Question

A series RLC circuit containing a resistance of 12Ω, an inductance of 0.15H and a capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuit’s current, and the power factor.

Answer 1

Answer:

Impedance = 19.44ohms

Current = 5.14A

Power factor = 0.62

Explanation:

Impedance in an RLC AC circuit is defined as the total opposition to the flow of current in the resistor, inductor and capacitor.

Impedance Z = √R²+(Xl-Xc)²

Where R is the resistance = 12Ω

Inductance L = 0.15H

Capacitance C = 100uF = 100×10^-6F

Since Xl = 2πfL and Xc = 1/2πfC where f is the frequency.

Xl = 2π×50×0.15

Xl = 15πΩ

Xl = 47.12Ω

Xc = 1/2π×50×100×10^-6

Xc = 100/π Ω

Xc = 31.83Ω

Z =√12²+(47.12-31.83)²

Z = √144+233.78

Z = 19.44Ω

Impedance = 19.44ohms

To calculate the circuit current, we will use the expression V=IZ where V is the supply voltage = 100V

I = V/Z = 100/19.44

I = 5.14Amperes

To calculate the power factor,

Power factor = cos(theta) where;

theta = arctan(Xl-Xc)/R

theta = arctan(47.12-31.83)/12

theta = arctan(15.29/12)

theta = arctan1.27

theta = 51.78°

Power factor = cos51.78°

Power factor = 0.62

Answer 2

Answer:

The circuit impedance $=19.4 \Omega$

The circuit's current $=5.14 A$

Circuit Power Factor $=0.62$

Explanation:

Given:

Resistance $R=12 \Omega$

Inductance $=0.15H$

Capacitance $=100uF$

Voltage $=100V$

Step 1:

To calculate the inductive reactance, $X_{L}$.

$X_{L}=2 \pi f L=2 \pi \times 50 \times 0.15=47.13 \Omega$

To calculate the Capacitive reactance,

$X_{C}=\frac{1}{2 \pi f C}$

$=\frac{1}{2 \pi \times 50 \times 100 \times 10^{-6}}$

$=31.83 \Omega$

Step 2:

Circuit impedance,

$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$

where R is the resistance,

$&Z=\sqrt{12^{2}+(47.13-31.83)^{2}}$

$&Z=\sqrt{144+234}=19.4 \Omega\end{aligned}$

Step 3:

Circuits Current, I

$I=\frac{V_{S}}{Z}$

$=\frac{100}{19.4}$

$=5.14 \ A$

Step 4:

Voltages across the Circuit, $\mathrm{V}_{\mathrm{R}}, \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$

$V_{R}=I \times R=5.14 \times 12=61.7 volts$

$V_{L}=I \times X_{L}=5.14 \times 47.13=242.2 volts$

$V_{C}=\ I \times X_{C}=5.14 \times 31.8=163.5 volts$

Step 5:

Circuits Power factor

$=\frac{R}{Z}=\frac{12}{19.4}=0.619$

Therefore,

The circuit impedance $=19.4 \Omega$

The circuit's current $=5.14\ A$

Power Factor $=0.62$

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