a series rlc circuit consists of a 60 resistor a 26 mh inductor and a 610 nf capacitor it is con

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A series rlc circuit consists of a 60 resistor a 26 mh...

A series RLC circuit consists of a 60 Ω resistor, a 2.6 mH inductor, and a 610 nF capacitor. It is connected to an oscillator with a peak voltage of 5.2 V. Determine the impedance at a frequency of 4000 Hz. Determine the peak current at a frequency of 4000 Hz. Determine the phase angle at a frequency of 4000 Hz

Transcript

00:02 Hi, here in the given series rlc circuit, the value of resistance is 41 .0.

00:25 Om, self -inductance is 3 .30 millie henry.

00:33 Capacetance, that is 480, nanofaridates, nanofarret and pgmf applied that is 5 .00 volt in the part d of this problem frequency of the applied source voltage is 4 ,000 hertz and at this frequency we have to find impedance of the circuit so first of all we find inductive reactants means opposition offered in the path of ac by inductor coil.

01:14 Its expression is 2 pi fl.

01:18 So this is 2 times of 3 .14 multiplied by the frequency 4 ,000 into l which is 3 .30 millie henry or into 10 to the power minus 3 henry.

01:32 It will be measured in on.

01:36 So it comes out to be equal to 82 .9 om.

01:46 Now capacity reactants means opposition offered in the path of ac by a capacitor.

01:53 Its expression is 1 by 2 pi fc so plugging in the known values here also this is 2 times of 3 .14 into frequency 4 ,000 into capacitance 480 nanofarret or into 10 to the power minus 9 ferret and this capacitive reactance comes out to be equal to and this is also a t2 .9 om so impedance of such a circuit will be given by it is given as z is equal to vector sum of net reactants xl minus x c square plus r square and as xl is equal to x c so it will come out to be equal to zero net reactance is 0 plus r square means this is r only or we can say impedance is 41 .0 om answer for this part d of the problem then in the next part, part e of the problem we have to find peak current.

03:22 So using homes law it will be given as v0 by z.

03:28 Means for z this is r only, so this is 5 .0 volt divided by 41 .0 oom.

03:37 And this peak current here comes out to be equal to 0 .1 to 2amper.

03:44 Here.

03:45 Answer for part e of this problem here.

03:51 Then in the next part, part f of the problem, we have to find phase angle between current maximum and voltage maximum but as the circuit is in resonance because we know the circuit lcr circuit will be in resonance when inductive reactants becomes equal to capacity reactants.

04:28 So the circuit here is in resonance.

04:31 So net impedance is equal to resistance only.

04:37 And current will be in phase with emf applied...

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