Question

All of the questions in this problem are based on the circuit below. R=10Ω,L=5mH,C=500μF. The source voltage is 10cos(200t+45LaTeX: ^{^\circ}∘). Round all of your answers to two decimal places if necessary. Omit the units. What is the inductor impedance value in ohms? First, what is the REAL part.

Answer 1

Answer:

1) The real part of the inductor impedance is 0 Ω and

2) the imaginary part of the inductor impedance 1 Ω

3) The real part of the impedance of the capacitor is 0 Ω and

4) the imaginary part of impedance of the capacitor  is -10 Ω

5) The real part of total impedance is 10Ω

6) The Imaginary part of total impedance is -9j Ω

Explanation:

Given that R=10Ω,L=5mH,C=500μF and The source voltage is 10cos(200t+45°)

The voltage in an AC circuit is given by:

$V=V_mcos(wt+\theta)$

Comparing $V=V_mcos(wt+\theta)$ with 10cos(200t+45°), we get that the angular frequency w = 200 rad/s

A complex number given by x + jy has a real part of x and an imaginary part of y

The inductor impedance (Z) is given by: $Z_L=jwl = j*200*5*10^{-3}=j = 0 \ \Omega \ + \ j\ \Omega$

1) The real part of the inductor impedance is 0 Ω and

2) the imaginary part of the inductor impedance 1 Ω

The impedance of the capacitor is given by:

$Z_c=\frac{1}{jwC} =-j*\frac{1}{wC}=-j*\frac{1}{200*500*10^{-6}} =0\ \Omega \ - j10 \ \Omega$

3) The real part of the impedance of the capacitor is 0 Ω and

4) the imaginary part of impedance of the capacitor  is -10 Ω

The total impedance of the circuit is the sum of the resistance, capacitive impedance and inductive impedance. It is given by:

$Z=R+Z_L+Z_C=10+(0+j)+(0-j10)=10+j-j10=10\ \Omega - \ j9\ \Omega$

5) The real part of total impedance is 10Ω

6) The Imaginary part of total impedance is -9j Ω