Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a battery of 100V. If the energy stored in two combinations is 0.045 J and 0.25 J respectively, determine the value of C1 and C2. Also calculate the charge on each capacitor in parallel combination.
When the capacitor are connected in parallel,
Equivalent Capacitance,
CP=C1+C2
The energy stored in the combination
EP=12CPV2
⇒0.25=12(C1+C2)(100)2
⇒C1+C2=5×10−5⟶(i)
Similarly when the capacitor are connected in series
Cs=C1C2C1+C2
Energy Stored Es=12CsV2=12C1C2C1+C2(100)2=0.045
⇒C1C2=4.5×10−10
Now, (C1−C2)2=(C1+C2)2−4C1C2
=(5×10−5)2−4(4.5×10−10)
=7×10−10
C1−C2=√7×10−10=2.64×10−5⟶(ii)
on Solving (i) and (ii) we get,
C1=38μF, C2=1.2μF
charges
Q1=C1V=38×10−6×100=38×10−4C
Q2=C2V=12×10−6×100=12×10−4C