Two capacitors, C1 = 27.0 µF and C2 = 45.0 µF, are connected in series, and a 15
ID: 2050124 • Letter: T
Question
Two capacitors, C1 = 27.0 µF and C2 = 45.0 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 16.88µF
total energy stored 1.899e^-3J
(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
I figured part A but I cannot get the rest. Thank you.
Explanation / Answer
a) C1=27 F==>C2=45 F==>V=15 V==>Ceq=1/(1/C1+1/C2)16.88 F
Ceq*V=Qtotal=16.88 F==>16.88 F*15 V253.13 C
WC=1/2*C*V2=1898.44 Joules total==>
b) 253.13 C/27 F9.375 V for C1
253.13 C/45 F5.625 V for C2==>1/2*27 F*9.3752
1186.5 Joules for C1 or 0.0011865 Joules
1/2*45 F*5.6252711.9 Joules for C2 or 0.0007119 Joules
0.0011865+0.00071190.00189844 Joules check ok!
c) (0.00189844/(72 F)*2)7.262 V
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