Respuesta :
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/[tex]\sqrt{LC}[/tex]
w=1/[tex]\sqrt{8*10^{-3} *12.5*10^{-6} }[/tex]
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*[tex]10^{-3}[/tex]
VL=1.79 V
Vc=IX
Vc=I/wC
Vc=0.071/(3162.27*12.5*[tex]10^{-6}[/tex])
Vc=1.79 V
1. [tex]\omega=3160 \ rad/s[/tex]
2. [tex]R=175\Omega[/tex]
3. [tex]i_{max}=0.143A[/tex]
4. [tex]t=0.000331[/tex]
L-R-C series circuit: An LRC circuit, also known as a resonant circuit, tuned circuit, or an RLC circuit, is an electrical circuit consisting of an inductor (L), capacitor (C), and resistor (R) connected in series or parallel. The LCR circuit analysis can be understood better in terms of phasors.
1. The angular frequency where the impedance is smallest is:
[tex]\omega=\frac{1}{\sqrt{LC}}= \ \\=\frac{1}{\sqrt{\left(\left( 8.0\times 10^{-3} \right)\left( 12.5\times 10^{-6} \right) \right)}}=3160 \ rad/s[/tex]
2. The impedance at this frequency is:
[tex]Z=\sqrt{R^{2}+(X_{L}-X_{c})^{2}}[/tex]
[tex]=\sqrt{R^{2}}[/tex]
[tex]R=175\Omega[/tex]
3. The maximum current is:
[tex]i_{max}=\frac{V}{Z}=\frac{25}{175}=0.143A[/tex]
4. Given that:
[tex]i=i_{max}cos\omega t[/tex]
[tex]\frac{i_{max}}{2}=i_{max}cos\omega t[/tex]
[tex]cos\omega t=\frac{1}{2}[/tex]
[tex]t=0.000331[/tex]
Thus,
The voltage across the ac source is:
[tex]v=Vcos\left( \omega t \right)=12.5V[/tex]
The voltage across the resistance is:
[tex]V_{R}=iRcos\left( \omega t \right)=12.5V[/tex]
The voltage across the capacitor is:
[tex]v_{c}=iX_{c}cos\left( \omega t-\frac{\Pi}{2} \right)[/tex]
[tex]=i\frac{1}{\omega C}cos\left( \omega t-\frac{\Pi}{2} \right)=3.13V[/tex]
The voltage across the inductor is:
[tex]v_{L}=iX_{L}cos\left( \omega t+\frac{\Pi}{2} \right)[/tex]
[tex]=i\omegaLcos\left( \omega t+\frac{\Pi}{2} \right)=-3.13V[/tex]
Learn more about the topic L-R-C series circuit: https://brainly.com/question/15595203
