Capacitance homework problem help

[ranju]

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..??

 

[Drakkith]

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?

 

[Delta2]

Why capacitance behaves as shortcircuited in uncharged state and why as open circuited when its fully charged..?? I am reallt doubtful..

Well for example when we charge a capacitor with a voltage source V via a resistance R, then we have from kirchhoffs 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 doesn't 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=Cdv/dt...so if we r appyling a dc voltage ..in that case i should be zero.>! thn why short-circuited..??

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.

 

[ranju]

Drakkith..and Delta your ans. is quite helpful. thanxx..

 

[sardar]

Capacitance behaves as short-circuited in the uncharged state because there is no voltage difference across the plates of the capacitor. In this state, the capacitor acts as a conductor and allows current to flow through it easily. This is because there is no electric field present between the plates to resist the flow of current.

On the other hand, when the capacitor is fully charged, there is a large voltage difference across the plates. This creates a strong electric field between the plates, which resists the flow of current. As a result, the capacitor behaves as an open circuit and does not allow current to pass through it.

The equation i=Cdv/dt represents the relationship between current, capacitance, and voltage. In the case of a DC voltage, the current is constant and does not change with time. This is because there is no change in the voltage over time. Therefore, the derivative of voltage with respect to time (dv/dt) is equal to zero, and the equation becomes i=0. This does not mean that the capacitor is short-circuited, but rather that there is no current flowing through it due to the constant voltage.

I hope this helps to clarify your doubts. If you have any further questions, please feel free to ask. Remember, as a scientist, it is important to question and seek understanding in order to deepen our knowledge and understanding of the world around us.