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Capacitance homework problem helpby ranju
Tags: capacitance 
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#1
Jul514, 06:37 AM

P: 85

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 shortcircuited..?? 


#2
Jul514, 06:55 AM

Mentor
P: 11,891

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? 


#3
Jul514, 06:55 AM

P: 436

VI*Rq/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 VIR=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. 


#4
Jul514, 08:03 AM

P: 85

Capacitance homework problem help
Drakkith..and Delta your ans. is quite helpful. thanxx..



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