t=41.267sin鈦�\left[\right(250rad/s\left)t\right]$

Thus, the summation of all the value gives at t = 20 ms can be given as:

$\begin{array}{r}{v}_R+v_L+v_C=12.38\cos 鈦�\left[(250\mathrm{rad}/\mathrm{s})\left({20}^{鈭�}{10}^{鈭�3}\right)\right]鈭�13.928\sin 鈦�\left[(250\mathrm{rad}/\mathrm{s})\left(20鈭�{10}^{鈭�3}\right)\right]\\ +41.267\sin 鈦�\left[(250\mathrm{rad}/\mathrm{s})\left({20}^{鈭�}{10}^{鈭�3}\right)\right]\\ =3.51+13.35鈭�39.55\\ =鈭�22.70\mathrm{V}\end{array}$

Therefore, the net instantaneous voltage across resistor, inductor, and capacitor is -22.70 volts.

The instantaneous supply voltage at t = 20 ms can be given as:

$\begin{array}{r}v=V\cos 鈦�[蝇t+蠒]\\ =30\cos 鈦�\left[250鈭�20鈭�{10}^{鈭�3}鈭�65.64鈭�\frac{蟺}{180}\right]\\ =鈭�22.7\mathrm{V}\end{array}$

From step 4 and 5 we can conclude that

$v_R+v_L+v_C=v$

Therefore, on comparing the instantaneous supply voltage to the summation of instantaneous voltage across individual electric components, we can conclude that both of them are equal.

From step 5, amplitudes of voltage across each electric component can be listed as:

$\begin{array}{l}{V}_R=12.4V\\ V_L=139V\\ V_c=413V\end{array}$

The summation of all these voltages:

$$\begin{gathered} V_R+V_L+V_c=124+139+41.3 \\ =67.6V \end{gathered}$$

Thus, we can conclude that

$$\begin{gathered} V_R+V_t+V鈮�V \\ 67.6V鈮�30V \end{gathered}$$

Therefore, on comparing the amplitude of voltage of the source to the summation of amplitude of voltage across individual electric components, we can conclude that both of them are not equal.