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I'm trying to teach myself electronics from an old textbook, "Fundamentals of Electric Circuits", \$4\$th Edition, by Alexander and Sadiku. On page \$41\$, I can't figure out how they did example \$2.6.\$ See this diagram:

enter image description here

So I'm applying KVL around the loop. I get:$$-12 + 4i + 2v_0 - 4 - 6i = 0$$

But the book says it's \${+6i}\$. The current is flowing to the negative pole of the resistor, so it should be negative, right? Is the book wrong, or am I misunderstanding something?

Edit:

The book text for the problem reads:

Determine \$v_0\$ and \$i\$ in the circuit.
Solution:
We apply KVL around the loop as shown in the figure. The result is:\$-12 + 4i + 2v_0 - 4 + 6i = 0\$

Applying Ohm's Law to the 6 Ohm resistor gives:
\$v_0 = -6i\$

Substituting the previous equation into the first one yields: \$i = -8 A\$ and \$v_0 = 48 V\$


The way the book described to do this was to follow the current, and the sign of each voltage element is determined by the polarity. So \$-12 V, 4i\$, and so on. But continuing that pattern, we get to the \$-\$ pole of the \$6\$ Ohms resister before we get to the positive pole, so \$v_0\$ equals \$-6i V\$.

I'm not understanding something.

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  • \$\begingroup\$ What value do you need to find? "But the book says it's +6i." Which value should be +6i, and in which units? Is it Vo that is sought? [I don't have this book, as you can tell.] \$\endgroup\$ – Nick Alexeev Nov 9 '13 at 1:42
  • \$\begingroup\$ It's not clear what it is in the book that you disagree with. Is it an equation in the book? Is it possible that the book substituted -6i for v0? \$\endgroup\$ – Joe Hass Nov 9 '13 at 1:49
  • \$\begingroup\$ There has to be a constant too , with an i term(not 6i), as 12+4!=0. \$\endgroup\$ – Sherby Nov 9 '13 at 1:50
  • \$\begingroup\$ I updated the question. \$\endgroup\$ – Amy Nov 9 '13 at 5:43
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I think you may have an issue with one of the signs. Going around the loop, I get:

12 - 4 i - 2 Vo + 4 - 6 i = 0

Looks like you got the sign on the 6i term flipped.

So, if we take Vo = -6 i, you can calculate:

12 - 4 i + 12 i + 4 - 6 i = 0
16 + 2 i = 0
i = -8
Vo = 48

If you're stuck on understanding how to write this equation, the way to remember it is 'what goes up must come down'. Sum up the voltage across each element as you go around the loop. Since you go into the - side of the 12 volt source, you gain 12 volts. Then you pass through a resistor, and you lose 4 i volts. Then you go the wrong way through the dependent source and you lose 2 Vo volts. Then you go through the 4v source and gain 4 volts. Then you go through the final resistor and lose 6 i volts. You're back where you started, so all of that adds up to zero. Then you figure out what Vo is - if the current is flowing in the direction of the arrow, it will create a voltage drop in opposition to how Vo is marked, so Vo = -6 i. After that, finding i is just algebra.

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    \$\begingroup\$ I really like the gain/lose way of looking at this. This makes a lot of sense now. Thank you! \$\endgroup\$ – Amy Nov 9 '13 at 6:17
  • \$\begingroup\$ Glad to hear it! \$\endgroup\$ – alex.forencich Nov 10 '13 at 1:02
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disclaimer: I don't have the book, so I may be missing something. Here goes nothing.

First thing to note about this circuit is that \$V_7\$ (the rhombus with \$2V_o\$) is a dependent voltage source. It depends on \$V_o\$, the voltage across the 6Ω resistor. [I'll get to the signs soon.] The voltage across the \$V_7\$ is inverted with respect to voltage across 6Ω. That hints that the circuit has negative feedback.

The circuit doesn't have a pre-determined ground. I can choose any node as ground. For the sake of convenience, I choose this place as ground (bottom right corner).

enter image description here

Starting at that place and going clockwise. KVL.

\$V_o + 12 -4i - 2Vo + 4=0\$

For now, I'm using [like a dummy] the same sign for \$V_o\$ as shown in the drawing. Also, from Ohm's law

\$V_o = -6i\$

plug in

\$-6i + 12 -4i +12i + 4=0\$
\$2i + 16 = 0\$
\$i = -8\$
\$V_o = -6i = 48\$

Now we know everything that's going on in that circuit.

edit: In response to more text from the book.

Compared to the first equations in Alex's and mine posts, the book solution has inverted signs. In the book, both resistor drops are have a positive sign, while 12V and 4V voltage sources have a negative sign. But, this sign inversion is consistent. That's why the answer is the same.

In the beginning of the problem, we don't know that sign of the current will end up being negative. We don't know that the current will actually flow against the arrow in the drawing. But, the book authors knew that the current will be negative. May be, that's why they had inverted the sign. May be, he wanted to show that the choice of the sign is more a less arbitrary.

But, the sign choice in the book seems counter-intuitive to me too.

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  • \$\begingroup\$ How did you format your equations? \$\endgroup\$ – Amy Nov 9 '13 at 5:45
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    \$\begingroup\$ @Amy LaTeX is supported here. \$ V_7 \$ produces \$ V_7 \$ \$\endgroup\$ – Nick Alexeev Nov 9 '13 at 5:46
  • \$\begingroup\$ @Amy: See \$\LaTeX\$/MathJax sandbox, the RightClick \$\Rightarrow\$Show Math as\$\Rightarrow\$Tex Commands is useful to see how people wrote the equations. \$\endgroup\$ – RedGrittyBrick Nov 9 '13 at 12:15

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