We can apply the principle of exponential decay to find equations that model the charge remaining in a capacitor as it is discharged through a resistor. The circuit diagram is shown below...
When the switch is closed, current will start flowing from the capacitor, through the resistor until the capacitor is discharged (or the voltage in the capacitor is close enough to zero). This process is depicted below.
In the real world, this might be applicable to a camping torch where the resistor is the light source, and the capacitor is the battery.
Our question is:
When a capacitor is discharged through a resistor, the current I, the charge Q on the capacitor and the voltage V are related by Q = CV and V = IR. If a capacitor of 20 microfarads is discharged through a resistance of 3000 Ohms from 20 Volts, find the charge, voltage and current at time t.
Firstly, to ensure there's no short circuit, the voltage drop across the resistor must be the same as the voltage in the capacitor at any time. The voltage drop across the resistor is given by V = IR. Rearranging this, the current in the circuit at anytime is:
The charge in the capacitor is decreasing over time. Thus in the capacitor, dQ/dt must be negative. The rate of decrease is the same as the current that's flowing in the circuit. Therefore...
The relationship between the charge and voltage of the capacitor is Q = CV. Therefore by rearranging, we have...
The substituting this into equation (1), we have...
Equation (2) is the differential equation we'll solve to find the charge at any time. So, separating the variables and integrating...
All we need now are values for Q0 and RC. Given the initial voltage V0 = 20V...
Thus the charge remaining in the capacitor at any time is given by...
The voltage is also decreasing at the same rate as Q through the relationship Q = CV...
The current in the circuit, given by the relationship I = V/R at any time is then...
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