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I would like to know why the sign for \$4Ω\$ resistor is positive as the equation in the image shows

\$-I1∙8Ω+I2∙4Ω=0\$

Even though the equation is formed using KVL why the sign for \$4Ω\$ resistor is positive when it actually should be negative since it dissipates heat? Circuit Analysis

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  • \$\begingroup\$ Negative voltage x Negative current = Positive power. The -I1 x 8ohm term is a negative voltage, not a negative power. \$\endgroup\$ – MarkU Apr 7 '16 at 7:16
  • \$\begingroup\$ @MarkU:Sorry I couldn't see anything related to power here.Did you mean to say voltage? \$\endgroup\$ – justin Apr 7 '16 at 7:18
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    \$\begingroup\$ You confuse yourself and others by coupling the sign of the current to heat dissipation while the two are unrelated. The magnitude of the current relates to the heat dissipation, not the sign or direction. \$\endgroup\$ – Bimpelrekkie Apr 7 '16 at 7:26
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    \$\begingroup\$ shouldn't both resistors have the same sign? The sign is just that a sign it gives no indication of the actual direction of current flow. You're linking the mathematical variables to physical entities already why none have been determined yet. Imagine that the current is zero and no power is dissipated ! Does the KVL then suddenly not hold anymore ? No it doesn't. Because it relates to the circuit or network. Not to how much current is actually flowing. \$\endgroup\$ – Bimpelrekkie Apr 7 '16 at 7:38
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    \$\begingroup\$ The sign of the current does matter when determining the voltage, V=IR. The sign of the current does not matter when determining the power dissipation, P=II*R. The square of a negative value is a positive value. \$\endgroup\$ – MarkU Apr 7 '16 at 7:38
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I think you're a little confused, dissipation of power as heat isn't used anywhere in KVL. That comes into play when you're calculating things like efficiency and losses.

Firstly I think that question is laid out in a bit of a strange way. What they're trying to say is that the voltage drop across the \$8\Omega \$ resistor is the same as the voltage drop across the \$4\Omega \$ resistor. A more sensible way to write this would be \$I_{1}*8\Omega = I_{2}*4\Omega\$ or \$I_{1}*8\Omega - I_{2}*4\Omega = 0\$

The way it's written the way it is, is due to the position of the 'node' they've used (which kindly they haven't put on the circuit for us)

enter image description here

As for why the resistor doesn't have a negative sign, think about it in terms of a real world example. If you measure a voltage with a multimeter one way you get +12V, you reverse the leads you get -12V. Now if you do the same with a resistor you're going to get the same resistance either way.

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  • \$\begingroup\$ :Do you mean to say if the resistor is held in '-' to '+' direction you get an negative voltage and vice versa? \$\endgroup\$ – justin Apr 7 '16 at 7:41
  • \$\begingroup\$ Again you assume that a negative current gives a negative voltage ! Let go of that value of the current ! For KVL it does not matter what the value or the sign is of the current. Here I1 is just the mathematical variable assigned to the current through the 8 ohm resistor. And VAB is the voltage across it. If you reverse the direction of the arrows (for the currents) and/or the + and - of VAB, it will make no difference for the KVL. \$\endgroup\$ – Bimpelrekkie Apr 7 '16 at 7:45
  • \$\begingroup\$ @Justin if you're referring to the last paragraph I'm talking about a real world example, it doesn't matter if you put +12V or -12V through a resistor, it doesn't ever change it's value. Just by reversing the voltage and/or current doesn't make a 430R resistor a -430R resistor \$\endgroup\$ – Doodle Apr 7 '16 at 7:47
  • \$\begingroup\$ What is needed for KVL is a "reference direction" so you can just pick one at random meaning the direction of the arrows (for the currents) and the + and - signs of VAB are ARBITRARY CHOICES You can choose them to be any direction and KVL will still give the same result. \$\endgroup\$ – Bimpelrekkie Apr 7 '16 at 7:47
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    \$\begingroup\$ For example, if I was to reverse the direction of my arrow so that the loop was going counter clockwise then you'd get \$-I2∙4Ω+I1∙8Ω=0\$ but the overall answer doesn't change. The signs are dependant on where your node is and the direction you choose (CW/CCW). But this only affects voltage and current. If you need an even more real world example, think of squeezing a hose with your hand. It doesn't matter which way the water flows through it, just because the water is flowing in reverse doesn't mean your grip on the hose has changed. Think of that as your resistance \$\endgroup\$ – Doodle Apr 7 '16 at 7:50

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