The relationship between the voltage and current in AC circuit can be written in the form analogous to Ohm's Law:
`V=IZ`
, where I and V are effective, or rms, values of the current and voltage, and Z is the impedance. For the RC circuit, the impedance is
`Z=sqrt(R^2 + X_C^2)`
, where `X_C=1/(omegaC)` .
Since we don't know the values of R and C (and therefore ` ` ), we have to use the known values of V, I and ` ` and write the system of two equations with two variables. It is easier to use the Ohm's Law with the both sides squared:
`V^2=I^2(R^2 + X_c^2)`
For the frequency f = 20 kHz, the angular frequency is
`w=2pif = 126*10^3 (rad)/s`
and the current is `I = 38*10^(-3) A` .So the equation becomes
`5^2 = (38*10^(-3))^2(R^2 + 1/((126*10^3)^2*C^2))`
Divide by the coefficient on the right side in order to isolate the parenthesis:
`0.017*10^6 = R^2 + 1/(15,876*10^6*C^2)`
Similarly, for the frequency f = 28 kHz, the angular frequency is
`w = 2pif = 176*10^3 (rad)/s` and the equation becomes, after plugging in the current of 50 mA:
`5^2 = (50*10^(-3))^2(R^2 + 1/((176*10^3)^2*C^2))`
This becomes, after dividing by the coefficient in front of the parenthesis
`0.01*10^6 = R^2 + 1/(30,976*10^6*C^2)`
So we have two equations with two unknown variables, R and C. We can solve it by eliminating R. Subtract the second equation from the first one. `R^2` will cancel out and we will get
`0.007*10^6 = 1/C^2(1/15876-1/30976)*10^(-6)`
`7*10^3 = 1/C^2*3.07*10^(-5)*10^(-6)`
Finally, from here `C^2 = (3.07*10^(-11))/(7*10^3)`
and `C=0.66*10^(-7) = 6.6*10^(-8)` Farad.
The resistance then can be found from one of the equations. Using the second equation,
`R^2 = 0.01*10^6-1/(30,976*10^6*C^2)`
Plugging in C results in
`R^2=0.01*10^6-7.4*10^3 = 10*10^3-7.4*10^3=2.6*10^3``<br data-mce-bogus="1">`
`R=51` Ohm
So the values of R and C are 51 Ohm and 6.6*10^(-8) Farad, respectively.
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