An initially uncharged 4.41-F capacitor and a 6.25-k? resistor are connected in

Question

An initially uncharged 4.41-F capacitor and a 6.25-k? resistor are connected in series to a 1.50-V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes? Calculate the circuit's time constant in milliseconds. How much time, in milliseconds, must elapse from the closing of the circuit for the current to decrease to 2.15% of its initial value? Initial current: Number 0.24 mA Time constant: Number 2.756% 1 0-1 ms Elapsed time: Number ms Incorrect.

Explanation / Answer

time constant = 6.25 k x 4.41 µ = 27.56 ms

2.15% x 240 µA = 5.16 µA
that current causes a voltage across 6.25 k? of 32.25 mV
which means the voltage across the cap is 1.5–32.25 mV = 1.468

voltage on a cap, charging
v = v?[1–e^(–t/?)]
v? is the final voltage
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = ? = time constant

v = v?[1–e^(–t/?)]
1.468 / 1.5 = 1–e^(–t/?)
e^(–t/?) = 1– 0.9785 = 0.0215
–t/? = –3.839
t = 105.8 ms

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