A 8.0-mu F unchanged capacitor is connected in series with a 6.0-k ohm resistor,

Question

A 8.0-mu F unchanged capacitor is connected in series with a 6.0-k ohm resistor, an ideal 20-V de source, and an open switch. If the switch is closed at time t = 0.0s, what is the charge on the capacitors at t = 9.0 ms? 0 C 37% of the minimum charge 17% of the maximum charge 68% of the minimum charge 96%of the maximum charge A 2.0-mu F capacitor that is initially uncharged is charged through a 50-k ohm resistor. How long does it take for the capacitor to reach 90% of its full charge? 0.90s 0.23 s 2.2 s 2.3s A A fully charged 37-mu F capacitor is discharged through a 1.0-k ohm resistor. If the voltage across the capacitor is reduced to 7.6 volts after just 20 ms, what was the original potential across the capacitor? 16 V 13 V 11 V 9.0 V 8.0 V A 2.0-mu F capacitor is charged to 12 V and then discharged through a 4.0-M ohm resistor. How long will it take for the voltage across the capacitor to drop to 3.0 V? 8.0s 11s 22 s 24s

Explanation / Answer

RC circuit

charging and discharging

time constant is T = R*C

charging condition is Q(t) = Q0(1-e^(-t/T)

discharging condition is Q(t) = Q0(e^(-t/T)

149)

C = 8.0*10^-6 F, R = 6.0*10^3 ohm, V0 = 20 V

t = 9.0 ms

here total charge Q0 = C*V = 8.0*10^-6*20 C = 160*10^-6 C

Q(t) = Q0(1-e^(-t/T)

substituting the values

Q(t) = 160*10^-6(1-e^(-(9*10^-3)/(6*10^3*8*10^-6))) C

Q(t) =2.7355341091136*10^-5 C

tha tis 27.355341091136*10^-6 = x%of Q0

x = 27.355341091136*10^-6*100 /(160*10^-6) = 17.09

which is 17.09 % of total charge

so the answer C is correct

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150)

C= 2*10^-6 F, R = 50 kohm

Q(t) = 90% of Q0, t = ?

T = R*C = 50000*2*10^-6 s = 0.1 s

Q(t) = Q0(1-e^(-t/T)

0.9 Q0 = Q0(1-e^(-t/0.1))

t = 0.230 26 s

so the answer is option B

151)

C = 37*10^-6 F,R = 1000 ohm, v(t) = 7.6 V, t = 20 ms, T = R*C = 1000*37*10^-6 s = 37 ms

from the equation

V(t) = V0(e^(-t/T)

7.6= V0*e^(-0.02/0.037)

V0 = 13.05 V

so ther answer is option B

152) C = 2*10^-6 F, R = 4*10^6 ohm, V0 = 12 V,V9t) = 3.0 V

T = R*c = 4*10^6*2*10^-6 s = 8 s

from the relation

V(t) = V0(e^(-t/T)

3 = 12*e^(-t/8)

solving for t , t = 11.09s

so the answer is optionn B 11s

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