Answer:
(a) 3.0 μF
(b) 75.0 μF
Explanation:
The equivalent capacitance when the capacitors are connected all in series is:
[tex]\frac{1}{C_{T} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} } +\frac{1}{C_{4} } +\frac{1}{C_{5} }[/tex]
Where [tex]C_{T}[/tex] is the equivalent capacitance, [tex]C_{1}[/tex], [tex]C_{2}[/tex], [tex]C_{3}[/tex], [tex]C_{4}[/tex] and [tex]C_{5}[/tex] are the capacitance of each capacitors. In this case, they all have 15.0 μF. So we can replace as:
[tex]\frac{1}{C_{T} } =\frac{1}{15.0} +\frac{1}{15.0} +\frac{1}{15.0} +\frac{1}{15.0} +\frac{1}{15.0}[/tex]
[tex]\frac{1}{C_{T} } =\frac{5}{15.0}[/tex]
The solve for [tex]C_{T}[/tex]:
[tex]C_{T}[/tex]=3.0 μF
The equivalent capacitance when the capacitors are connected all in parallel is:
[tex]C_{T}=C_{1}+C_{2}+C_{3}+C_{4}+C_{5}[/tex]
Replacing values we get:
[tex]C_{T}[/tex]=15.0 μF+15.0 μF+15.0 μF+15.0 μF+15.0 μF
[tex]C_{T}[/tex]=75.0 μF
