a) Two capacitors of 26F and 4.44 F are connected in parallel and charged with a

Question

a) Two capacitors of 26F and 4.44 F are connected in parallel and charged with a150 V
power supply. Find the total energy stored in thecapacitors. Answer inunits of J.

b) What potential difference would be required across the same twocapacitors connected in
series in order for the combination to store the same amount ofenergy?
Answer in units of V.
a) Two capacitors of 26F and 4.44 F are connected in parallel and charged with a150 V
power supply. Find the total energy stored in thecapacitors. Answer inunits of J.

b) What potential difference would be required across the same twocapacitors connected in
series in order for the combination to store the same amount ofenergy?
Answer in units of V.

Explanation / Answer

Given C1 = 26F = 26 *10-6  F    C2 = 4.44 F   = 4.44*10-6   F voltage   V = 150 V     equivalent capacitance of thecapacitors                C = C1 + C2   = 26   + 4.44 =30.44 *10-6   F    energy stored in the capacitors                U    = 1 / 2 C V2                      = 0.5 * 30.44*10-6   F * 150*150                       = 0.3 J     Capacitors connected in seriescombination          equivalentcapacitance    C   = C1 C2 / C1 +C2                                                     = 26*10-6   F *4.44*10-6   F   / (26 + 4.44 )10-6   F                                                     = 3.79* 10-6   F              energy stored in the capacitor   U = 1 / 2 C V2   required potential energy V2 = 0.3*2    / 3.79*10-6   F                   V    =397.88 V

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