| Themes > Science > Physics > Electromagnetism > Electrostatics > Capacitors & Dielectrics > Resistor-Capacitor Circuits > Charging a Capacitor |
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When the switch in the above circuit is closed, and current begins to flow, the flow will not be constant. In fact, the current will decrease with time. This occurs because the charge builds up on the plates of the capacitors until the charge flow ceases. The rate of charge flow is not the only thing varying with time, however. The capacitor's potential V and charge Q on the capacitor also vary with time. Since the definition of capacitance tells us that Q=CV, we know that V and Q will increase at the same rate. The rate of charge and electric potential increase of the capacitor is the same as the rate of electric current decrease in the circuit. Let us consider the ciruit above with the switch just having been closed. Right as the switch is closed, and time t equals zero, our equation V=IR holds true. At this point, the circuit is treated like the capacitor is not present. The voltage in the circuit equals E (of the battery) and current is at its maximum (I=E/R). As time progresses, the capacitor affects the flow of charge as described above. Current will decrease according the equation: also written as, Further, the electric potential of the capacitor and the charge on the capacitor will increase according to the equations:
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