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Using the following circuit:

enter image description here

How do you find the current i, using source transformation? What is the principle of it, and how to use it?

All I know about it, is that you must simplify the circuit, but I'm not sure how.

This was my attempt, this was the one thing in class I really struggled with:

enter image description here

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  • \$\begingroup\$ First, you could try to change the 4 amps current source plus 5 ohms resistor parallel to it into a voltage source and a series resistor. Can you do it? \$\endgroup\$ – G36 Oct 12 '19 at 11:53
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Let's take the circuit and remember the two rules to make source transformation:

1- A current source sourcing a current \$I\$ in parallel with a resistor \$R\$ transforms into a voltage source with voltage \$R\,I\$ and the same resistor \$R\$ in series with this voltage source. That is the principle of finding the Thévenin Equivalent.

2- A voltage source with voltage \$V\$ in series with a resistor \$R\$ transforms into a current source with current \$\dfrac{V}{R}\$ and the same resistor \$R\$ in parallel with this voltage source. That is the principle of finding the Norton Equivalent.

Illustrating what has been aforementioned:

schematic

simulate this circuit – Schematic created using CircuitLab

Given this explanation the circuit then boils down to some steps.

First step: Voltage source to the left into a current source and current source to the right into a voltage source.

schematic

simulate this circuit

Now transform the voltage source to the right into a current source:

schematic

simulate this circuit

Finally, the current is easily calculated from this last figure:

\$ i=3.5\,\dfrac{6.667}{20+6.667} \approx 875\, mA \$

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Source transformation means replacing a Thévenin source by an equivalent Norton source or vice versa.

Doing that may give you the opportunity to simplify the circuit in the next step, e.g. by combining two parallel or two series resistors into one equivalent single resistor.

You can repeatedly simplify the circuit until there is nothing left but a single source connected to the 20Ω resistor whose current you have to determine.

If you paid attention to your lectures you should be able to identify the Norton/Thévenin sources in your circuit and know how to replace them by equivalent ones of the other type (=do source transformation) as well as combine parallel/series resistors.

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  • \$\begingroup\$ I've tried to simplify it multiple times but kept getting stuck on both sides of the circuit. For example I got 4X5 = 20v on the right and then what? \$\endgroup\$ – Redsam121 Oct 12 '19 at 12:50
  • \$\begingroup\$ So, show us your attempt(s). \$\endgroup\$ – Chu Oct 12 '19 at 12:59
  • \$\begingroup\$ Okay, I added my attempt. \$\endgroup\$ – Redsam121 Oct 12 '19 at 13:21
  • \$\begingroup\$ Your conversion on the right side is incorrect. The component values are correct but you have not connected them properly. \$\endgroup\$ – Elliot Alderson Oct 12 '19 at 15:08
  • \$\begingroup\$ So the 5 ohm should be where the 4A was? \$\endgroup\$ – Redsam121 Oct 12 '19 at 16:05

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