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I am attempting to reduce the circuit shown into a form of a single voltage source and resistance in series using Thevenin and Norton conversions.

I approached the problem by first considering a Thevenin equivalence circuit to reduce the 6V source, 6 ohm resistor and 5 ohm resistor to a 30/11V source with a 30/11 ohm resistor in the position of the 6 ohm resistor.

I then did a Norton conversion for the 2A source, reducing the circuit to a 30/11 V source on the LHS and 60/11 V source in series with a 30/11 ohm resistor on the RHS, which reduced to a 30/11 V source in series with a 30/11 ohm resistance. This was incorrect.

Any advice on how to correctly reduce the circuit would be appreciated: I'm not sure which of the reductions I made were invalid.

[![enter image description here][1]][1]

The circuit should reduce to an 18V source with a 6 ohm resistance in series across the 2 wires on the right

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By Ohm's law, the \$\small 6\:\Omega\$ resistor must have \$\small 18\:V\$ on its right hand side in order to get \$\small 2\:A\$ through it. So the O/C voltage is \$\small V_{TH}=18\:V\$.

Turning both sources to zero, leaving their internal resistances, gives a resistance of \$\small 6\:\Omega\$ across the output, hence \$ \small R_{TH}=6\:\Omega\$.

Incidentally, for the current source to produce \$\small 2\:A\$, it must establish \$\small 28\:V\$ at its upper terminal (assuming \$\small 0\:V\$ ground reference is the junction of the two sources).

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A resistor in series with a current source does nothing to the circuit. (Same with a resistor in parallel to a voltage source.)

You can simply ignore it, then turn the voltage source and its series resistance into a current source with parallel resistance. Then determine the current through the resistor (that one from the voltage source) and thus, the voltage.

Take care of the direction of the voltage source!

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  • \$\begingroup\$ Thank you for your quick response. I'm a bit confused about how the reduction you mentioned would transfer into a single voltage source and resistance which would be able to be integrated into part of a larger circuit (with the two wires on the right of the diagram). Would it not be possible to reduce the circuit directly from the original state into two voltage sources ? \$\endgroup\$ – bebop Nov 27 '18 at 23:54
  • \$\begingroup\$ No, you don't have a single voltage source then but two current sources both driving the resistor from the voltage source. You have to add those currents through the resistor (mind their directions), then you have the voltage over the resistor. \$\endgroup\$ – Janka Nov 27 '18 at 23:56
  • \$\begingroup\$ You cannot turn a current source into a voltage source when there's no parallel resistance. \$\endgroup\$ – Janka Nov 27 '18 at 23:59
  • \$\begingroup\$ But would it be possible from the reduction you mentioned to say that across the wires on the right you essentially have a 18V source (ie. equivalent to the current through resistor * resistance of resistor) with a series resistance equivalent to the effective resistance across the wires. \$\endgroup\$ – bebop Nov 28 '18 at 0:04
  • \$\begingroup\$ No. Because the current through the resistor does nothing to the top-right node. The 10V over it mean the potential at the node between the 5Ω resistor and the current source is 10 V higher than the top-right node. That's all. (But noboby asked about that potential.) \$\endgroup\$ – Janka Nov 28 '18 at 1:06

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