When a series combination of two uncharged capacitors is connected to a 12 V battery

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?

1. Algebra 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

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)

Substitution and Calculation:

C₂=1/( V_b²/(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

2. 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

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

3. 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

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