Question

The equivalent capacity/between the points the circuit with C = 1 µF is

• A. 2µF
• B. 3µF
• C. 1 µF
• D. 0.5µF

Answer 1

The equivalent capacitance is 2/3 µF. There appears to be an error in the provided options, as none of them match the calculated value.

Answer 2

The circuit consists of three capacitors, each with capacitance $C = 1 \mu F$. Two capacitors are in parallel, and that combination is in series with the third capacitor.

1. Parallel Combination: The two capacitors in parallel have an equivalent capacitance of $C_p = C + C = 2C$.

2. Series Combination: The parallel combination ($2C$) is in series with the remaining capacitor ($C$). The equivalent capacitance of this series combination is given by:

$\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{1}{C} = \frac{1}{2C} + \frac{2}{2C} = \frac{3}{2C}$

Therefore, $C_{eq} = \frac{2C}{3}$.

3. Substituting C = 1 µF: $C_{eq} = \frac{2(1 \mu F)}{3} = \frac{2}{3} \mu F \approx 0.667 \mu F$.

Since none of the options match this calculated value, there is likely an error in the provided options.