Question

Three capacitors of capacitances $3\ \mu\text{F}$, $3\ \mu\text{F}$ and $3\ \mu\text{F}$ are connected in series. The equivalent capacitance is

• A. $1\ \mu\text{F}$
• B. $9\ \mu\text{F}$
• C. $3\ \mu\text{F}$
• D. $1.5\ \mu\text{F}$

Hint:

Remember, for series connections, the equivalent capacitance is always less than any individual capacitor's capacitance.

Answer 1

The Correct Option is A

Step 1: Recall series capacitors.
When capacitors join in series, the reciprocals add up. \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

Step 2: Put in the values.
All three are $3\,\mu$F. \[ \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \]

Step 3: Add the fractions.
\[ \frac{1}{C_{eq}} = \frac{3}{3} = 1 \] (in units of per microfarad).

Step 4: Take the reciprocal.
\[ C_{eq} = 1\,\mu\text{F} \]

Step 5: Use a quick shortcut.
For equal capacitors in series, the result is one value divided by the number of capacitors. \[ C_{eq} = \frac{3}{3} = 1\,\mu\text{F} \] This matches our work.

Step 6: State the answer.
The equivalent capacitance is smaller than any single one, as expected for series. \[ \boxed{C_{eq} = 1\,\mu\text{F}} \]