Question

In the following circuit, the equivalent capacitance between terminal A and terminal B is :

• A. 2 µF
• B. 1 µF
• C. 0.5 µF
• D. 4 µF

Answer 1

Answer: (A)

2 µF

Answer 2

The equivalent capacitance is calculated as:

- First, combine series capacitors:

$C_s = \frac{1}{\frac{1}{1} + \frac{1}{\frac{1}{2}}} = 1 \, \ muF$.

**- Combine with parallel capacitors: **

$C_p = 1 + 1 = 2 \, \mu F$.