Why capacitance behaves as shortcircuited in uncharged state and why as open circuited when its fully charged..?? I am reallt doubtful.. i=Cdv/dt...so if we r appyling a dc voltage ..in that case i should be zero.>!!! thn why short-circuited..??
A capacitor stores electrical energy by moving charges off of one plate and onto the other. Initially each plate is electrically neutral, just like a normal conductor, so the first few charges require very little work to move and current flows as if the capacitor were short circuited. As more and more charges are moved from one plate to the other, an electric field develops that opposes the voltage source, meaning that each additional charge requires more work to move. When the electric field is equal to the applied voltage, there is no longer any difference in electric potential (voltage) between the voltage source and the plates so you have no more current flow, just like an open. Does that help?
Well for example when we charge a capacitor with a voltage source V via a resistance R, then we have from kirchoff's Voltage Law V-I*R-q/C=0 (1). When the capacitor is uncharged then q=0 the term q/C is also zero, hence what is left from (1) is V-IR=0, it is like the capacitor has been short circuited (or it doesnt exist). When capacitor is fully charged then q=C*V hence from (1) follows that -I*R=0 thus I=0, it is like the circuit is open and no current can flow. i=0 in the steady state. Before the steady state, there will be a transient state of time dt where the current has very high value (theoretically infinite ) and it charges the capacitor very rapidly (within time dt) to final charge q=C*V. So again the capacitor passes from the "short circuit" stage where the current is very high to the open circuit stage where the current is zero.