5. When a series
combination of two uncharged capacitors is connected to a 12 V battery, 96 mJ of energy is drawn from the battery. If one of the capacitors has a capacitance of
4 mF, what is the capacitance of the
other?
- Algebra Based Solution
Given Data:
Vb=12V
U=96 mJ
C1=4mF
C1 and C2 are series
C2=?
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Useful formulas:
U=QV/2=CV2/2
C=Q/V
1/Ceq=1/C1+1/C2
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Algebra Job
U=CeqVb2/2
1/Ceq= Vb2/(2U)
1/C1+1/C2 = Vb2/(2U)
1/C2 = Vb2/(2U) - 1/C1
C2=1/( Vb2/(2U) - 1/C1 )
or
1/C2 = Vb2/(2U) - 1/C1
1/C2 = (Vb2/2 – U/C1) /U
C2 = U/(Vb2/2 -
U/C1)
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Substitution and Calculation:
C2=1/( Vb2/(2U) - 1/C1
) = 1/( (12V)2/(2·96 mJ) - 1/(4mF) )
Expression for Google Scientific Calculator
1/( (12Volt)^2/(2*96microjoule) – 1/(4microfarad)) in microfarad
Google result: 2 microfarad
Result: C2 = 2mF
- Arithmetic Based Solution
Given Data:
Vb=12V
U=96 mJ
C1=4mF
C1 and C2 are series
C2=?
|
Useful formulas:
U=QV/2=CV2/2
C=Q/V
1/Ceq=1/C1+1/C2
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U=CeqVb2/2
96mJ =Ceq(12V)2/2
(4·4·2·3)mJ =Ceq ·4·3·4·3V2/2
(2)mJ =Ceq ·3V2/2
4/3 (mJ/ V2)
= 4/3 mF = Ceq
1/Ceq = ¾ ·
1/mF
1/C2=1/Ceq-1/C1
1/C2 = ¾ ·
1/mF – ¼ ·1/mF
1/C2 = 2/4 ·
1/mF = 1/2 ·
1/mF
C2= 2mF
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- Another Arithmetic Based
Solution
Given Data:
Vb=12V
U=96 mJ
C1=4mF
C1 and C2 are series
C2=?
|
Useful formulas:
U=QV/2=CV2/2
C=Q/V
Q1=Q2=Qeq=Q
Vb=V1+ V2
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Q=2U/V=2·96
mJ/12V
Q= 2·4·4·2·3
/(4·3) mJ/V
Q= 2·4·2 mC
V1=Q/C1=2·4·2 mC/4mF=4V
V2=Vb -V1=12V-4V=8V
C2=Q/V2=2·4·2 mC/8V=2mF
C2= 2mF
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