A145 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

0.922 A ,133.761 volt ,426.09 volt,438.90 volt

Explanation:

We have given resistance R=145 OHM

Inductance

Capacitance

Frequency f =1115 Hz

Emf equation = 190 sin(2πft)

So rms voltage

Inductive reactance

Capacitive reactance

Impedance

RMS current

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

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~