If the resistor is 9v, the capacitor is 3 volts.
Vc=q/C=12/(5E6*6E-6)*e^-t/rc
3*30/12=e^-t/30
take the ln of each side, solve for t.
check my typing, my eyes are blurry this am.
Now for the ac part, Z=5E6+1/(2PI60)*6E-6 figure that, it is nearly all resistive.
Then I=12/Z
I can check your work on this.
A 12 V battery is connected to 5 mega ohm resistor and a 6 micro farad capacitor and an open switch. After how many seconds will the voltage across the resistor reach 9 V?
Suppose the battery is replaced by 12 V ac with a frequency of 60 Hz.Calculate the Irms in the circuit
2 answers
a. RC = 5*10^6 * 6*10^-6 = 30 s.
Vr = 12/e^x = 9
e^x = 12/9 = 1.333
X = 0.288 = t/RC.
t/30 = 0.288
t = 8.63 s.
b. Xc = 1/(2pi*F*C) = 1/(6.28*60*6*10^-6) = 442 ohms.
Z = 5*10^6 - j442 = 5*10^6 0hms.
I = E/Z = 12/(5*10^6) = 2.4*10^-6 A = 2.4 uA.
Vr = 12/e^x = 9
e^x = 12/9 = 1.333
X = 0.288 = t/RC.
t/30 = 0.288
t = 8.63 s.
b. Xc = 1/(2pi*F*C) = 1/(6.28*60*6*10^-6) = 442 ohms.
Z = 5*10^6 - j442 = 5*10^6 0hms.
I = E/Z = 12/(5*10^6) = 2.4*10^-6 A = 2.4 uA.
1 answer