A 145 Ω resistor is connected in series with a 66 mH inductor and a 0.3 μF capac- itor. The applied voltage has he form E 190 V sin(2r/t), where the frequency is f 1115 cycles/s. Find the rms current Answer in units of A 005 (part 2 of 4) 10.0 points Find the ms voltage across R. Answer in units of V 006 (part 3 of 4) 10.0 points Find the ms voltage across L Answer in uts of V 007 (part 4 of 4) 10.0 points Find the rms voltage across C Answer in units of V

Respuesta :

Answer:

0.922 A ,133.761 volt ,426.09 volt,438.90 volt

Explanation:

We have given resistance R=145 OHM

Inductance [tex]L=66 mH=66\times 10^{-3}H[/tex]

Capacitance [tex]C=0.3\mu F=0.3\times 106{-6}F[/tex]

Frequency f =1115 Hz

Emf equation = 190 sin(2πft)

So rms voltage [tex]=\frac{190}{\sqrt{2}}=\frac{190}{1.414}=134.37volt[/tex]

Inductive reactance [tex]X_L=\omega L=2\times \pi \times 1115\times 66\times 10^{-3}=462.1452ohm[/tex]

Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 1115\times 0.3\times 10^{-6}}=476.04ohm[/tex]

Impedance [tex]Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{145^2+(476.04-462.145)^2}=145.66ohm[/tex]

RMS current [tex]i=\frac{V}{Z}=\frac{134.37}{145.66}=0.922A[/tex]

RMS voltage across resistor = 0.922×145=133.761 volt

RMS voltage across inductor =0.922×462.145=426.09 volt

RMS voltage across capacitor =0.922×438.90 volt