Three capacitors C1-10.9 F, C2 = 20.0 F, and C3-28.9 F are connected in parallel

Question

Three capacitors C1-10.9 F, C2 = 20.0 F, and C3-28.9 F are connected in parallel. To avoid breakdown of the capacitors, the maximum charge any one of them can individually hold is 1500 HC. (a) Determine the total maximum charge in the parallel combination. 8.228 Let one of the capacitors have the maximum charge of 1500 C. Based on the values of C, to which capacitor can you allocate this value? How is the potential difference across each capacitor related in a parallel combination? mC (b) Determine the maximum energy stored in the parallel combination. 0.57 How is the energy stored related to the capacitance and the charge?

Explanation / Answer

we know that

Q = C*V

in parallel combination V is same on all three capacitors .

so, maximum charge on the capacitor with highest capacitance .

C3 is highest so given charge in on C3. so,

we can calculate V for the combination .

V = Q3/C3 = 1500*10-6/ 28.9*10-6

V = 51.90 V

(a)

Ceq = C1 + C2 +C3

Ceq = (10.9 + 20 + 28.9)*10-6

Ceq = 59.8*10-6 C

so,

Q = Ceq *V

Q = 59.8*10-6 * 51.90

Q = 3.10*10-3 C

(b)

E = 1/2*Q*V

E = 0.5*3.10*10-3 *51.90

E = 0.080 J

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