Five capacitors each of 10 μF are connected in series, the equiva
| Five capacitors each of 10 μF are connected in series, the equivalent capacitance of the system is;
A. 2 μF
B. 20 μF
C. 30 μF
D. 50 μF
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors’ capacitances.
\({C_{eq,parallel}} = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)
When capacitors are connected in series, the total capacitance is less than anyone of the series capacitors’ individual capacitances.
\(\frac{1}{{{C_{eq,series}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)
Calculation:
\(\frac{1}{{{C_{eq,series}}}} = \frac{1}{{10}} + \frac{1}{{10}} + \frac{1}{{10}} + \frac{1}{{10}} + \frac{1}{{10}}\)
\(= \frac{5}{{10}} = \frac{1}{2}\)
\(\therefore {C_{eq,series}} = 2\;\mu F\)