Three 2-µF capacitors are arranged as seen in the circuit above . The effective capacitance between E and F is

Answer Details

To find the effective capacitance between points E and F, we need to analyze the circuit and determine which capacitors are in series and which are in parallel. Capacitors in series add reciprocally, while capacitors in parallel add directly.

Starting from point E, the two capacitors connected to it are in series. Their combined capacitance is:

C₁ + C₂ = 2µF + 2µF = 4µF

This 4µF equivalent capacitor is in parallel with the third 2µF capacitor, so the total capacitance between E and F is:

C_total = C₁ + C₂ || C₃

C_total = C₁ + C₂ + C₃ / (C₁ + C₂) || C₃

C_total = (C₁ + C₂) * C₃ / (C₁ + C₂ + C₃)

C_total = (4µF) * (2µF) / (4µF + 2µF)

C_total = 8µF / 6µF

C_total = 1.33µF

Therefore, the effective capacitance between E and F is 1.33µF.