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enter image description hereI know that in a circuit with DC sources a capacitor(steady state) can be replaced with an open circuit and an inductor(steady state) can be replaced with a short circuit.

My understanding is that since there will be no current through the capacitors in steady state I could just remove them from the circuit.

I've simplified the circuit using those information given above but I am unsure what to do first to find ix and Vx.

I tried combining the two 2 ohm resistors and doing a current division using the 4 amp source but ix turned out to be different from the answer (ix = -1/4A and Vx = 9/2V)

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Next step in a systematic approach like you have started would be to use superposition. In that technique you find the contribution of each source separately to the $i_x$ and $V_x$. Be sure that you pay careful attention to all signs. A missed negative can really mess you up.

First, find the (signed) contribution to $i_x$ and $V_x$ from the 5V supply by replacing the current supply with an open circuit.

Then, find the (signed) contribution to $i_x$ and $V_x$ from the 4A current supply by replacing the voltage supply with an short circuit.

If you ever have any dependent supplies, you cannot replace them.

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  • $\begingroup$ "If you ever have any dependent supplies, you cannot replace them." Does this mean that I will have to solve for ix and Vx with out replacing both the current and voltage sources since they are both dependent sources? $\endgroup$ – user114027 Mar 5 '15 at 1:26
  • $\begingroup$ Your problem does not have dependent sources. $\endgroup$ – Bill N Mar 5 '15 at 3:39
  • $\begingroup$ Indeed. Sorry, I confused dependent for indpendent sources. I did the superposition method and a combination of KVL and Ohm's Law to find ix and Vx and I'm still not getting the right answer. $\endgroup$ – user114027 Mar 5 '15 at 14:52
  • $\begingroup$ You have sign errors. $\endgroup$ – Bill N Mar 5 '15 at 21:45
  • $\begingroup$ I solved the circuit again this time keeping track of ALL the signs as you've mentioned and I got the right answer. Thanks so much for the help! $\endgroup$ – user114027 Mar 7 '15 at 23:24

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