a capacitor 'c' is being charged by connecting it across a dc source along with an ammeter. will the ammeter show any deflection momentary during the process of charging? if so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? write the expression for the current inside the capacitor.
Asked by vasundhra khatter  6th Mar, 2012, 03:11: PM
a capacitor, in short circuit with a battery, will act like a battery. It will 'charge' up to whatever voltage the battery is willing to put out. The internal 'resistance' of the capacitor will become evident as it charges. As the plates get their charge, they start to oppose the incoming charge. When the internal voltage of the capacitor reaches the same voltage as its source, it will have infinite resistance (in a charging sense), and no more current will flow. The difference between a battery and a capacitor (of similar voltage ratings and physical size) is that the capacitor holds 'less' energy (in all but a few cases)  but that capacitor can dump its entire charge in a few nanoseconds, where the battery couldn't possibly dump that much charge.
Graphs of charge vs time and current vs time for charging capacitors are shown below. Mathematically, both of these graphs are exponential functions  current is an exponential decay, while charge is an exponential growth.
Charging capacitor graphs
charge vs time
current vs time
When the switch is initially closed, the capacitor does not have carry any charge and Kirchoff's loop rule would result in the equation:
 R V_{c} = 0
= R
= / R
Graphs of charge vs time and current vs time for charging capacitors are shown below. Mathematically, both of these graphs are exponential functions  current is an exponential decay, while charge is an exponential growth.

When the switch is initially closed, the capacitor does not have carry any charge and Kirchoff's loop rule would result in the equation:
 R V_{c} = 0
= R
= / R
Answered by  7th Mar, 2012, 12:29: PM
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