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This is the original question: enter image description here

and we were supposed to find all current in the circuit. I was told we mesh analysis should be able to solve the problem. But, the voltage across each resistor was 5V according to mesh analysis, the current values are also different from the solution.

This is my result using mesh analysis: enter image description here

while is the instructor's solution: enter image description here

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2 Answers 2

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For the analysis I'm going to assume the current flows in a clockwise direction, (and as you know if the sign comes out positive that means the actual current is flowing the same direction as the current arrow. If the sign comes out negative the current is flowing the other way, opposite the direction of the current arrow).

Also I'm going to keep the 3A current source.

enter image description here

For mesh 1: $$ -R_1I_1 - R_2(I_1 - I_2) = 0\\ -R_1I_1 - R_2I_1 + R_2I_2 = 0\\ $$ Or $$ -(R_1 + R_2)I_1 + R_2I_2 = 0 \tag{1} $$

For mesh 2 $$ -R_2(I_2 - I_1) - R_3(I_2 - I_3) = 0\\ -R_2I_2 + R_2I_1 - R_3I_2 + R_3I_3 = 0 $$

$$ R_2I_1 - (R_2 + R_3)I_2 + R_3I_3 = 0 \tag{2} $$

However $$ I_3 = -1A \tag{3} $$

Replacing (3) in (2) we obtain

$$ R_2I_1 - (R_2 + R_3)I_2 = R_3 \tag{4} $$

So, from (1) and (4) $$ \begin{cases} -(R_1 + R_2)I_1 + R_2I_2 = 0\\ R_2I_1 - (R_2 + R_3)I_2 = R_3 \end{cases} $$ $$ \begin{cases} -30I_1 + 10I_2 = 0\\ 10I_1 - 14I_2 = 4 \end{cases} $$

This gives us $$ I_1 = -125mA\\ I_2 = -375mA $$

enter image description here

From picture , the voltage across \$R_2\$ is $$ V_2 = R_2(I_1 - I_2)\\ V_2 = (10\Omega)(-125mA + 375mA) V_2 = 2.5V $$

(Or could have been taken from \$R_1\$). The current \$I_a\$ is $$ I_a = \frac{V_2}{R_2} = \frac{2.5 V}{10 \Omega}\\ I_a = 250 \,mA $$

Note: If we apply KCL on node \$V_2\$ we obtain $$ I_1 + I_x = I_a + I_z\\ I_z = I_1 + I_x - I_a\\ I_z = -125mA + 3 - 250mA\\ I_z = 2.625A $$ End of note

Voltage \$V_3\$ is $$ V_3 = R_3(I_2 - I_3)\\ V_3 = 4(-375mA + 1A)\\ V_3 = 2.5V $$

And current \$I_b\$ is $$ I_b = \frac{V_3}{R_3} = \frac{2.5V}{4\Omega}\\ I_b = 0.625A $$

Thus, when a ideal current source is placed with a short circuit the whole current will go there. It neither absorb power nor deliver power (voltage across current source is zero) and the current in the circuit will be equal to value of current source, i.e., you get the constant current at zero volts.

A short circuit applied directly to the output of a current source is no problem, it does not interact with the rest of the circuit, just circulates current back to itself.

Here a simulation enter image description here

With this in mind we can delete it from the analysis because the analogous of a shorted ideal current source is an open circui, and as you see. voltage across dthe elements and currents through these ones are the same. enter image description here

The only thing that changes is the voltage across the cable, however, we are not interested in this in the analysis, only the voltages and currents of the elements

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Your mistake was treating the line that I3 is connected to as if it was a mesh element. It's not. Only components can be mesh elements, not lines. If you insist on adding a mesh element there, it'll have to be an explicit zero-ohm resistor.

So, to analyze the circuit using mesh method, you need to either:

  1. Replace current source shorts with 0-ohm resistors, or
  2. Remove shorted current sources.

Thus:

schematic

simulate this circuit – Schematic created using CircuitLab

If any division-by-zero occurs due to zero-ohm resistors, you'll need to take limits (from calculus) to resolve that.

So - there's not much to analyze. The 3A source is shorted, so it's effectively not in the circuit even though it's drawn that way. All you got is a parallel combination of resistors and the 1A current source.

The parallel combination of resistors has resistance

$$ R=\frac{1}{1/4 + 1/10 + 1/20}=\frac{5}{2}=2.5{\,\Omega}. $$

Driving 1A through this resistor drops 2.5V. Now you know the voltage across the resistors, and can figure out the individual currents.


There's another form of the parallel resistor formula that only needs one division, and for "reasonable" resistor values can be done completely using integer arithmetic:

$$\begin{aligned} R_{1||2} &= \frac{R_1 R_2}{R_1 + R_2} \\ R_{1||2||3} &= \frac{R_1 R_2 R_3}{R_1 R_2 + R_1 R_3 + R_2 R_3} \\ R_{1||2||3||4} &= \frac{R_1 R_2 R_3 R_4}{R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4} \\ ... \end{aligned}$$

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  • \$\begingroup\$ but if you use mesh analysis for diagram 2, it gives the wrong answer. Does that mean mesh analysis does not apply to this problem? \$\endgroup\$
    – user352000
    Commented Oct 12, 2023 at 21:46
  • \$\begingroup\$ @user352000 It does not give the wrong answer :) If you think it does, you'll have to edit your question and add your attempted solution. Then someone can point out the problem. As it is, you claim it doesn't work, I claim it does work, but I'm not doing your homework for you, so show your work first :) \$\endgroup\$ Commented Oct 13, 2023 at 13:24

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