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My question is after steady state(ie. when capacitor is fully charged and behaves as an open circuit) is achieved what would be the potential difference across the dependent current source? How do I find it?

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    \$\begingroup\$ " independent current source " or " dependent current source ". The two are completely different, but your title says one while question say the other. \$\endgroup\$ – Tom Carpenter Feb 21 '17 at 8:13
  • \$\begingroup\$ 1) draw your circuit more clearly by separating the componnets more so the text does not overlap 2) you assume a steady state is achieved, is that always so ? What is required for a steady state to be achieved ? Think about a current flowing into a capacitor. \$\endgroup\$ – Bimpelrekkie Feb 21 '17 at 8:19
  • \$\begingroup\$ @TomCarpenter Sorry ,I have edited it, my question here is specific to the given circuit. \$\endgroup\$ – GeneX Feb 21 '17 at 8:41
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    \$\begingroup\$ For a capacitor, \$I = C\frac{\mathrm{d}V}{\mathrm{d}t}\$. What happens if \$I\$ is constant? \$\endgroup\$ – Tom Carpenter Feb 21 '17 at 8:44
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Assuming your v1 is constant and the current source 2v1 (times some suitable dimensions) is also constant your capacitor voltage will charge forever and the voltage across the current source will be approximately equal to the negative of that. It will climb until the current source headroom is reached or the capacitor (or other circuit component or connection) suffers a high voltage breakdown.

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  • \$\begingroup\$ So there would be no such state where in capacitor gets fully charged and behaves like an open circuit? \$\endgroup\$ – GeneX Feb 21 '17 at 9:02
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    \$\begingroup\$ A capacitor is never "fully" charged. If connected to a voltage limited source it will approach the voltage source asymptotically at a rate dependant on the circuit resistance. If fed from a current source it will charge indefinitely until some other limit is reached. A capacitor does not have a "full" voltage rating, it has working voltages and breakdown voltages that need to be respected. \$\endgroup\$ – KalleMP Feb 21 '17 at 9:30
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    \$\begingroup\$ A chemical battery does reach full charge when the reactions have run to completion, they too will continue to draw current (with various negative consequences) if it is not voltage limited in some way to remain within the cell voltage as determined by the electro chemistry. \$\endgroup\$ – KalleMP Feb 21 '17 at 10:02

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