Any electric circuit that contains only resistors, current generators or voltage generators can be replaced by a practical current generator at any two chosen points, thus simplifying the network for further calculations.
Any electric circuit that contains only resistors, current generators or voltage generators can be replaced by a practical voltage generator at any two chosen points, thus simplifying the network for further calculations.
The equivalent circuit connected the same load resistor has the same values for voltage, current and power. With a single practical generator instead of a complex circuit, further calculations are much simpler.
All voltage generators are replaced by a short circuit and the current generators by an open circuit, and then the resultant resistance is calculated from the two points. This is the Norton or Thevenin resistance.
The output voltage can be calculated, for example, using the voltage divider formula, or by any other method. It needs to determined what voltage a voltmeter would show if placed between the two points in the circuit.
Knowing the voltage and resistance, the current is easy to calculate.
$ I_n = \dfrac{V_{th}}{R_n} $
$ V_1 = 10 \, [V] $
$ R_3 = 16 \, [Ω] $
$ R_1 = 12 \, [Ω] $
$ R_4 = 18 \, [Ω] $
$ R_2 = 14 \, [Ω] $
Following the steps above, first calculate the resultant resistance between the two points without the voltage source, then calculate the output voltage in the original circuit, and finally calculate the Norton current.
$ R_{AB} = (R_1 + R_2) \times (R_3 + R_4) $
$ R_{AB} = R_{th} = R_n = 14.73 \, [Ω] $
$ V_{out} = V_1 ⋅ \dfrac{R_3 + R_4}{R_1 + R_2 + R_3 + R_4 } $
$ V_{out} = V_{th} = 5.667 \, [V] $
$ I_{n} = \dfrac{V_{th}}{R_n} = 0.385 \, [A] $
$ V_{th} = 5.667 \, [V] $
$ R_{th} = 14.73 \, [Ω] $
$ I_n = 0.385 \, [A] $
$ R_n = 14.73 \, [Ω] $
$ V_1 = 10 \, [V] $
$ R_3 = 16 \, [Ω] $
$ R_1 = 12 \, [Ω] $
$ R_4 = 18 \, [Ω] $
$ R_2 = 14 \, [Ω] $
$ R_5 = 20 \, [Ω] $
The steps are the same, except that the resistance on the output wire of point A is not included in the voltage divider formula. This is because a voltmeter has infinite resistance, the output voltage is not affected by R5.
$ R_{AB} = (R_1 + R_2) \times (R_3 + R_4) + R_5 $
$ R_{AB} = R_{th} = R_n = 34.73 \, [Ω] $
$ V_{out} = V_1 ⋅ \dfrac{R_3 + R_4}{R_1 + R_2 + R_3 + R_4 } $
$ V_{out} = V_{th} = 5.667 \, [V] $
$ I_{n} = \dfrac{V_{th}}{R_n} = 0.163 \, [A] $
$ V_{th} = 5.667 \, [V] $
$ R_{th} = 34.73 \, [Ω] $
$ I_n = 0.163 \, [A] $
$ R_n = 34.73 \, [Ω] $