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

Question

A R-C series circuit is connected to a 12 volt battery. Initially the capacitor hat zero charge. Assume R = 26 Ohm, C = 2.680 mu F. When the circuit is closed, calculate the time (in Millisecond) when the capacitor voltage will be 63.21% of its final voltage. Your Answer: In the following circuit. C_1 = 8 F, C_3 = 8/2 F and C_4 = 8/2 F. Equivanent Capacitace is given 0.473 F. What it C_2 in F? Your Answer: A parallel plate capacitor it connected with a 1.044 volt battery and each plate contains 2,192 micro Coulomb charge. How much energy it stored in the capacitor? Your Answer:

Explanation / Answer

(7)
R = 26 ohm
C = 2680 uF

Time will be equal to 1 time constant,
t = R*C
t = 26 * 2680 * 10^-6 s
t = 69.7 ms

(8)
c1 = 8 F
c2 = ?
c3 = 4 F
c4 = 4 F
ceq = 0.473 F

c3 & c4 are in paralle, C34 = c3 + c4
c34 = 4 + 4 = 8 F

c1 , c2 & c34 are in series,
Ceq = 1/(1/cx1 + 1/c2 + 1/c34)
0.473 = 1/(1/8 + 1/c2 + 1/8)
c2 = 0.536 F

(9)
Q = 2192 * 10^-3 C
V = 1044 V
Energy stored, U = 1/2*Q*V
U = 1/2 * 2192 * 10^-3 * 1044
U = 1144.2 J

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