calculating the discharging time in RC circuit !!?
I'm having a hard time to solve this question. I will really thank you for helping me.
So, the problem is I have to calculate the discharging time of a capacitor in an RC circuit.
1. If the experimental set up in the picture below has a 12V battery and a 0.22microF capacitor,
determine the time it would take for the fully charged capacitor to decrease the voltage
from 12.0V to 8.0V (switch will be in the discharging position A-C).
2. The two resistors, 100Ω and 1MΩ, in this circuit are different by orders of
magnitude. Explain why
I firstly added these two resistors up so R = 100 + 10^6 = 1000,100(ohm)
and the time constant(T) = RC = (1000,000)*(0.22* 10^-6) = 0.22
The equation for discahrging the capacitor is v = v0e^(-t/T)
hence the time(t) = -T*ln(v/v0) and I got 0.0892s.
But I think this number is too small and I got confused. Can someone answer me these two questions? Thanks in advance!
- billrussell42Lv 73 weeks agoFavorite Answer
no reason to add the two resistors, the 100 Ω is there to just limit the current into the cap when charging. When discharging, it is out of the circuit.
τ = RC = 1M x 0.22 µF = 0.22 sec
v = v₀e^(–t/τ)
8 = 12e^(–t/0.22)
e^(–t/0.22) = 8/12 = 2/3
–t/0.22 = ln 2/3 = -0.405
t = 0.0892 sec
why? the circuit designer wanted the charge time to be short compared to the discharge time. But you could use 1 MΩ for both, it would just take longer for the cap to charge up.
voltage on a cap, discharging
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant