A 2.8-k resistor is in series with a 2.1-µF capacitor in an ac circuit. The rms

Question

A 2.8-k resistor is in series with a 2.1-µF capacitor in an ac circuit. The rms voltages across the two are the same. (a) What is the frequency? Hz (b) Would each of the rms voltages be half of the rms voltage of the source? Yes No If not, what fraction of the source voltage are they? (In other words, VR/script Em = VC/script Em = ?) [Hint: Draw a phasor diagram.] (If so, enter 0.5.) (c) What is the phase angle between the source voltage and the current? ° Which leads? I script E (d) What is the impedance of the circuit?

Explanation / Answer

a) as they are in series, they will have same current flowing through them

hence magnitude impedance of the capacitor =resistance of the resistor

if frequency is f,

then 1/(2*pi*f*2.1*10^(-6))=2.8*10^3

==>f=1/(2*pi*2.1*10^(-6)*2.8*10^3)=27.067 Hz

b)

capacitive reactance=-j 2800 ohms

then net impedance=2800 - j2800 ohms

hence if voltage is V volts,

then current=V/(2800-j2800)

so voltage across resistance=current*resistance=V*2800/(2800-j2800)

=V/(1-j)

magnitude of voltage=V/sqrt(1^2+1^2)=V/sqrt(2)

hence voltage across resistance and capacitance is 1/sqrt(2) of the rms voltage supplied.

c)as we found in the previous part,

if voltage is V (angle =0 degree)

then current=V/(2800-j2800)

angle=45 degrees

hence the current leads the voltage by 45 degrees.

d)impedance of the circuit=2800 - j2800 ohms

magnitude of impedance=sqrt(2800^2+2800^2)=3959.8 ohms

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