Figure 2 shows a circuit with a resistor an inductor and a...

Figure 2 shows a circuit with a resistor, an inductor, and a capacitor in parallel. Kirchhoff’s rules can be used to express the impedance of the system as 1/Z=√(1/R^2 +(ωC-1/ωL)^2 ) where Z = impedance (Ω) and ω = the angular frequency. Find the ω that results in an impedance of 75 Ω with initial guesses of 1 and 1000 for the following parameters: R = 225 Ω, C = 0.6 × 10-6 F, and L = 0.5 H.

Transcript

00:01 Okay, in this problem, we're told we have an rlc parallel circuit, and this is the known impedance of such a circuit.

00:14 And we want to know what frequency will, or actually really what angular frequency will have when z is 75 oms.

00:26 So this problem is really just a lot of algebra from this point on.

00:33 Because we can set z equal to 75, and then with a little bit of work, we can turn this into a quadratic for omega, which we can then solve.

00:56 Alternatively, i think what they really want us to do is to go through, i would recommend using a spreadsheet, you plug in your omega of different values of omega until you find one where z is equal to 70.

01:12 And that's actually not hard to do because you can do that.

01:17 But i want to give you the exact answer that it's not that hard to get.

01:25 So what i'm going to do, i'm going to square out both sides of this equation.

01:30 So we get 1 over z squared is equal to 1 over r squared plus this combination, 1 over xl minus 1 over xc squared.

01:44 So i'm using xl and xc because those are the kind of the standard reactances we have for ac circuits, just because they're familiar.

01:56 Okay.

01:57 So we get 1 over z squared minus 1 over r squared on the left is equal to, and then xl is omega -l.

02:10 So it's 1 over omega -l minus omega -c quantity squared.

02:17 Excuse me to the one half no it is squared yeah that's correct so if we raise this to the one half power let's do it this way let's do it this way so take the thing on the left we'll take the square root because that'll get rid of that square so we get one over z squared minus one over r squared to the one half is equal to one over omega l minus omega c so that's a that can be rearranged into a quadratic for omega the thing on the left which i'm going to call why we know what it is and it's 1 .26 times 10 of the minus two and the unit of that is reciprocal oms so it's it's mo's but we'll call it reciprocal oms okay and we can solve that basically we can write this out as a quadratic so it's going to look like um omega squared c plus y omega minus 1 over l is equal to 0.

04:05 So that's a quadratic for omega...

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