Two capacitors, C1 = 19.0 µF and C2 = 31.0 µF, are connected in series, and a 15
ID: 2050318 • Letter: T
Question
Two capacitors, C1 = 19.0 µF and C2 = 31.0 µF, are connected in series, and a 15.0 V battery is connected across them.(a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor.
equivalent capacitance:___ µF
total energy stored:___ J
(b) Find the energy stored in each individual capacitor.
energy stored in C1:___ J
energy stored in C2:___ J
Show that the sum of these two energies is the same as the energy found in part (a). Will this equality always be true, or does it depend on the number of capacitors and their capacitances?
(c) If the same capacitors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy as in part (a)?
V:___
Which capacitor stores more energy in this situation?
Explanation / Answer
Q = CV
energy = (1/2) C V^2
a) In series,
Equivalence capacitance : Ceq = C1C2/(C1+C2)
=19*31/(19+31)
=11.78F
total energy stored =1/2Ceq V^2
=0.5*11.78*15^2
=1325.25*10^-6 jule
b) due to series combination the potentials across capacitors are
V1 = 9.3V and V2 = 5.7V, with these values the energies of capacitors are
U1 = 1/2 C1V1^2 =8.21*10^-4 J and U2 = 1/2 C2V2^2 = 5.03*10^-4 J
and sum of their energies U1+U2 =13.24 *10^-4J
this will be applicble for any number of capacitors.
c)when the capacitors are in parallel for the same energy as in (a) the potential needed will be U = 1/2 (C1+C2) V^2, from which V =5.625*10^-4.the energy stored will be more in C2.
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