Figure 1 above shows two
capacitors C1 and C2 connected in parallel across a
voltage source of V. From figure 1 above, it is noted that the capacitors
connected in parallel is like increasing the area of the plates of a single
capacitor. Therefore the total capacitance is greater than that of either one.
Since both capacitors have to charge from the same voltage source, the charge
drawn from the voltage source (battery) is
QT = Q1 + Q2
But Q = CV
Therefore CTVT
= C1V1 + C2V2
= C1V1 + C2V2
Since figure 1 is a simple
parallel circuit,
parallel circuit,
VT = V1 = V2
Therefore VT, V1
and V2 can be replaced by V
and V2 can be replaced by V
CTV = V(C1
+ C2)
+ C2)
CT = C1 + C2
So in a general term, when a
given number of capacitors are connected in parallel, the total capacitance is
the sum of all the individual capacitances.
given number of capacitors are connected in parallel, the total capacitance is
the sum of all the individual capacitances.
CT = C1 + C2
+ C3 + … e.t.c.
+ C3 + … e.t.c.
Example 1
Two capacitors C1 and
C2 of capacitances 0.01µF and 0.05µF respectively are connected in
parallel in a circuit. What is the capacitance of a single capacitor that can
be used to replace C1 and C2
C2 of capacitances 0.01µF and 0.05µF respectively are connected in
parallel in a circuit. What is the capacitance of a single capacitor that can
be used to replace C1 and C2
Solution
CT = C1 + C2
= 0.01 + 0.05 = 0.06µF
