Question

An L-R-C series circuit is constructed using a 175Ω resistor, a 12.5μF capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V.
1-At what angular frequency will the impedance be smallest?
2-What is the impedance at this frequency?
3-At the angular frequency in part A, what is the maximum current through the inductor?
4-At the angular frequency in part A, find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value.

Enter your answers numerically separated by commas.

Answer 1

Answer:

3162.27 rad/sec, 175Ω, 0.142 A, VR= 12.5V, VL=1.79V, Vc=1.79V

Explanation:

1.

Impedance is smallest at resonant frequency. At resonant

XL = Xc

⇒ wL= 1/wC

⇒ w²= 1/LC

⇒w= 1/$\sqrt{LC}$

w=1/$\sqrt{8*10^{-3} *12.5*10^{-6} }$

w=3162.27 rad/sec

2.

At resonance XL = XC, i.e inductive  is cancelled by capactive reactance and the circuit behaves as pure resistive circuit.

so Z=R= 175 Ω

3.

Since all the components are connected in series so same current will flow in the circuit. Maximum current will flow when the voltage is maximum in the circuit i.e. 25 V.

Vmax=25 V

so I=Vmax/Z= Vmax/R ( at angular frequency in part A Z=R)

Imax= 25/175

Imax=0.142 A

Maximum current through inductor is 0.142A

4.

Current in the circuit will be

I= 0.142/2 A

I=0.071 A

Now

VR= IR= 0.071*175

VR= 12.5 V

VL=I X

VL= IwL

VL= 0.071*3162.27*8*$10^{-3}$

VL=1.79 V

Vc=IX

Vc=I/wC

Vc=0.071/(3162.27*12.5*$10^{-6}$)

Vc=1.79 V