Answer:
<em>All the calculations are shown in the explanation</em>
Explanation:
<u>RLC Circuit
</u>
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20e%3D3.4%5C%20sin%5C%2010%2C000t)
We must find the RMS voltage as
![\displaystyle v_e=\ \frac{3.4}{\sqrt{2}}=2.4042\ V](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_e%3D%5C%20%5Cfrac%7B3.4%7D%7B%5Csqrt%7B2%7D%7D%3D2.4042%5C%20V)
The given parameters of the circuit are
![\displaystyle R=47\Omega](https://tex.z-dn.net/?f=%5Cdisplaystyle%20R%3D47%5COmega)
![\displaystyle L=3.3mH=0.0033H](https://tex.z-dn.net/?f=%5Cdisplaystyle%20L%3D3.3mH%3D0.0033H)
![\displaystyle C=3.3\mu\ F=3.3\ 10^{-6}\ F](https://tex.z-dn.net/?f=%5Cdisplaystyle%20C%3D3.3%5Cmu%5C%20F%3D3.3%5C%2010%5E%7B-6%7D%5C%20F)
![\displaystyle w=10,000\ rad/s](https://tex.z-dn.net/?f=%5Cdisplaystyle%20w%3D10%2C000%5C%20rad%2Fs)
(a)
Let's find the reactances
![\displaystyle X_L=wL=10,000(0.0033)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20X_L%3DwL%3D10%2C000%280.0033%29)
![\displaystyle X_L=33\Omega](https://tex.z-dn.net/?f=%5Cdisplaystyle%20X_L%3D33%5COmega)
![\displaystyle X_C=\frac{1}{wC}=\frac{1}{10,000(3.3150)}=30.30\Omega](https://tex.z-dn.net/?f=%5Cdisplaystyle%20X_C%3D%5Cfrac%7B1%7D%7BwC%7D%3D%5Cfrac%7B1%7D%7B10%2C000%283.3150%29%7D%3D30.30%5COmega)
Now the impedances are
![\displaystyle Z_R=47+j0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Z_R%3D47%2Bj0)
![\displaystyle Z_L=j33](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Z_L%3Dj33)
![\displaystyle Z_C=-j30.03](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Z_C%3D-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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Z_%7Beq1%7D%3D%5Cfrac%7BZ_L%5C%20Z_C%7D%7BZ_L%2BZ_C%7D%3D%5Cfrac%7B%28j33%29%28-j30.03%29%7D%7Bj33-j3003%7D%3D%5Cfrac%7B1000%7D%7B2.9697%7D%3D-j336.73%5COmega)
Computing the total impedance of the circuit
![\displaystyle Z_t=Z_R+Z{eq1}=47+j0-j336.73](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Z_t%3DZ_R%2BZ%7Beq1%7D%3D47%2Bj0-j336.73)
![\displaystyle Z_t=47-j336.73](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Z_t%3D47-j336.73)
Converting to phasor form
![\displaystyle Z_t=(340,-82.054^o)\Omega](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Z_t%3D%28340%2C-82.054%5Eo%29%5COmega)
The given voltage of the source is
![\displaystyle V_s=(2.4042,0^o)\ V](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V_s%3D%282.4042%2C0%5Eo%29%5C%20V)
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)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%3D%5Cfrac%7BV_s%7D%7BZ_t%7D%3D%5Cfrac%7B%282.4042%2C0%5Eo%29%7D%7B%28340%2C-82.054%5Eo%29%7D)
![\displaystyle I=(0.00707,82.054^o)\ A](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%3D%280.00707%2C82.054%5Eo%29%5C%20A)
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)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V_R%3DZ_R.I%3D%2847%29%280.00707%2C82.054%5Eo%29)
![\displaystyle V_R=(0.3323,82.054^o)\ V](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V_R%3D%280.3323%2C82.054%5Eo%29%5C%20V)
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)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I_C%3D%5Cfrac%7BZ_L%7D%7BZ_L%2BZ_C%7D%5C%20I%3D%5Cfrac%7Bj33%7D%7Bj2.9897%7D%280.00707%2C82.054%5Eo%29)
![\displaystyle I_C=(0.0786,82.054^o)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I_C%3D%280.0786%2C82.054%5Eo%29)
![\displaystyle I_L=\frac{Z_C}{Z_L+Z_C}\ I](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I_L%3D%5Cfrac%7BZ_C%7D%7BZ_L%2BZ_C%7D%5C%20I)
![\displaystyle I_L=(0.0715,-97.946^o)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I_L%3D%280.0715%2C-97.946%5Eo%29)
(b) The aparent power from the source is the product of the voltage by the total current
![\displaystyle P_s=V_s\ I](https://tex.z-dn.net/?f=%5Cdisplaystyle%20P_s%3DV_s%5C%20I)
![\displaystyle p_s=(2.4042,0^o)(0.007,82.054^o)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20p_s%3D%282.4042%2C0%5Eo%29%280.007%2C82.054%5Eo%29)
![\displaystyle P_s=(0.017,82.054)\ VA](https://tex.z-dn.net/?f=%5Cdisplaystyle%20P_s%3D%280.017%2C82.054%29%5C%20VA)
Finally, the power factor is
![\displaystyle P_f=cos\ 82.054^o](https://tex.z-dn.net/?f=%5Cdisplaystyle%20P_f%3Dcos%5C%2082.054%5Eo)
![\displaystyle P_f=0.1382](https://tex.z-dn.net/?f=%5Cdisplaystyle%20P_f%3D0.1382)
As mentioned before, since the current leads the voltage, the circuit is primarily capacitive