Two capacitors, C1 = 25.0 µF and C2 = 44.0 µF, are connected in series, and a 15

Question

Two capacitors, C1 = 25.0 µF and C2 = 44.0 µF, are connected in series, and a 15.0-V battery is connected across the two capacitors. (a) Find the equivalent capacitance. µF (b) Find the energy stored in this equivalent capacitance. J (c) Find the energy stored in each individual capacitor. capacitor 1 J capacitor 2 J (d) Show that the sum of these two energies is the same as the energy found in part (b). This answer has not been graded yet. (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)? V (g) Which capacitor stores more energy in this situation, C1 or C2? C1 C2 The capacitors store the same amount of energy.

Explanation / Answer

a)    equivalent capacitance. =   15.942 µF

b)    the energy stored in this equivalent capacitance = 0.5 * 15.942* 10-6 * 15 *15

                                                                                      = 1.793 * 10-3 J

c)     Voltage across C1 = 15.942 * 15/25

                                        = 9.56 V

      => energy stored in C1 = 0.5 * 25 * 10-6 *9.56 *9.56

                                               = 1.142 * 10-3 J

        => energy stored in C2 = 0.5 * 44 * 10-6 *5.44 *5.44

                                               = 0.651 * 10-3 J

d)   sum of these energies =   1.142 * 10-3   + 0.651 * 10-3

                                                           =    1.793 * 10-3 J

    So, sum of these two energies is the same as the energy found in part (b)

e)    This equality will always be true for series combination of capacitors .

f)     equivalent capacitance for parallel combination =   69 µF

    potential difference would be required = sqrt( 2 * 1.793 * 10-3/69 *10-6)

                                                                  = 7.21 V

g)    energy stored by C1 =   0.5 * 25 * 10-6 *7.21 *7.21

                                          = 0.6498 * 10-3 J

       energy stored by C2 =   0.5 * 44 * 10-6 *7.21 *7.21

                                          = 1.143 * 10-3 J

=>   C2 capacitor store more energy .

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.