Can someone give an explanation to why there no current in i0 (or why does i0 equal 0)? It was a step I had to know before using Kirchoff's laws.
Also, is there a voltage across the nodes that i0 goes between?
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3\$\begingroup\$ There's no return path. The entire lower conductor is one node. \$\endgroup\$– ChuCommented Mar 2, 2016 at 1:06
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1\$\begingroup\$ Also, if you remember the current into a node = current out of a node, then you can see that the net current between those two nodes must be zero. \$\endgroup\$– Keegan JayCommented Mar 2, 2016 at 1:16
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\$\begingroup\$ Related: Kirchhoff's Current Law at the ground node? \$\endgroup\$– The PhotonCommented Mar 2, 2016 at 1:36
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\$\begingroup\$ Specifically, look up the cut-set version of the Kirchoff Current Law. \$\endgroup\$– The PhotonCommented Mar 2, 2016 at 1:38
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2 Answers
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Assuming ideal circuit elements and conductors:
Simply because there is no return path for \$i_o\$.
The entire lower line (call it ground or common) in the schematic is considered to be one node, therefore there is considered to be no voltage difference across it either.
(Suppose that there was a current flowing in \$i_o\$. Then the only way back to the starting point would be trough the same node but in opposite direction, meaning that the sum of the currents is \$i_o + (- i_o) = 0\$.)
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This would be like taking a battery and connecting only one terminal of it to your circuit. You need a path for the current to return
