Question

Find the equivalent capacitance of FIVE capacitors each having 15.0 μF and connected: (a) all in series; (b) all in parallel.

Answer 1

Answer:

(a) 3.0 μF

(b) 75.0 μF

Explanation:

The equivalent capacitance when the capacitors are connected all in series is:

$\frac{1}{C_{T} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} } +\frac{1}{C_{4} } +\frac{1}{C_{5} }$

Where  $C_{T}$ is the equivalent capacitance, $C_{1}$, $C_{2}$, $C_{3}$, $C_{4}$ and $C_{5}$ are the capacitance of each capacitors. In this case, they all have 15.0 μF. So we can replace as:

$\frac{1}{C_{T} } =\frac{1}{15.0} +\frac{1}{15.0} +\frac{1}{15.0} +\frac{1}{15.0} +\frac{1}{15.0}$

$\frac{1}{C_{T} } =\frac{5}{15.0}$

The solve for $C_{T}$:

$C_{T}$=3.0 μF

The equivalent capacitance when the capacitors are connected all in parallel is:

$C_{T}=C_{1}+C_{2}+C_{3}+C_{4}+C_{5}$

Replacing values we get:

$C_{T}$=15.0 μF+15.0 μF+15.0 μF+15.0 μF+15.0 μF

$C_{T}$=75.0 μF