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Both the KCL and KVL equations are required to completely model a resistor network. However, the two main methods to analyze these circuits Nodal analysis and Mesh analysis only use KCL and KVL respectfully. In other words,these methods don't use the "complete model of the circuit" (i.e. Nodal analysis ignores the KVL part of the circuit model). Yet somehow they're able to completely analyze the entire circuit - how? How is something that doesn't take the entire model into account able to completely analyze the circuit?

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  • \$\begingroup\$ KVL and KCL are not the only ways of looking at a circuit. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 9 '14 at 14:47
  • \$\begingroup\$ @IgnacioVazquez-Abrams True, but if you do choose to look at the circuit through them, you gotta take both of them into account right? \$\endgroup\$ – dfg Sep 9 '14 at 14:48
  • \$\begingroup\$ Unless you can replace one or the other with an alternate. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 9 '14 at 14:48
  • \$\begingroup\$ What would the alternatives be in the case of Nodal and Mesh analysis? \$\endgroup\$ – dfg Sep 9 '14 at 14:49
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    \$\begingroup\$ ... Ohm's Law . \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 9 '14 at 14:50
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In other words,these methods don't use the "complete model of the circuit" (i.e. Nodal analysis ignores the KVL part of the circuit model

That is not correct.

  • Nodal analysis appears to be using KCL only, where it actually uses KVL implicitly, absorbed in Ohm's law in the currents equations.

  • Mesh analysis appears to be using KVL only, where it actually uses KCL implicitly, absorbed in Ohm's law in the voltages equations.

By the way, KVL and KCL are just a consequence of the lumped-element abstraction of linear circuits at low frequencies (when the wavelength of the signal is much more larger than the physical lengths of the wires and components). You can derive it easily from Maxwell equations (which are more accurate models of the circuit behavior.)

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  • \$\begingroup\$ Are KVL and KCL (the laws not the equations) derivable from each other? Are they different ways of describing the same abstract concept? \$\endgroup\$ – dfg Sep 21 '14 at 0:05
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Both the KCL and KVL equations are required to completely model a resistor network.

This is true, but often we don't need to know everything about a network to solve our problem.

The complete solution of a network would give the voltage at every node and the current through every branch.

If we use the nodal analysis, we get the node voltages but not the branch currents. If we use the mesh analysis we get only the branch currents (or mesh currents, from which we can easily find the individual branch currents), and not the node voltages.

If we have one set of variables (currents or voltages) it's trivial to obtain the remainder. For example, if we know the node voltages we can get the current through any particular branch by applying the characteristic rule for the component on that branch. But we often don't even need to do that.

If we are only interested in one set of variables (voltages say, and maybe one or two of the currents) it wouldn't make sense to solve a larger matrix to get the currents we're not interested in.

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I am using a simple circuit here to apply KVL and then prove to you that KCL can be derived from KVL with the help of ohm’s law.

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First we apply Ohm's law:

\$V_{R_1} =I_1R_1\$

\$ or, I_1=\frac{V_{R_1}}{R_1}\$

\$V_{R_2} =I_2R_2\$

\$ or, I_2=\frac{V_{R_2}}{R_2}\$

\$ E =I(R_1||R_2) = \frac{IR_1R_2}{R_1+R_2} \$

\$or, I=\frac{E (R_1+R_2)}{R_1R_2}----(1)\$

Now we apply KVL:

For loop at the left:

\$E =V_{R_1}\$

For loop at the right:

\$ V_{R_1}=V_{R_2} \$

Therefore, we get-

\$E=V_{R_1}=V_{R_2}\$

Now we use the equation (1)

\$ I=\frac{E (R_1+R_2)}{R_1R_2}\$

\$or, I=\frac{ER_1}{R_1R_2}+\frac{ER_2}{R_1R_2}\$

\$or, I=\frac{E}{R_2}+\frac{E}{R_1}\$

\$or, I=I_2+I_1\$

If you apply KCL to node “a” you will get this same equation. Therefore applying KVL only doesn’t mean we are neglecting KCL.

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  • \$\begingroup\$ Are KVL and KCL (the laws not the equations) derivable from each other? Are they different ways of describing the same abstract concept? \$\endgroup\$ – dfg Sep 21 '14 at 0:06
  • \$\begingroup\$ @dfg - Short answer is yes and equation is the law. These laws are internally connected. \$\endgroup\$ – Amit Hasan Sep 21 '14 at 4:20

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