I refer to Figure 3.82 in this text book and shown below:

enter image description here

Why does iN = 2A? The book says 'Since all of the 2A current flows through the short, iN = 2A'. However, why doesn't some of the current flow through the 2Ohm resistor?


1 Answer 1


What's the voltage across a short?

So as the resistor is connected between the same points, what's the voltage across the resistor?

Given Ohm's Law, \$ I=\frac{V}{R}\$, how much current flows through a finite resistor with that voltage across it, regardless of the actual value of the resistance.

Note that although the short circuit carrying \$i_N\$ has zero volts across it, it also has zero resistance. This means we cannot calculate the current flowing through it by Ohm's Law, as 0/0 is indeterminate. To save stressing about what 0/0 is, we can flip Ohm's Law into its product form, \$V=IR\$. Now we see that in a short circuit where R=0, any value of current will produce a zero volt drop. We note that the current coming from the current source is 2A, so that's what \$i_N\$ is. If we increased the current source to 10A, then it would be 10A.

An ideal short circuit is defined to always have 0v across it, regardless of the current through it. Similarly, an open circuit is defined to always have zero current through it, regardless of the voltage across it.

In summary, if we have a parallel connection of several components, and one of them is a short circuit, then it will hold the voltage across all the other components to 0v, which will result in a zero current flow across components with a finite resistance.

Anticipating your next question, it is meaningless ask what proportion of the impressed current flows though any of two or more ideal short circuits in parallel, all parallel short circuits can be replaced by a single short circuit. If there is some difference between them, let's say you've 'shorted circuited' your power supply by a thin and thick wire in parallel, then they are not ideal short circuits, they are resistors with a finite resistance, you would draw them as resistors, and you can do Ohm's Law on them.

What's drawn there is an ideal circuit. Those things drawn as wires are ideal wires, they have zero voltage drop across them.

  • \$\begingroup\$ that's helpful but then how come the current in In isn't zero? Because the voltage across In is zero, then I = 0/0 = 0? \$\endgroup\$
    – edd91
    Commented May 6, 2017 at 6:22
  • \$\begingroup\$ @edd91 answered your question in my text \$\endgroup\$
    – Neil_UK
    Commented May 6, 2017 at 7:44
  • \$\begingroup\$ 0/0 isn't 0. It's undefined. So we can't find the current In in the way you've just mentioned. Wjat we can do however,is what Neil mentioned. First prove that the current skips the resistor. Them remove it. Then apply the current rule to any point in the resulting circuit. \$\endgroup\$ Commented May 6, 2017 at 7:47
  • \$\begingroup\$ 0/0 is indeterminate; it may be zero, it may be infinite, in fact it may be anything. \$\endgroup\$
    – Chu
    Commented May 6, 2017 at 10:14

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