2.) Consider an RL series circuit. R is still 887 ? and you measure L to be 805

Question

2.) Consider an RL series circuit. R is still 887 ? and you measure L to be 805 mH. You again set the signal generator output to produce a sinusouidal output with an amplitude of 2.05 V.

i) At what frequency will XL equal R?

ii) What will the voltage across the inductor be at this frequency?

We use the compound signal described above.

iii) Answer the following: What will the ratio of the amplitude of the voltage across the resistor due to the 52.8 Hz component of the signal to that of the 762 Hz component?

Explanation / Answer

1.

Given

XL=R

since Inductive reactance

XL=2pifL

so

2pi*f*L=R

f=R/2pi*L =887/(2pi*0.805)

f=175.37 Hz

2.

Given

XL=R =887 ohms

so impedance

Z=sqrt[R^2+XL^2] =sqrt[887^2+887^2]

Z=1254.4 ohms

Total current in the circuit is

I=V/Z =2.05/1254.4

I=1.634*10^-3 A

so Voltage across inductor

VL=I*XL =1.634*10^-3*887

VL=1.45 Volts

3.

at f=52.8 Hz

XL=2pi*f*L=2pi*52.8*0.805 =267.06 ohms

Z=sqrt[R^2+Xl^2]=sqrt[887^2+267.06^2]

Z=926.33 ohms

I=V/Z =2.05/926.33 =2.213*10^-3 A

Voltage across resistor is

V1=I*R =2.213*10^-3*887=1.963 Vilts

at f=762 Hz

XL=2pi*f*L=2pi*762*0.805 =3854.17 ohms

Z=sqrt[R^2+Xl^2]=sqrt[887^2+3854.17^2]

Z=3954.92 ohms

I=V/Z =2.05/3954.92 =5.183*10^-4 A

Voltage across resistor is

V2=I*R =5.183*10^-4*887

V2=0.46 Volts

so

V1/V2 =1.963/0.46 =4.267

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