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------------------------------------------------ Determine the equivalent capacitance, \( C_{AB} \), of this circuit if \( C_1 = 150 \, \mu F \), \( C_2 = 150 \, \mu F \), \( C_3 = 150 \, \mu F \).

A. 50 \(\mu F\)

B. 100 \(\mu F\)

C. 150 \(\mu F\)

D. None of the above

The equivalent capacitance AB of a circuit with C1, C2, and C3 each at 150µF, where C1 and C2 are in series and in parallel with C3, is 225µF. The correct answer is 'None of the above' since this value is not listed as an option.

Option d.

Explanation:

The student asks to determine the equivalent capacitance, AB, of a circuit when C1=150µF, C2=150µF, and C3=150µF. In this circuit, C1 and C2 are in series, and their combination is in parallel with C3. To find the equivalent series capacitance (C's) of C1 and C2, we use the formula for capacitors in series:

C's = (C1 × C2) / (C1 + C2)

Since C1 and C2 have the same value, we can simplify this to C's = C1 / 2 = 150µF / 2 = 75µF. Next, we find the total capacitance by adding C's and C3, which are in parallel:

Ctotal = C's + C3. With C's being 75µF and C3 also being 150µF, the total equivalent capacitance AB is Ctotal = 75µF + 150µF = 225µF. Therefore, the correct answer is d) None of the above, as 225µF is not listed as an option.

Option d.