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

1) 0 Ω

2) jΩ

3) 0 Ω

4) -10j Ω

5) 10Ω

6) -9j Ω

Step-by-step explanation:

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

The source voltage is given by
`V=V_mcos(wt+\theta)`

Therefore comparing with the source voltage, the angular frequency w = 200 rad/s

a) The impedance of the inductor is given by:

`Z_L=jwL=j*200*5*10^(-3)=j = 1\angle 90\ \Omega`

The real part is 0 Ω and

the imaginary part is j Ω

b) The impedance of the capacitor is given by:

`Z_C=(1)/(jwC)=(1)/(j*200*500*10^(-6))=-10j\ \Omega=10\angle -90\ \Omega`

The real part is 0 Ω and

the imaginary part is -10j Ω

c) The total impedance of the circuit is given by:

`Z=R+jwL+(1)/(jwC)=10+j-10j=10-9j`

The real part is 10Ω

The Imaginary part is -9j Ω