Three capacitors have the capacitances C 1 = 15.0mF, C 2 = 16.8mF, and C 3 = 11.

Question

Three capacitors have the capacitances C1 = 15.0mF, C2 = 16.8mF, and C3 = 11.4mF.

What is the maximum possible capacitance that these three capacitors could have (in mF)? What is the minimum possible capacitance that these three capacitors could have (in mF)?

If a voltage of 20 V is applied, what is the maximum energy that could be stored in the capacitors? If C1 is charged by itself using the 20 V battery and a resistance of 1000 ?, what time would be required to charge to 80% of a full charge?

Explanation / Answer

in parallel connection we get maximum capaciatnce

Cmax = c1 + c2 + c3

= 15 + 16.8 + 11.4

= 43.2 mF

in series connection we get minimum capaciatnce

1/Cmin = 1/c1 + 1/c2 + 1/c3

1/Cmin = 1/15 + 1/16.8 + 1/11.4

Cmin = 4.67 mF

Qmax = 0.5*Cmax*V^2

= 0.5*43.2*10^-3*20^2

= 8.64 J
===============================

Time constant, T = R*C1 = 1000*15*10^-3 = 15 s

Apply, Q = Qmax*(1 - e^(-t/T))

0.8*Qmax = Qmax*(1 - e^(-t/T))

0.8 = 1 - e^(-t/T)

e^(-t/T) = 1 - 0.8

t = -T*ln(0.2)

= -15*ln(0.2)

= 24.14 s

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