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If the resistor or impedance is being connected in parallel to supernode, then what current value will be across that resistor or impedance?

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closed as unclear what you're asking by Turbo J, Bence Kaulics, uint128_t, Autistic, Transistor Jul 10 '16 at 10:34

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    \$\begingroup\$ diagram????????? \$\endgroup\$ – Neil_UK Jul 9 '16 at 13:47
  • \$\begingroup\$ Do you understand ohms law? \$\endgroup\$ – Andy aka Jul 9 '16 at 14:23
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    \$\begingroup\$ Define supernode... Never heard that in 20 years experience in electrical.... \$\endgroup\$ – soosai steven Jul 9 '16 at 15:36
  • \$\begingroup\$ @soosai_steven, it's a very well-known term in circuit theory. \$\endgroup\$ – The Photon Jul 9 '16 at 15:38
  • \$\begingroup\$ @soosaisteven. en.wikipedia.org/wiki/Supernode_(circuit). Vote to close has been denied. \$\endgroup\$ – Sparky256 Jul 9 '16 at 17:14
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Generally we define a supernode around two nodes connected by a voltage source, in order to facilitate a solution to a circuit by the modified nodal analysis.

If you connect a resistor across a voltage source, you get a current through it determined by Ohm's Law, regardless of whether we're analyzing the circuit with MNA, mesh analysis, or any other method.

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If a resistor is connected across the terminals of a supernode, then the resistor terminals will remain at the voltage across the terminals of the supernode. The resistor will allow current to flow between the terminals of the supernode. The current is dependent on the value of the resistor. The current can be calculated using Ohm's Law: I = V/R

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  • \$\begingroup\$ "leak power across the terminals" ? ? \$\endgroup\$ – Marla Jul 9 '16 at 14:53
  • \$\begingroup\$ If I am correct, a voltage applied across a resistor will create a current that's size is dependent on the size of the resistor. Power is dependent on current and voltage. \$\endgroup\$ – supermitchell2 Jul 9 '16 at 15:00
  • \$\begingroup\$ that could make an improvement to your answer (edit your answer). Use the word Value rather than "size". \$\endgroup\$ – Marla Jul 9 '16 at 15:03

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