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I specifically want to know how does the current flow in the schematic attached. There are two voltage sources (one dependent, the other independent) in the figure. Correct me if I'm wrong, but it appears to me that the current generated by 35 V and 2vx will collide each other though the magnitude of their currents are equal since all the elements are in a series connection.

This circuit I'm working on is one of the practice problems of Kirchhoff's Laws in Fundamentals of Electric Circuits by Alexander and Sadiku. The ones asked in this problem is vo and vx. I tried solving the problem but it does not match the given answer in the book (the solution is not shown though).

I would like to know how the current flows across 5 Ω resistor.

@Ignacio Vazquez-Abrams: That was precisely why I asked the question. Though many here are saying this is a simple circuit, I had been confused with this problem because I was imagining how the current will flow due to the two voltage sources.

@The Photon: No, this is not a homework question. I was reviewing for a quiz the other day by picking questions on our textbook.

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    \$\begingroup\$ Current going in the reverse direction through a voltage source is not a physical impossibility; it "charges" the voltage source. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 4 '13 at 17:30
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it appears to me that the current generated by 35 V and 2vx will collide each other

It may be that you are assuming that a voltage source, whether independent or controlled, must source current, i.e., supply power to the circuit.

But, at least in ideal circuit theory, there's nothing "wrong" with a voltage source sinking current, i.e., receiving power from the circuit.

For a real world example, consider that, when a battery is being charged, the current is in the opposite direction than when the battery is being discharged.

I would like to know how the current flows across 5 Ω resistor.

If you're planning to be an EE, don't write or say things like "current across"; current is through, voltage is across.

Now, this circuit is very easy to solve. There are two unknowns so you need two independent equations.

For the 1st, write a KVL equation clockwise 'round the loop:

$$35V = v_x + 2v_x - v_o \rightarrow 3v_x = 35V + v_o$$

Now, you need one more independent equation. Can you find one?

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Since all the components are connected in series, the current will be the same in all of them. So, pick a ground reference, write out your KVL or KCL equations and solve for I or Vx. If you are doing a KVL and you get a negative current, all it means is that the current flows in the opposite direction. If I did the algebra correctly, Vx should be 10V and I should be 1 A in the CW direction

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Notice the second source is a dependent voltage source, not an independent source.

You could equally show a resistor as a dependent voltage source with voltage proportional to it's input current.

In fact, if you look you'll see that the way this source is connected, it's voltage is proportional to to the 10 Ohm resistor voltage. That voltage is proportional to the 10 Ohm resistor current. And the 10 Ohm resistor current is the same as the current through the dependent source. So this source has the exact same effect as a third resistor in the circuit.

Even without noticing those connections, you can simply solve this problem by writing KVL in the usual way, as other answers have said.

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  • \$\begingroup\$ Indeed, it is equal to 20ohm resistor: U = 2.V = 2.I.10ohm => U/I = 2.10Ohm = 20ohm; So, the whole current is 36V/(10+20+5) = 1A \$\endgroup\$ – johnfound Nov 7 '13 at 13:09
  • \$\begingroup\$ @johnfound, sure, but this looks a lot like a homework question, so I prefer not to just hand the answer to OP. \$\endgroup\$ – The Photon Nov 7 '13 at 17:39
  • \$\begingroup\$ This wasn't actually a homework question, I was reviewing for a quiz the other day. \$\endgroup\$ – ellekaie Nov 10 '13 at 2:59
  • \$\begingroup\$ Either way, if we hand you the answer it's easy to read it and move on without internalizing the reasoning behind it. If we explain it without giving the final answer you have to understand the reasoning to generate the final answer, and you're more likely to remember when you get to the final test. \$\endgroup\$ – The Photon Nov 10 '13 at 20:16
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You can do this in your head. The 10 ohm has \$v_x\$ across it. Since the same current flows through the 5 ohm resistor, this means that the 5 ohm has \$\frac{1}{2}v_x\$ (and this is equal to \$-v_o\$. And so, \$\left(1 + \frac{1}{2} + 2\right)v_x = 35\$. In other words, the two resistors and dependent source have to add up to 35. Once you know \$v_x\$ use \$\frac{1}{2}v_x = v_o\$ to get \$v_o\$.

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Use a ratio of the resistors to solve, i.e., drop across the 10 ohm resistor:

$$V_x = \frac{10}{10+5} \times (35-2V_x)$$

Should end up with \$V_x = 9.33\$, \$2V_x = 18.67\$.

The results are:

  • drop across the 10 ohm resistor = 9.33V
  • drop across the 5 ohm resistor = 7V
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