Question

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.

Answer 1

All the calculations are shown in the explanation

Step-by-step 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=\ (3.4)/(√(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=(1)/(wC)=(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)=(Z_L\ Z_C)/(Z_L+Z_C)=((j33)(-j30.03))/(j33-j3003)=(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=(V_s)/(Z_t)=((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=(Z_L)/(Z_L+Z_C)\ I=(j33)/(j2.9897)(0.00707,82.054^o)`

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

`\displaystyle I_L=(Z_C)/(Z_L+Z_C)\ I`

`\displaystyle I_L=(-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