An uncharged capacitor with C = 96 ?F and a resistor with R = 45 ? are connected

Question

An uncharged capacitor with C = 96 ?F and a resistor with R = 45 ? are connected in series with a battery of ? = 8.5 V.

Part (a) Express the time constant ? in terms of R and C.

Part (b) Calculate the numerical value of ? in ?s.

Part (c) Express the maximum charge Q on the capacitor in terms of C and ?.

Part (d) Calculate the numerical value of Q in ?C.

C = 96 ?F R = 45 ? ? = 8.5 V

9.) Two capacitors of capacitance 3C and 5C (where C = 0.105 F) are connected in series with a resistor of resistance R = 6.5 ?.

Part (a) How long will it take the amount of charge in the circuit to drop by 75% in seconds?

Part (b) If the circuit was charged by a 10.0 V source how much total charge (in C) did both capacitors have in them to begin with?

R = 6.5 ? C = 0.105 F

Explanation / Answer

Given

8)

C = 96 ?F and a resistor with R = 45 ? are connected in series with a battery of ? = 8.5 V.

Part a

time constant is T = R*C

Part b

T = 45*96*10^-6 s

T = 0.00432 s

Part C

the maximum charge on the capacitor is  

from the relation between Q, C and V

Q = C*V

Part d

Q = 96*10^-6*8.5 C

Q = 0.000816 C

9)

C1 = 3C = 3*0.105 = 0.315 F,

C2 = 5C = 5*0.105 = 0.525 F

R = 6.5 ohm

when C1,C2 are connected in series then the net capacitance is

C = C1*C2/(C1+C2)

C = (0.315*0.525)/(0.315+0.525) F

C = 0.196875 F

discharging of a capacitor in RC circuit is  

q(t) = Qe^-t/T

here T = R*C = 6.5*0.196875 s = 1.2796875 s

0.75*Q = Qe^-(t/1.2796875)

solving for t , t = 0.3681 s

time taken take the amount of charge in the circuit to drop by 75% in seconds is take the amount of charge in the circuit to drop by 75% in seconds is 0.3681 s

Part (b)

Charge Q = C*V

Q = 0.196875 *10 = 1.96875 C

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