A 40 kHz sinusoidal voltage has zero phase angle and a...

A $40 \mathrm{kHz}$ sinusoidal voltage has zero phase angle and a maximum amplitude of $2.5 \mathrm{m}$ V. When this voltage is applied across the terminals of a capacitor, the resulting steady-state current has a maximum amplitude of $125.67 \mu \mathrm{A}$ a) What is the frequency of the current in radians per second? b) What is the phase angle of the current? c) What is the capacitive reactance of the capacitor? d) What is the capacitance of the capacitor in microfarads? e) What is the impedance of the capacitor?

Transcript

00:01 In this problem, we have been given that there is a capacitor and across the ends of this capacitor, an alternating current source is applied, which is having frequency of 40 kilohertz.

00:15 So let's represent frequency using f, and that's 40 kilohertz.

00:19 So that's 40 times 10 raise to 3 hertz.

00:22 And it is observed that the maximum voltage that this source supplies is 2 .5 millivolds.

00:33 So let's represent the maximum voltage using v -0, and that's 2 .5 millivolts, which is 10 -raised to minus 3 volts.

00:42 And it is observed that across the terminals of the capacitor, when the source voltage is connected, the current is seen in the circuit, which is having maximum amplitude of 125 .67 milli -amper.

01:04 It's in micro -ampier, so it will be 10 -raised to minus -6 -ampers.

01:08 This is the maximum current that is observed here and we need to figure out the rest things dependent on what information we have been given.

01:17 So first we will figure out the frequency of the current in terms of radiance per second.

01:23 So basically we have to determine the angular frequency and in the next case we have to determine here the phase angle of the current.

01:35 And in another case we have to figure out the capacitive reactants represented by a xc which is obtained as 1 by omega c so we will use this expression to get the capacitative reactants and finally we have to determine the capacitance of this capacitor represented by c here and at last we have to determine the impedance of the capacitor so let's represent the impedance using v which we have to determine so it can be observed that the expression of the voltage that is v0 sine omega t so omega here is the angle here is the angle frequency which will be seen for both the current and the voltage and to get angular frequency we just multiply two times pi into the given frequency so as we are given the frequency 40 ,000 and we multiply this given frequency with two times of the value of pi which is 3 .1 for approximately so putting this value we're gonna get the angular frequency coming out to be 2 .51 into 10 raise to five radiance per second.

02:46 So we can get the expression of current from here.

02:51 And the expression of current will be i .0 sign omega t minus pi by 2.

02:57 Because current, we know in a capacitative circuit that lacks behind the voltage by a phase angle of pi by 2.

03:03 So if we talk about the phase angle of current, it will be minus pi by 2 radiance...

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