I\'m lost... Consider the circuit of Figure 6.19. The AC supply has a peak volta

Question

I'm lost...

Consider the circuit of Figure 6.19. The AC supply has a peak voltage of 10 V and a frequency of l kHz. The resistor has a value of l k Ohm, and the capacitor has a capacitance of 0.1 1mu F Find the peak value of the current through the circuit. Explain how the current flows through the capacitor. What is the peak voltage between points A and B? We change the frequency of the source so that the peak voltage through the capacitor is the same as the peak voltage through the resistor What is the value of the frequency? What are the peak voltages across the resistor and capacitor? Explain why they are not 5 V.

Explanation / Answer

the voltage across the capacitor for a RC series circuit is

where Vs(t)= 10sin(2000*pi*t) = 10/_0 Volts

Z = R-jXc = 1000-j(1/(2*pi*1000*0.1*10^-6) =1000-j1591 ohms = 1879/_-57.85 ohms

Ic(t) = Vs(t)/Z = 5.32/_57.85 m A = 5.32sin(2000pi*t+57.85) mA

a) peak value of current = 5.32mA

b)

Ic(t) = Vs(t)/Z = 5.32/_57.85 m A = 5.32sin(2000pi*t+57.85) mA

c)

Vc(t) =Ic(t)*-jXC = 8.46/_-32.15 V

maximum voltage is 8.46 V

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