1. A 58.0-?F capacitor is connected to a generator operating at a low frequency.

Question

1.A 58.0-?F capacitor is connected to a generator operating at a low frequency. The rms voltage of the generator is 5.50 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches 11.0 A. As the generator frequency is increased, at what frequency will the fuse burn out?

2. A capacitor is connected across an ac generator whose frequency is 900 Hz and whose peak output voltage is 140 V. The rms current in the circuit is 4.2 A. (a) What is the capacitance of the capacitor? (b) What is the magnitude of the maximum charge on one plate of the capacitor? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.

Explanation / Answer

1)

for the circuit,

net impedance, Z = 1/(2*pi*f*C)

where, f = frequency

C = capacitance = 58*10^-6 F

pi = 3.14

Volatge , V = 5.5 V

So, net current = V/Z = 5.5/(1/(2*pi*f*C)) = 5.5*2*pi*f*C

So, as per the question, when current = 11 A

So, 11 = 5.5*2*pi*f*C

So, 11 = 2*3.14*f*(58*10^-6)

So, f = 3.02*10^4 Hz <------------answer

2)

a)

Vmax =140 V

We know , Vmax = Vrms*sqrt(2)

So, Vrms = 140/sqrt(2) = 99 Hz

Irms = 4.2 A

f = 900 Hz

From the equation in question 1,

current,I = V/Z = V/(1/(2*pi*f*C))

So, I = V*2*pi*f*C <------note: here V = Vrms, I = Irms

So, C = I/(2*V*pi*f) = 4.2/(2*99*3.14*900)

= 7.5*10^-6 F = 75 uF <--------------answer

b)

maximum charge,Q = C*Vmax = 75*10^-6*140

= 0.0105 C <-------answer

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