Question

 The capacitors have values C1 = 2.5 μF, C2 = 4.0 μF, C3 = 6.0 μF and C4 = 8.0 μF. Assume that the capacitors are connected in series. Find the equivalent capacitance of the circuit.

Answer 1

Explanations & Calculations

• If the equivalent capacitance is $\small C_e$, then,

$\qquad\qquad \begin{aligned} \small \frac{1}{C_e}&=\small \frac{1}{2.5}+\frac{1}{4.0}+\frac{1}{6.0}+\frac{1}{8.0}\\ \small C&=\small 1.1\,\mu F \end{aligned}$

• When connected in series, the equivalent capacitance drops to a value even lower than the lowest capacitance in the arrangement.