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schematic

simulate this circuit – Schematic created using CircuitLab

The KVL equation for the right loop is: $$-(I-gV_c)(\frac{10}{s}+10)-5I+V_s=0$$

But my professor use: $$(I-gV_c)(\frac{10}{s}+10)+5I+V_s=0$$

Which sign convention is right?

If I solve for the transfer function $$H=\frac{V_0}{V_s}$$ using my equation $$-(I-gV_c)(\frac{10}{s}+10)-5I+V_s=0$$.

First I find an equation relating Vc to I: $$gV_c-I=\frac{sV_c}{10}$$. This is the Ohm law for the capacitor. s is Laplace variable. So $$V_c=\frac{10I}{10g-s}$$. Plug this into the my KVL equation $$(I-gV_c)(\frac{10}{s}+10)+5I=V_s$$, you get $$(I-\frac{10gI}{10g-s})(\frac{10}{s}+10)+5I=V_s$$. $$V_s=\frac{50gI-10I-15sI}{10g-s}$$.

I found an equation relating V0, the voltage at the top node, to I, this is basically the Ohm law for the center branch: $$gV_c-I_1=\frac{V_0}{\frac{10}{s}+10}$$ So $$V_0=I\frac{10+10s}{10g-s}$$.

Finally, we divide V0 by Vs:$$H=\frac{10+10s}{50g-10-15s}$$

Now using my professor's equation: $$(I-gV_c)(\frac{10}{s}+10)+5I+V_s=0$$. My professor has the same equation: $$V_0=I\frac{10+10s}{10g-s}$$ and $$V_c=\frac{10I}{10g-s}$$

Finally, my professor got: $$H=\frac{2(s+1)}{3s+2-10g}$$ The denominator is mine multiplied by -1.

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  • \$\begingroup\$ Correct the diagram - capacitor should be 1/10 F, not s/10 F. \$\endgroup\$
    – Chu
    Dec 17, 2017 at 18:19
  • \$\begingroup\$ @Chu Or rather \$\frac{1}{sc}=\frac{1}{10s}\$. \$\endgroup\$ Dec 17, 2017 at 18:31
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    \$\begingroup\$ @HarrySvensson, looking at the analysis, I think it's 0.1F. \$\endgroup\$
    – Chu
    Dec 17, 2017 at 19:07
  • \$\begingroup\$ I corrected it. The capacitance is 1/10. So the laplace of it is 10/s \$\endgroup\$ Dec 17, 2017 at 19:34

2 Answers 2

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Your professor is correct.

Current down through \$\ C_1 ,\: R_2\$ is: \$\ (gV_C-I)\$, hence \$V_0\$ for this branch is:

$$V_0=(gV_C-I)(10+\frac{10}{s})$$

Voltage across \$R_3\$ is \$\ 5I\$, hence \$V_0\$ for the right branch is:

$$V_0=5I+V_S$$

Equating these:

$$(gV_C-I)(10+\frac{10}{s})-5I-V_S=0$$

Multiply through by -1:

$$(I-gV_C)(10+\frac{10}{s})+5I+V_S=0$$

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  • \$\begingroup\$ But how do you know the actual current goes down through the middle branch? \$\endgroup\$ Dec 17, 2017 at 19:43
  • \$\begingroup\$ Ok maybe the direction of the actual current does not matter. But I dont get why $$(I-gV_c)(\frac{10}{s}+10)$$ does not have a negative sign in front of it. If we are assuming the actual current goes down through the middle branch but $$(I-gV_c)$$ is going up, so against the current. Thus negative. \$\endgroup\$ Dec 17, 2017 at 19:51
  • \$\begingroup\$ gVc is down, I is up, therefore (gVc - I) is down. (I-gVc) would be up. This is a basic relationship. If there is 2A flowing down and 5A flowing up then there is (2-5)=-3A flowing down. The sign indicates the flow direction relative to the specified direction. \$\endgroup\$
    – Chu
    Dec 18, 2017 at 0:54
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It doesn't matter which direction you choose for the current, it should all work out if you stay consistent with the passive sign convention (psc).

What I would do is add sign reference to the relevant resistors here—\$R_2\$ and \$R_3\$. This is arbitrary.

schematic

simulate this circuit – Schematic created using CircuitLab

Now, if you start from the reference node, and go around the loop through \$R_2\$, \$C_1\$, \$R_3\$, and \$V_S\$, you write KVL as follow:

$$ V_{R_2}+V_{C_1}-V_{R_3}-V_S=0 $$

I gave the voltage rises (from - to +) a positive sign and a negative one to the drops (from + to -). Fair enough. Since I have assumed the current goes up from the reference node around that loop, I run into an issue with the psc. That is because the current going up enters through the negative terminals of \$R_2\$ and \$C_1\$ and this is what happens according to the psc:

enter image description here

That means that the previous KVL equations becomes:

$$ -(R_2)(I-gV_c)-(\dfrac{1}{sC_1})(I-gV_c)-(R_3)(I)-V_S=0 $$

Notice that the terms for \$R_2\$ and \$C_2\$ have a negative sign, because they don't follow the passive sign convention. That is why, if you have a choice, you choose the current to go in the direction of the voltage drops (that is the current enters the positive terminal), that way you don't have to 'manually' insert a negative sign when plugging in for the values of the voltages.

That equation can be rewritten as:

$$ -(I-gV_c)(R_2 + \dfrac{1}{sC_1})-(R_3)(I)-V_S=0 $$

That matches your professor's equation.

I could have alternatively chosen the opposite reference for the voltage across \$R_2\$

schematic

simulate this circuit

If I write the same KVL loop, assuming the current is going up from the reference node,

$$ -V_{R_2}+V_{C_1}-V_{R_3}-V_S=0 $$

Notice the negative sign in front of \$V_{R_2}\$, because now I have chosen it to be a drop in the direction of the current and as per my previous rule, drops are negative.

When you plug in Ohm's law, you don't need to worry about putting a negative sign in front of it, because it now follows the psc. \$C_1\$, however, does not, you'd still need to put a negative sign:

$$ -(R_2)(I-gV_c)-(\dfrac{1}{sC_1})(I-gV_c)-(R_3)(I)-V_S=0 $$

The equation is still the same.

You could do the same experiment by changing the reference for polarity of \$C_1\$. It's all about being consistent with the references and the convention.

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