A capacitance of 100 uF is connected in
series with a resistance of 8000 ohm. The
time constant of the circuit is
(a) 0.2 s
(6) 0.4 s
(c) 0.6 s
(d) 0.8 s
Answers
Answer:
Half, about 0.7RC in seconds. I have that memorized because that is the formula when using a 555 timer. When it is charging, it is going from 1/3 of Vcc to 2/3 of Vcc, that is doubling the voltage, and when discharging it goes from 2/3 of Vcc to 1/3 of Vcc, that is half. And 0.7RC = t will get you pretty close.
If you want to know more precisely, the formula for capacitor discharge through a resistor is:
Vc = Vs*e^(-t/RC)
Vc is the capacitor voltage at time t, Vs is the starting voltage, e is a constant derived using calculus (about 2.718), and of course R is the resistance and C is the capacitance.
A bit of algebra:
Vc/Vs = e^(-t/RC)
ln(Vc/Vs) = -t/RC (note: “ln” means log base e)
t = -RC * ln(Vc/Vs) and ln(0.5) = -0.69314718 which is within 1% of 0.7 (ln is the natural log, the logarithm to the base of the constant e)
t = 0.693RC
t = 0.693 * 10^6 * 10^-7 = 6.93*10^-2 or 69.3ms
Pls mark as branliest
Answer:
0.8 sec hoga ,lets do this