An uncharged 3.20-μF capacitor is in series, through a switch, with a 5.55-M Ω resistor and a 36.0- V battery (with negligible internal resistance.) The switch is closed at t = 0 and a current of Ii immediately appears. Determine Ii.
How long will it take for the current in the circuit to drop to 0.37Ii?
To determine the initial current, we can use Ohm's Law, which states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is provided by the battery and the resistance is given by the resistor. Therefore, Ii = V/R.
Substituting the values given in the question, we have:
Ii = 36.0 V / 5.55 MΩ = 6.49 × 10^(-6) A (rounded to 3 significant figures)
Now, to find out how long it will take for the current to drop to 0.37Ii, we need to consider the time constant (τ) of the circuit. The time constant is given by the product of the resistance and the capacitance (RC).
τ = R * C
= 5.55 MΩ * 3.20 μF
= 5.55 × 10^6 Ω * 3.20 × 10^(-6) F
= 17.76 s
The time constant represents the time it takes for the current to drop to approximately 36.8% of its initial value. We can use this information to solve for the time it takes for the current to drop to 0.37Ii.
0.37Ii = Ii * e^(-t / τ) (where e is Euler's number)
Rearranging the equation to solve for time (t):
t = -τ * ln(0.37)
Using a scientific calculator, we can find:
t ≈ -17.76 s * ln(0.37)
≈ 17.76 s * 0.9957
≈ 17.673 s
Therefore, it takes approximately 17.673 seconds for the current in the circuit to drop to 0.37Ii.