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; at the frequency of 28.0 kHz, the rms current is...
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; at the frequency of 28.0 kHz, the rms current is 50.0 mA.
What are the values of R and C in this circuit?
So far I've been doing problems where R and XC are given, but I'm not sure how to do this.
The relationship between the voltage and current in AC circuit can be written in the form analogous to Ohm's Law:
, 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">`
So the values of R and C are 51 Ohm and 6.6*10^(-8) Farad, respectively.