Here is a circuit with a battery, a resistor and a capacitor. The switch allows the battery to be included to the circuit, as shown. The battery can also be removed from the circuit. Recall the voltage across a capacitor is the charge on the capacitor, Q, divided by its capacitance, C. The voltage across the resistor is the current times the resistance. Let's suppose that the capacitor is fully discharged, and the battery is connected. Apply the loop rule to this case. The direction assumed for any current is as indicated. We assumed that the capacitor is discharged, that is, that Q is equal to zero. Therefore, a current equal to the e m f divided by the resistance flows through the resistor, as if the capacitor were not present. Because electric charge cannot cross the gap between the capacitor plates, charge accumulates on them. Note that the negative charge carriers move in the direction opposite that of the current. The positive charges on one capacitor plate represent the deficit of negative charge there. The voltage across the capacitor builds up. The current decreases as this charge increases… …and eventually approaches a steady state. In this steady state, there is no current and therefore no voltage drop across the resistor. The e m f is balanced by the voltage across the capacitor. The charge on the capacitor is the capacitance multiplied by the e m f. Suppose the battery is disconnected and a circuit containing only the resistor and the capacitor is established. We can apply the loop rule to this situation. The voltage across the fully charged capacitor is equal to the e m f, as we found previously. The sign convention for the direction of any current remains the same, as shown. Solving this for the current, we find that its magnitude is the same as when the battery is first connected to the uncharged capacitor, but it is flowing in the opposite direction. The current is flowing in the direction opposite to that of the current arrow I. The capacitor is now discharging through the resistor. Eventually the capacitor will be fully discharged. The capacitor in the circuit shown has been fully charged by the battery and the switch has just been closed. What is the voltage change across the resistor in the indicated direction of I? ... The net voltage around the circuit is zero. ... The capacitor is fully charged. ... This a picture of an oscilloscope screen that is displaying voltage in a resistor-capacitor circuit as a function of time. An electronic switch repeatedly switches an e m f in the circuit on and off. We can derive the behavior of a resistor-capacitor (or R C) circuit as a function of time. First we will consider the case of a capacitor being charged by a battery. Recall the equation for the voltages around the circuit. According to the definition of the electric current, the rate of change of the charge on the capacitor is equal to the current. Make this substitution in the equation for the voltages. This can be solved for the charge on the capacitor as a function of time. To see a detailed derivation of this equation, click on "Show Derivation". If we divide both sides of this equation by C, we obtain the voltage across the capacitor as a function of time. Here is a plot of the voltage across the capacitor as a function of time. At time t equals zero - the moment the switch is closed - the voltage across the capacitor is zero. As time t becomes large, this voltage approaches the value of the e m f. The characteristic time required for the capacitor to charge, R C, is called the capacitive time constant. The voltage across the resistor decreases with time as the capacitor charges. This voltage is shown as a dashed line. The solution for the case of a discharging capacitor initially charged by an e m f of magnitude epsilon can be solved in a similar way. Click on the "Show Derivation" button to see the details. The solution for the voltage across the capacitor and a plot of this solution is shown here. A circuit contains a charged 5 microfarad capacitor and a 400 ohm resistor. How long will it take for the capacitor to lose 99 percent of its charge through the resistor? ... The time to lose 99% of the charge is greater than the capacitive time constant. ... Check your calculation of the capacitive time constant. ... A circuit contains a 200 ohm resistor and a discharged capacitor. A 4 volt potential is suddenly applied and the potential across the capacitor rises to 2 volts in 10 microseconds. What is the capacitance of the capacitor? ... Remember that one is 10-6 F. ... The final voltage is only 50% of the applied value. ...

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