Two capacitors, C_1 = 28.0 mu F and C_2 = 43.0 mu F, are connected in series, an
ID: 1599090 • Letter: T
Question
Two capacitors, C_1 = 28.0 mu F and C_2 = 43.0 mu F, are connected in series, and a 15.0-V battery is connected across the two capacitors. (a) Find the equivalent capacitance. (b) Find the energy stored in this equivalent capacitance. (c) Find the energy stored in each individual capacitor. capacitor 1 capacitor 2 (d) Show that the sum of these two energies is the same as the energy found in part (b). (e) Will this equality always be true, or does it depend on the number of capacitors and their capacitances? This equality will always be true. This equality depends on the number of capacitors. (f) 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)? (g) Which capacitor stores more energy in this situation, C_1 or C_2? C_1 C_2 The capacitors store the same amount of energy.Explanation / Answer
In series combination
Cs = (c1*C2)/(C1+C2)
Cs = (28*43)/(28+43)
Cs = 16.9 uF <<<====answer
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(b)
energy stored U = (1/2)*Cs*v^2 = (1/2)*16.9*10^-6*15^2 = 0.0019 J
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(c)
charge on c1 = c2 = q = Cs/E = 16.9*15 uC = 253.5*10^-6 c
energy of capacitor 1 = U1 = (1/2)*Q^2/c1
u1 = (1/2)*( 253.5*10^-6)^2/(28*10^-6)
u1 = 0.00115 J
energy of capacitor 2 = U2 = (1/2)*Q^2/c2
u2 = (1/2)*( 253.5*10^-6)^2/(43*10^-6)
u1 = 0.00074 J
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(d)
U1 + U2 = 0.0012 + 0.0007
U = 0.00190125
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(e)
This equality will always be true
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(f)
In parallel equivalent capacitance Cp = C1 + C2 = 28 + 43 = 71 uF
energy stored U = (1/2)*Cp*V^2
U = 0.0019 J
0.0019 = (1/2)*71*10^-6*v^2
V = 7.32 V
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(g)
energy stored = (1/2)*C*v^2
as v is same
C2 > C1
C2 stores more energy
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