A capacitor of capacitance C= 10mF, is charged using a battery of potential V, through a resistor of resistance R = 250 ohms.
the charge of the capacitor is given by
Q(t)=CV[1− e ^ (−t/RC)]
Find an expression for the current passing through the resistor as a function of time, and make a graph clearly indicating the initial values. how long will it take for the current passing through the resistor to drop to 37% of its max value?
Current is the derivative of the capacitor charge.
I(t) = dQ/dt = CV/(RC) exp(-t/RC)
= (V/R) exp(-t/RC)
Current is 1/e = 37% of the initial value when t/RC = 1. That is called the time constant.
You will have to draw you own graph.
To find the expression for the current passing through the resistor as a function of time, we can use Ohm's Law, which states that the current (I) through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R). In this case, the voltage across the resistor is equal to the voltage of the battery (V) since the capacitor is fully charged.
Therefore, the expression for the current passing through the resistor is given by:
I(t) = V / R
Now, let's calculate the time it takes for the current passing through the resistor to drop to 37% of its maximum value. To determine this, we need to find the time (t) such that I(t) = 0.37 * Imax, where Imax is the maximum current passing through the resistor.
Given the expression for the current (I), we substitute I(t) and Imax:
0.37 * Imax = V / R
Now, rearrange the equation to solve for time (t):
t = -RC * ln(1 - 0.37)
Using the given values for capacitance (C = 10mF = 10 * 10^-3 F) and resistance (R = 250 Ω), and assuming the battery potential (V) is known, you can substitute these values into the equation to calculate the time it takes for the current passing through the resistor to drop to 37% of its maximum value.