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Why does clockwise direction is giving one answer and anticlockwise giving different when applying KVL?

Fig E2.49 io (clockwise) = 13.33 mA and io (anticlockwise) = 0.02 A

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Please help me point out the mistake arising here. Same thing happened with another question below:

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See fig E2.48 when i is anticlockwise value of V is -36.7 V redrawn in fig E2.48(a) but when taken clockwise its -3.3 V. The magnitude shouldn’t change like that.

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    \$\begingroup\$ Please show us your clockwise and anticlockwise KVL. So that we can point out where the mistake is. \$\endgroup\$
    – G36
    Commented Jun 13, 2023 at 20:25
  • \$\begingroup\$ It's clearly 20 mA so I don't know what you did to get 13.33 mA. \$\endgroup\$
    – Andy aka
    Commented Jun 13, 2023 at 20:36
  • \$\begingroup\$ @Andyaka The 500 I0 is exactly the behavior of a resistor. There's no difference. Given the direction of I0, the positive end of a resistor would be exactly as shown. And the voltage difference would be exactly as computed. \$\endgroup\$ Commented Jun 13, 2023 at 21:04
  • \$\begingroup\$ Tabreek, thanks for the addition. The photo and writing are clear enough. So +1 for the question. Keep in mind that when you choose to reverse the current direction you must also reverse the sign in the dependent source, since that source depended upon a particular current direction. \$\endgroup\$ Commented Jun 13, 2023 at 21:05
  • \$\begingroup\$ Got it! Thank you so much. For other question with independent sources they specified in answer that direction must be kept anticlockwise because the question was asked in a specific way in terms of charges being injected to 50V source. Previously i didn’t read the question thoroughly. \$\endgroup\$ Commented Jun 13, 2023 at 21:16

2 Answers 2

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On first glance, you should be able to see that the dependent source, the one showing as \$500\cdot I_0\$ is exactly what a resistor of \$500\:\Omega\$ would look like, given the direction of \$I_0\$ in the loop. If you look at your \$100\:\Omega\$ resistor, then the voltage drop across it would be \$100\cdot I_0\$ and the sign arrangement would be the same as your dependent source. (Using the passive sign convention.)

In short, your dependent source behaves just as a \$500\:\Omega\$ resistor would act.

So the current is \$I_0=\frac{8\:\text{V}}{100\:\Omega\,+\,500\:\Omega}=13.\overline{3}\:\text{mA}\$.

Working the loop starting at the bottom and moving clockwise with the direction of \$I_0\$, the KVL would be:

$$0\:\text{V}-500\:\Omega\cdot I_0+2\:\text{V}-100\:\Omega\cdot I_0+6\:\text{V}=0\:\text{V}$$

Note that when I wrote out \$500\:\Omega\cdot I_0\$ above, I used the \$\Omega\$ sign with the 500. That's because it must be in Ohms, since that's the only way to convert a current into a voltage. Keep in mind that units for each term must be consistent, when summed. This dimensional analysis requirement can also be a useful clue.

If you want to reverse the current direction, this is the same thing as changing the sign of the current. To make this explicit and clear, let's call the counter-clockwise current \$I_0^{\:'}\$. That current goes in the opposite direction with respect to \$I_0\$. So take note that \$I_0^{\:'}=-I_0\$; or that \$I_0=-I_0^{\:'}\$.

To complete the circuit alteration to use this new counter-clockwise current, anywhere you see \$I_0\$ in a dependent source you must then substitute for the newly defined counter-current direction. The dependent source is \$500\cdot I_0\$. So, without changing its meaning but using the newly re-directed current, the dependent source is also \$500\cdot -I_0^{\:'}=\left(-500\right)\cdot I_0^{\:'}\$. This is still the same source. But since you are applying a differently directed current, you need to consider the change of sign here.

The KVL is now, working counter-clockwise:

$$0\:\text{V} - 6\:\text{V} -100\:\Omega\cdot I_0^{\:'}-2\:\text{V}+\left(-500\:\Omega\right)\cdot I_0^{\:'}=0\:\text{V}$$

From which you should get \$I_0^{\:'}=-13.\overline{3}\:\text{mA}\$. The opposite sign, only. Same magnitude. It's just in the other direction, now.

I'll add a Spice run to close the issue:

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The discrepancy in the answers comes from the fact that you keep switching the polarity of the resistors (in both examples) whenever you redraw the circuit.

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    \$\begingroup\$ Do resistors have polarity? Like capacitors? Or is the direction of the current flow? \$\endgroup\$
    – Solar Mike
    Commented Dec 19, 2023 at 9:48
  • \$\begingroup\$ I actually mean the + and - signs on the resistors. Switching their positions will affect the result. \$\endgroup\$
    – Jovanny
    Commented Dec 19, 2023 at 13:55

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