#### Answer

Maximum equivalent capacitance= $6.5\,\mu F$
Minimum equivalent capacitance= $0.67\,\mu F$

#### Work Step by Step

Maximum equivalent capacitance is obtained when the capacitors are connected in parallel. In that case, the equivalent capacitance
$C_{p}=C_{1}+C_{2}+C_{3}=1.5\,\mu F+2.0\,\mu F+3.0\,\mu F$
$=6.5\,\mu F$
When the connection is in series, the equivalent capacitance is minimum which is equal to
$C_{s}=\frac{1}{\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}}=\frac{1}{\frac{1}{1.5\,\mu F}+\frac{1}{2.0\,\mu F}+\frac{1}{3.0\,\mu F}}=0.67\,\mu F$