Question

Need help ASAP. PLEASE!!!

Answer 1

Answer:

All the calculations are shown in the explanation

Explanation:

RLC Circuit

The circuit proposed in the problem consists in one resistor R in series with the parallel of a capacitor C and an inductor L. All the impedances, voltages, currents and powers must be expressed as complex numbers since they all have an active and a reactive component. The formulas are very similar to those of the Ohm's law, as will be shown below.

The source has a time function expressed as

$\displaystyle e=3.4\ sin\ 10,000t$

We must find the RMS voltage as

$\displaystyle v_e=\ \frac{3.4}{\sqrt{2}}=2.4042\ V$

The given parameters of the circuit are

$\displaystyle R=47\Omega$

$\displaystyle L=3.3mH=0.0033H$

$\displaystyle C=3.3\mu\ F=3.3\ 10^{-6}\ F$

$\displaystyle w=10,000\ rad/s$

(a)

Let's find the reactances

$\displaystyle X_L=wL=10,000(0.0033)$

$\displaystyle X_L=33\Omega$

$\displaystyle X_C=\frac{1}{wC}=\frac{1}{10,000(3.3150)}=30.30\Omega$

Now the impedances are

$\displaystyle Z_R=47+j0$

$\displaystyle Z_L=j33$

$\displaystyle Z_C=-j30.03$

The equivalent impedance of the parallel of the capacitor and the inductor is

$\displaystyle Z_{eq1}=\frac{Z_L\ Z_C}{Z_L+Z_C}=\frac{(j33)(-j30.03)}{j33-j3003}=\frac{1000}{2.9697}=-j336.73\Omega$

Computing the total impedance of the circuit

$\displaystyle Z_t=Z_R+Z{eq1}=47+j0-j336.73$

$\displaystyle Z_t=47-j336.73$

Converting to phasor form

$\displaystyle Z_t=(340,-82.054^o)\Omega$

The given voltage of the source is

$\displaystyle V_s=(2.4042,0^o)\ V$

It has an angle of 0 degrees since it's the reference. Let's compute the total current of the circuit

$\displaystyle I=\frac{V_s}{Z_t}=\frac{(2.4042,0^o)}{(340,-82.054^o)}$

$\displaystyle I=(0.00707,82.054^o)\ A$

We can see the current leads the voltage, so our circuit has a capacitive power factor, as shown ahead .

The voltage acrosss the resistor is

$\displaystyle V_R=Z_R.I=(47)(0.00707,82.054^o)$

$\displaystyle V_R=(0.3323,82.054^o)\ V$

The currents through the capacitor and inductor will be computed with the formula of the current divider .

$\displaystyle I_C=\frac{Z_L}{Z_L+Z_C}\ I=\frac{j33}{j2.9897}(0.00707,82.054^o)$

$\displaystyle I_C=(0.0786,82.054^o)$

$\displaystyle I_L=\frac{Z_C}{Z_L+Z_C}\ I$

$\displaystyle I_L=\frac{-j30.03}{j2.9697}(0.00707,82.054^o)$

$\displaystyle I_L=(0.0715,-97.946^o)$

(b) The aparent power from the source is the product of the voltage by the total current

$\displaystyle P_s=V_s\ I$

$\displaystyle p_s=(2.4042,0^o)(0.007,82.054^o)$

$\displaystyle P_s=(0.017,82.054)\ VA$

Finally, the power factor is

$\displaystyle P_f=cos\ 82.054^o$

$\displaystyle P_f=0.1382$

As mentioned before, since the current leads the voltage, the circuit is primarily capacitive