Saturday, May 17, 2014

An ac generator supplies an rms voltage of 5.00 V to an RC circuit. At the frequency of 20.0 kHz, the rms current in the circuit is 38.0 mA;...

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...

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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