enter image description hereI'm currently studying electronics for the first time. The objetive of this exercise is to calculate the current I and the dissipated power in the dependent source. However, I'm having a hard time because I believe this circuit to be a current divider, but the solution states two values (I=2A and P=560W) that are incompatible with my interpretation. I am aware of how basic this question is, but I've searched everywhere for an answer with no sucess.

  • \$\begingroup\$ You can use analytic techniques, by rote, and come up with an answer. This is usually a good approach, because once you get used to it then it is fairly bullet-proof and just hauls you in the right direction. But in this case, you could just ask yourself: "If \$I\$ flows though \$70\:\Omega\$ and if \$2 I\$ flows through the dependent source, both of which have the exact same voltage across them, is there an easy resistor value to use in order to replace the \$2 I\$ dependent source that would seem easy to work out?" Learning to apply such questions is also a good skill. \$\endgroup\$ – jonk May 24 '20 at 21:52
  • \$\begingroup\$ I can imagine a resistor to replace the dependent \$2 I\$ source, call it \$R\$, where I could then work out the voltage across all three resistors by \$16 \: \text{A}\cdot \left( 14\: \Omega \mid\mid R \mid\mid 70\: \Omega \right)\$ and therefore the current you want as \$I=16\: \text{A} \cdot\frac{ 14\: \Omega \,\mid\mid\, R \:\mid\mid\, 70\: \Omega }{70\: \Omega}\$. Yes? \$\endgroup\$ – jonk May 24 '20 at 21:58
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    \$\begingroup\$ Yes, you're absolutely right. It gets quite easy that way. Thank you! \$\endgroup\$ – Luís Pires May 24 '20 at 22:30
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    \$\begingroup\$ Keep working on flexible thinking, which is developed by learning to "see" things in various ways. A good practice idea is to take a circuit you've grown comfortable with, but one that is complex enough to avoid being trivial, and then trying to find four or five different ways to analyze it to get similar results. Best wishes! \$\endgroup\$ – jonk May 24 '20 at 22:51

An easy way is to realize that both resistors are across the same voltage since their ends are directly connected. If 70 Ohm is 5 times 14 Ohm, then the current through the smaller resistor is 5 times I. The total current, 16A, consists of 5I + 2I + I = 8I, which means I = 16A / 8 = 2A.

Now the voltage dropped across the larger resistor is U = I times 70 Ohm = 140 V. The power dissipation is then P = 140 V times 2 times 2A, thus 560 W.

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    \$\begingroup\$ Generally we try not to give actual answers to homework questions, but rather guidance on strategy in order to support rather than defeat the educational goal of homework. So your first point is great. Your second clever. But working it to a solution is traditionally "left as an exercise to the reader" \$\endgroup\$ – Chris Stratton Dec 8 '20 at 17:20
  • \$\begingroup\$ Good point. Just felt enticed by a certain elegance in the solution. Thanks! \$\endgroup\$ – Carsten B. Dec 10 '20 at 9:25

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