a 1.05 mF capacitor is connected in series with a 1.0 kW resistor. This combinat

Question

a 1.05 mF capacitor is connected in series with a 1.0 kW resistor. This combination is connected in series with a 6 V AC signal generator with 500 Hz signal. The capcitor starts off fully discharged.

(A) What is the maxium charge that the capcitor can attain after the switch is closed?

(B) What is the capacitive reactance?

(C) What is the impedence of the circuit?

(D) Calculate the phase angle between the current and the voltage across the circuit.

Thanks in advance if you are able to answer all of these questions!

Explanation / Answer

A) c = q/v ===> q = c*v = 1.05*10^-3 * 6 = 6.3 mC.......

B) capacitive reactance = 1/wc = 1/(2*314*500*1.05*10^-3) = 3.03*10^-3 ohm............

C) impedance = R + 1/wc = 1 k ohm............

D) theta = arc tan (1/wcR) ........substitute w = 2*pi*f , c and R values in this aqun and get theta.

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