When an ac source with a peak potential difference of 38 V...

When an ac source with a peak potential difference of 38 V is connected to a series RLC circuit, the peak current is 5 A. The capacitor is $C = 25 \mu F$, the frequency is 100 Hz, and the current leads the potential difference by $55^o$. Find the resistance and the inductance.

Transcript

00:01 Hello, here we have to solve the following problem which describes an rlc circuit.

00:06 There is an ac source with a big potential difference of 32 volts and the peak current is 3 ampires.

00:19 The capacitor has a capacitance of 35 microferrets and the frequency is 110 hertz.

00:29 The current leads the potential by the difference of 40 degree and here we have to calculate the resistance and the inductance let's do this first let's start with a phase diagram so here that is let's start with vr which is this guy and it leads it and i are leads the impedance leads leads leads the potential different potential difference by 40 degrees so it means that this vector which describes that which is total impedance times i so this difference is angle 5 so there is a capacit there is a inductance resistance which is upwards and capacitance resistance which is downwards so first we can find the total impedance and total impedance equals to the peak voltage divided by big current which is 32 volts divided by 3 amperes that is 10 .67 oms which is so therefore that is roughly 10 .7 oms now from here are equals to that times cosine phi therefore r equals to 10 .7 oms times cosine of 40 degree that is 8 .20 oms so let me rewrite it more clear so here r is 8 .20 oms so here r is 8 .20 oms so this part is done now we have to determine the inductance let's do this so let's first find the frequency omega that is 2 pi f which is 6901 radiance per second and now let's use the formula which describes the impedance so that squared equals to r squared plus xel minus xc squared therefore here xl minus xc squared equals to that's squared minus r squared okay here i probably need to clean up some part of the screen so let's keep writing so here as we can see xc which we can get let's let's denote the same in the same way so here xc is greater than xl therefore xc minus xl equals to square root of z squared minus r squared on the other hand that equals to 1 over omega c minus omega l therefore 1 over omega c equals to omega l plus square root of z squared minus r squared so let's calculate it oh actually here we have to determine l not c so therefore let's rewrite it a little bit omega l equals to 1 over omega c minus square root of z squared minus r squared therefore l equals to 1 over omec minus omega squared, actually 1 over omega times 1 over omega c minus square root of that squared minus r squared.

06:03 So let's calculate it.

06:07 That is 1 over 791 radiance per second times 1 over 691 radiance per second times 1 over 691 radiance 10 power by negative 6 ferrets minus and here i have to clean up some space.

06:44 So basically let me rewrite what that is.

06:47 That is 10 .7 oms...

Question

When an ac source with a peak potential difference of 38 V is connected to a series RLC circuit, the peak current is 5 A. The capacitor is $C = 25 \mu F$, the frequency is 100 Hz, and the current leads the potential difference by $55^o$. Find the resistance and the inductance.

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