Need help ASAP. PLEASE!!!

Please note that there's a figure containing following.

Figure: RLC series circuit with a Resistor and a Capacitor in parallel with an Inductor.

In the circuit the source delivers a sinusoidal output. For a particular frequency, the impedances are in the proportions of R : XC : XL is as 3 : 4 : 5.

R = 47 Ω, L = 3.3 mH, C = 3.3 µF. The input provided by the source is e = 3.4 sin 10000t V

a) Calculate the voltages across and currents through the other components – including phase angles taking the EMF from the source as reference.

b) Calculate the (apparent) Power input from the source and the power factor of the 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

[tex]\displaystyle e=3.4\ sin\ 10,000t[/tex]

We must find the RMS voltage as

[tex]\displaystyle v_e=\ \frac{3.4}{\sqrt{2}}=2.4042\ V[/tex]

The given parameters of the circuit are

[tex]\displaystyle R=47\Omega[/tex]

[tex]\displaystyle L=3.3mH=0.0033H[/tex]

[tex]\displaystyle C=3.3\mu\ F=3.3\ 10^{-6}\ F[/tex]

[tex]\displaystyle w=10,000\ rad/s[/tex]

(a)

Let's find the reactances

[tex]\displaystyle X_L=wL=10,000(0.0033)[/tex]

[tex]\displaystyle X_L=33\Omega[/tex]

[tex]\displaystyle X_C=\frac{1}{wC}=\frac{1}{10,000(3.3150)}=30.30\Omega[/tex]

Now the impedances are

[tex]\displaystyle Z_R=47+j0[/tex]

[tex]\displaystyle Z_L=j33[/tex]

[tex]\displaystyle Z_C=-j30.03[/tex]

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

[tex]\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[/tex]

Computing the total impedance of the circuit

[tex]\displaystyle Z_t=Z_R+Z{eq1}=47+j0-j336.73[/tex]

[tex]\displaystyle Z_t=47-j336.73[/tex]

Converting to phasor form

[tex]\displaystyle Z_t=(340,-82.054^o)\Omega[/tex]

The given voltage of the source is

[tex]\displaystyle V_s=(2.4042,0^o)\ V[/tex]

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

[tex]\displaystyle I=\frac{V_s}{Z_t}=\frac{(2.4042,0^o)}{(340,-82.054^o)}[/tex]

[tex]\displaystyle I=(0.00707,82.054^o)\ A[/tex]

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

[tex]\displaystyle V_R=Z_R.I=(47)(0.00707,82.054^o)[/tex]

[tex]\displaystyle V_R=(0.3323,82.054^o)\ V[/tex]

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

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

[tex]\displaystyle I_C=(0.0786,82.054^o)[/tex]

[tex]\displaystyle I_L=\frac{Z_C}{Z_L+Z_C}\ I[/tex]

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

[tex]\displaystyle I_L=(0.0715,-97.946^o)[/tex]

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

[tex]\displaystyle P_s=V_s\ I[/tex]

[tex]\displaystyle p_s=(2.4042,0^o)(0.007,82.054^o)[/tex]

[tex]\displaystyle P_s=(0.017,82.054)\ VA[/tex]

Finally, the power factor is

[tex]\displaystyle P_f=cos\ 82.054^o[/tex]

[tex]\displaystyle P_f=0.1382[/tex]

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