1. CAPACITORS IN PARALLEL By J. Jaffrey at Long Bay College, NZ

2. CAPACITORS IN PARALLEL
If we now add another capacitor C2 here in parallel with the first one it will also have V across it. Q1 V
This will produce a charge Q2 on the second capacitor.
If we apply a potential difference V to this capacitor C1 we will get a charge Q1 stored on it.
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3. CAPACITORS IN PARALLELV Q1 Q2
So potential difference V gives a total charge stored of:
QT = Q1 + Q2
Potential difference V on capacitor C1 gives charge Q1 on it.
These capacitors could be replaced with CT which stores the same charge.
Another capacitor C2 in parallel with the first also has V across it.
By definition Q = CxV so here:
CTV = C1V + C2V
and so
This will produce a charge Q2 on the second capacitor.
CT = C1 + C2 + C3 + C4 +………
CT = C1 + C2
By adding more capacitors in parallel we will get
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4. CT = C1 + C2 + …. CAPACITORS IN PARALLEL V Q2 Q1 A2 A1
Another way to think about this uses the parallel plate construction formula.
So now our total area is A1 + A2 and….
CT = C1 + C2 + ….
If we keep ε and d the same…..
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5. CT = C1 + C2 +… Parallel capacitor problems Calculate: 1
VT = 12V
Calculate: 1
C1=8.0μF
(a) CT (b) Q1 (c) Q2 (d) QT
C2=16μF
VT = 25V
Calculate: 2
C1=6.0μF
(a) Q1 (b) QT (c) C2 (d) QT
CT=6.0μF 3
C1=6.0μF
Calculate:
CT=30μF
(a) C2 (b) V (c) Q1 (d) Q3
Q2=75μC
C3=9.0μF
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6. CT = C1 + C2 +… Parallel capacitor problems Calculate: 1
VT = 12V
Calculate: 1
CT=24μF
C1=8.0μF
(a) CT (b) Q1 (c) Q2 (d) QT
Q1=96μC
Q2=192μC
C2=16μF
QT=288μC
VT = 25V
Calculate: 2
C1=6.0μF
(a) Q1 (b) QT (c) C2 (d) Q2
Q1=150μC
QT=450μC
CT=18μF
C2=12μF
Q2=300μC 3
C1=6.0μF
Calculate:
CT=30μF
C2=15μF
(a) C2 (b) V (c) Q1 (d) Q3
V=5.0V
Q2=75μC
Q1=30μC
Q3=45μC
C3=9.0μF
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