A capacitor (capacitance C 1 ) is connected across the terminals of an ac genera

Question

A capacitor (capacitance C1) is connected across the terminals of an ac generator. Without changing the voltage or frequency of the generator, a second capacitor (capacitance C2) is added in series with the first one. As a result, the current delivered by the generator decreases by a factor of 7.00. Suppose the second capacitor had been added in parallel with the first one, instead of in series. By what factor would the current delivered by the generator have increased? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.

Explanation / Answer

Let:
v = generator voltage
i = current with C1 only
Xc1,Xc2 = reactance of C1,C2 respectively

Case 1.
v = i * Xc1

Case 2.
v = ( i/7) * (Xc1 + Xc2)

Substituting v, i disappears

Xc1 = (Xc1 + Xc2)/7

Xc2 = 2 * Xc1

Effective reactance of Xc1 and Xc2 in parallel

[1/Xp = (1/Xc1) + (1/Xc2)]

Xp = (Xc1 * Xc2)/(Xc1 + Xc2)
Substitute for Xc2 and simplify

Xp = 6 * Xc1 /7

The current will be v/(Xc1 * 6/7) = 7/6 * v/Xc1
= 7/6 * i
The current will increase by a factor of 7/6

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