Figure 1 above shows capacitors C1
and C2 connected in series across a voltage source (battery) V. In
this arrangement, the negative plate of C1 and the positive plate of
C2 will be charged by electrostatics induction and the effect is the
same as increasing the spacing between the plates of a single capacitor. The
total capacitance is less than that of either one, but the combination is
capable of withstanding a higher potential difference than either one by
itself.
Since figure 1 is a simple series
circuit,
circuit,
V = V1 + V2
Where V is the total voltage drop
from the source
V1 is the voltage drop
across capacitor C1
across capacitor C1
V2 is the voltage drop
across capacitor C2
across capacitor C2
Q is the charge accumulated by the
capacitors.
capacitors.
Since the current flowing through
the two capacitors is the same,
the two capacitors is the same,
QT = Q1 = Q2
Therefore the equivalent
capacitance of a number of capacitors connected in series is
capacitance of a number of capacitors connected in series is
Example 1
Two capacitors C1 and
C2 of capacitors 0.01µF and 0.05µF respectively are required to be
connected in parallel to an electronic component of a circuit. But due to
non-availability, what is the capacitance of a single capacitor required to
replace them.
C2 of capacitors 0.01µF and 0.05µF respectively are required to be
connected in parallel to an electronic component of a circuit. But due to
non-availability, what is the capacitance of a single capacitor required to
replace them.
Solution
= 8.33 x 10⁻³ µF
= 0.00833 µF
