How Do You Calculate Current Amplitude in a Parallel RC Circuit?

[nateastle]

A resistor with a resistance of 500 ohms and a capacitor with with a capacitance of 2 x 10^-6 f are connected in parrallel to an ac generator that supplies an rms voltage of 260 v at an angular frequency of 377 rad /s find the current amplitude in the resistor (Ir)peak.

I know I have to convert 377 from rad / s to hz when I do I get 60 hz. From there I am unsure of how to calculate the rest. I know in a traditional series lrc circuit you calculate Z = [(Xl -Xc)^2 + R^2]^.5 but I and with z you can find the i which would be V/Z.

 

[Andrew Mason]

A resistor with a resistance of 500 ohms and a capacitor with with a capacitance of 2 x 10^-6 f are connected in parrallel to an ac generator that supplies an rms voltage of 260 v at an angular frequency of 377 rad /s find the current amplitude in the resistor (Ir)peak.

I know I have to convert 377 from rad / s to hz when I do I get 60 hz. From there I am unsure of how to calculate the rest. I know in a traditional series lrc circuit you calculate Z = [(Xl -Xc)^2 + R^2]^.5 but I and with z you can find the i which would be V/Z.

Since V is the rms voltage then I = V/Z is the rms current. The peak voltage is $\sqrt{2}V_{rms}$ So peak current is $V_{peak}/Z$

AM

 

[nateastle]

would you calculate Z the same way with the capaicitor and reisitor in parrallel?

 

[Andrew Mason]

would you calculate Z the same way with the capaicitor and reisitor in parrallel?

No. The voltages are the same across R and C and the total current is the sum of the R and C currents (except there is a 90 degree phase difference between the two currents, of course). It is similar to parallel circuits with only resistance.

$$V_R = V_C = V; I_R = V/R; I_C = V/X_C$$

$$I_R + I_C = I = V/Z = V/R + V/X_C$$ so:

$$\frac{1}{Z} = \frac{1}{R} + \frac{1}{X_c} = \frac{1}{R} + j\omega C}$$

AM