A 37 mH inductor with 1.1 Ω resistance is connected in...

A 37 mH inductor with 1.1 $\Omega$ resistance is connected in series to a 26 $\mu$F capacitor and a 38 Hz, 55 V (rms) source. Calculate the rms current. Express your answer using two significant figures. $I_{rms}$ = Submit Request Answer Part B Calculate the phase angle. Express your answer using two significant figures. $\phi$ = Submit Request Answer Part C Calculate the power dissipated in this circuit. Express your answer using two significant figures.

Transcript

00:01 Hello, it is given in the question that in a lcr, this is an lcr series circuit is given, and in a lcs series circuit, it is given that the value of the phase angle is 5 equal to 53 degrees and the capacitive reactance that is equal to 350 oom and the value of r here, the resistance here is 170 om and the value of p, the power, average power, that is equal to 150 watts.

00:30 So the answer to the first part a of the question let us consider.

00:33 So this is the part of the question, the answer to that.

00:36 So we have to find the inductive reactants, that is we have to find.

00:40 So we already know that the 105 equal xl minus xc divided by r.

00:45 So let us put the values of the quantity into the expression and we get the value of xl that is equal to r 105 plus xc.

00:53 So putting the values of the quantity we get it is equal to 170 multiplied 1053 plus plus 350 and it is coming out to be about 575 .6 om so this is the value of inductive reactants that is the answer to the part a of the question let us now consider the part b of the question where we have to find i .rms that is the rms value of the current in the lcr series circuit so we know that i rms here that would be equal v rms divided by z the z here is the impedance and the impedance of the lcr circuit is given by the value r square plus xl minus xc whole square root over that and the value putting the values of the quantity we get it is equal to 170 squared plus 575 .6 minus 350 whole square and taking the square root of it so we get the value of impedance of the lcr circuit that is equal to 282 .5 om so this is the value of impedance now we know we can write that vrms equals 282 .5 om 282 0 .5 multiplied into irms we can write it from this equation over here putting the value of z here we can write this over here so let us consider this is equation 1 now what we can do here is that this is the value of r here r so we also know that the power p of the l cacet that is vrms multiplied i rmsm multiplied cos phi so we know that so now putting the value of equation 1 into the given equation and putting the value of phi into this given equation we can write otherwise as 282 .5 multiplied irms whole square multiplied cost of 53 that is equal to power p that is equal to 150 what already we know from the question.

02:43 Now what we can do is that we can find the value of irms from here...

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