A R-C series circuit is connected to a 12 volt batten.-. Initially the capacitor

Question

A R-C series circuit is connected to a 12 volt batten.-. Initially the capacitor has zero charge. Assume R= 42 ohm. C= 2.424 pF. When the circuit is closed, calculate the time (in Millisecond) when the capacitor voltage will be 63.21% of its final voltage. In the following circuit, Ci=8 F, C3=8/2 Fand C4=8/2 F. Equivalent Capacitate is given 0.380 F. What is C2 in F? A parallel plate capacitor is connected with a 1.897 volt battery and each plate contains 4.853 micro Coulomb charge. How much energy is stored in the capacitor? A certain capacitor stores 34 J of energy when it holds 3,485 uC of charge. What is the capacitance in nF?

Explanation / Answer

During charging a capacitor tha charge increases as

q = Q*(1-e^-(t/RC))

given R = 42

C = 2424 uF

q = 0.6321Q

0.6321 = 1-e^-(t/(42*2424*10^-6))

time t = 102 ms <<------answer

++++++++++++++++++

C1 , C2 are in series

C12 = (C1*C2)/(C1+C2) = (8*C2)/(8+C2)

C3, C4 are in parallel

C34 = C3 + C4 = 8/2 + 8/2 = 8 F

C12 & C34 are in series

Ceq = (C12*C34)/(C12 + C34)

0.38 = ((8*C2)/(8+C2)*8)/((8*C2)/(8+C2) + 8)

solving theabove equation

C2 =0.42 F <<<--answer

+++++++++++

energy stored in a capcitor E = (1/2)*Q*V = (1/2)*4853*10^-6*1897 = 4.61 F

_____________________

energy stored in a capcitor E = (1/2)*Q^2/C

34 = (1/2)*(3485*10^-6)^2/C = 178.6 nF <<<<---------answer

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