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I'm confused about two different approaches for current calculation through a 5 ohm resistor.

  1. The 5 ohm resistor is in parallel with a voltage source so the current through it should be 2 amperes.

  2. In a parallel connection, current follows the least resistance path, so all the current flows through the short circuited path (since its resistance is zero) and current through the 5 ohm resistance will be zero.

What will be the actual current through the 5 ohm resistance and how should I approach this problem?

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  • \$\begingroup\$ Current goes through, not across. Voltage is measured across two points, current is measured through one point. \$\endgroup\$
    – JRE
    Commented Aug 14, 2020 at 11:12
  • \$\begingroup\$ Thanks I edited it. \$\endgroup\$
    – user215805
    Commented Aug 14, 2020 at 11:14
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    \$\begingroup\$ There's no answer. The voltage across the resistor is both 10V (voltage source) and 0V (short circuit) in this schematic. Obviously both can not be true at the same time. In the real world, the voltage source would go up in smoke or blow a fuse. \$\endgroup\$
    – StarCat
    Commented Aug 14, 2020 at 11:25

2 Answers 2

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If indeed the wire labelled I1 is a short circuit, then the diagram represents something that is logically impossible. It's not possible to have a 10 V voltage source across a short circuit, which always has 0 V across it. It's as nonsensical as trying to make sense of the equation 10 = 0.

If the wire has some undrawn components on it that take a current of I1, then the power supply can deliver 10 V, at a current of 2+I1 amps, 2 A for the resistor, and I1 for the 'wire'.

If you build that circuit in the real world, then one of two things will happen, depending on which of the power supply and the short is 'stronger'. If it's heavy wire doing the short-circuiting, then the battery voltage will collapse as the current rises to as much as it can supply. The supply may destroy itself, or survive if it's been built to limit safely, either way it will deliver about 0 V. If it's a heavy supply like a car battery and a thin piece of wire, then the wire will burn through.

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Well, I am trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_\text{i}=\text{I}_1+\text{I}_2\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_\text{i}}{\text{R}_2} \end{cases}\tag2 $$

So, we can substitute \$(2)\$ into \$(1)\$:

$$\text{I}_\text{i}=\frac{\text{V}_\text{i}}{\text{R}_1}+\frac{\text{V}_\text{i}}{\text{R}_2}=\text{V}_\text{i}\left(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\right)\tag3$$

Now, when \$\text{V}_\text{i}>0\$ and \$\text{R}_2>0\$, what do we get when:

$$\lim_{\text{R}_1\to0}\text{I}_\text{i}=\lim_{\text{R}_1\to0}\text{V}_\text{i}\left(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\right)=\text{V}_\text{i}\left(\lim_{\text{R}_1\to0}\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\right)\tag4$$

It is not hard to see that \$\text{I}_\text{i}\to\infty\$ when \$\text{R}_1\to0\$.

The reasoning is the same and you'll find that \$\text{I}_1\to\infty\$ and \$\text{I}_2=\frac{\text{V}_\text{i}}{\text{R}_2}\$ such that \$\text{I}_1+\text{I}_2\to\infty\$.

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  • \$\begingroup\$ The question is about I2. \$\endgroup\$
    – StarCat
    Commented Aug 14, 2020 at 11:27
  • \$\begingroup\$ But you calculated Total current , what about current through 5ohm resistor? \$\endgroup\$
    – user215805
    Commented Aug 14, 2020 at 11:28
  • \$\begingroup\$ @user215805 see my edit. \$\endgroup\$ Commented Aug 14, 2020 at 11:45
  • \$\begingroup\$ Thanks for answer, but according to your calculation current through 5ohm should be 2ampere , but in Neil's answer it's not possible \$\endgroup\$
    – user215805
    Commented Aug 14, 2020 at 11:56
  • \$\begingroup\$ @user215805 Yes true, theoretically speaking the battery can provide an infinite amount of current. \$\endgroup\$ Commented Aug 14, 2020 at 11:59

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