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Prove that the capcitor will be open circuited when the dc source is applied?
Wiki User
∙ 13y ago
A capacitor will appear to be an open circuit to a DC source, but only after equilibrium is reached.
Proof:
A capacitor resists a change in voltage. The equation is ...
dv/dt = i/c
... which means that the rate of change of voltage in volts per second is equal to current in amperes divided by capacitance in farads.
A DC source has constant voltage. If you charge a capacitor to a constant voltage, then, at equilibrium, dv/dt is zero. This means that i must also be zero, since c is not zero.
Ohm's law states that resistance is voltage divided by current. The limit of this is that, when current is zero, then resistance must be infinity. Therefore, the capacitor will have infinite resistance and appear to be open circuited.
Wiki User
∙ 13y ago
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