Exercise 1: Two capacitors with capacitance C = 10 µF are connected in series in
ID: 1450824 • Letter: E
Question
Exercise 1: Two capacitors with capacitance C = 10 µF are connected in series in a circuit with resistance R = 150 . What is the time constant for this circuit?
Exercise 2: What would be the value for t1/2 for the circuit above?
Exercise 3: An AC power source supplies 12 V to a circuit and oscillates with a frequency of f = 60 Hz. What will the potential of the source be 12 ms after it is turned on?
Exercise 4: What happens to the value of t1/2 when the maximum potential of the source increases?
Note: this is for a lab on fast RC circuits
Explanation / Answer
1) Time constant, T = R*Cnet
= 150*(10/2)*10^-6
= 7.5*10^-4 s
2) Apply, Q = Qmax*(1 - e^(-t/T))
at t = t1/2, Q = Qmax/2
Qmax/2 = Qmax*(1 - e^(t1/2/T))
1/2 = 1 - e^(-t1/2/T)
e^(-t1/2/T) = 1 - 1/2
t1/2 = -T*ln(1/2)
= -7.5*10^-4*ln(1/2)
= 5.2*10^-4 s
3) V = Vmax*sin(w*t)
= Vmax*sin(2*pi*f*t)
= 12*sin(2*pi*60*12*10^-3)
= 12*sin(4.52)
= -11.8 volts
4) t1/2 does not depend on voltage of the source.
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