Chapter 18 Problem 34 - Norton Equivalent Circuit

Question 34

• The question asks to determine the Norton equivalent circuit for the network external to the 4 k\Omega inductive reactor in a given figure.

Finding Norton Equivalent Circuit

• To find the Norton equivalent circuit, we need to find two components: Norton impedance (ZN) and Norton current (IN).

1. Finding Norton Impedance (Z_N)

• Remove the independent sources:
  • Voltage sources are replaced with a short circuit.
  • Current sources are replaced with an open circuit.
• Leave the dependent sources as they are.
• Looking between points A and B (where the load was) to find Z_N. The circuit consists of:
  • A j4 \Omega inductive reactance.
  • A -j8 \Omega capacitive reactance.
  • An 8 \Omega resistor.
  • A 3 \Omega resistor.
• Z_N between points A and B will be the parallel combination of 8 - j8 in series, which is then in parallel with 3 \Omega, and finally in series with j4 \Omega.
• The parallel combination is calculated as: Z_{parallel} = \frac{(8 - j8) [3}{11 - j8} = \frac{24 - j24}{11 - j8}
• The total Norton impedance is: Z_N = j4 + \frac{24 - j24}{11 - j8}
• The calculated value for Z_N is 4.37 \Omega at an angle of 55.73 degrees, which represents a resistor and an inductor.

2. Finding Norton Current (I_N)

• Return all sources to the original circuit (without the load).
• Since there is a voltage source return it to the circuit.
• Short circuit points A and B. The current flowing through this short circuit is I_N.
• The circuit now has:
  • A j4 \Omega inductive reactance.
  • A -j8 \Omega capacitive reactance.
  • An 8 \Omega resistor.
  • A 3 \Omega resistor.
  • A 120V voltage source at 30 degrees.
• I_N is the current flowing through the short circuit between A and B.
• To find IN, determine the total current (I{total}) and then use the current divider rule.
• I_{total} is the source voltage divided by the total impedance seen by the source.
• First, calculates the impedance of j4 in parallel with the series combination of -j8 and 8 \Omega. This parallel combination is then in series with the 3 \Omega resistor. This is the total impedance (Z_{total}) seen by the voltage source.
• Z_{parallel} = \frac{(8 - j8) [j4}{8 - j4} = \frac{32 + j32}{8 - j4}
• Rationalize the denominator:
  • Z_{parallel} = \frac{(32 + j32)(8 + j4)}{(8 - j4)(8 + j4)} = \frac{256 - 128 + j(128 + 256)}{64 + 16} = \frac{128 + j384}{80}
• Z_{parallel} = 1.6 + j4.8 \Omega
• Z_{total} = 3 + 1.6 + j4.8 = 4.6 + j4.8 \Omega
• Convert to polar form: Z_{total} = 6 \angle 53.13^\circ \Omega
• Calculate the total current: I_{total} = \frac{120 \angle 30^\circ}{6 \angle 53.13^\circ} = 20 \angle -23.13^\circ A
• Use the current divider rule to find I_N. There are two approaches for current divider:
  • Long Way (using admittances):
    • Z_{total dash} = Impedance of the parallel combination of j4 and 8-j8 without the addition of the 3\Omega (i.e. before calculating the total impedence).
    • IN = I{total} \times \frac{\frac{1}{j4}}{\frac{1}{Z_{total dash}}}
    • Z{total} = Z{total dash} + 3\Omega
  • Short Way (using impedances, and only if you understand why):
    • IN = I{total} \times \frac{8-j8}{8 - j8 + j4}
• I_N = (20 \angle -23.13^\circ) \times \frac{8 - j8}{8 - j4}
• Convert to polar form:
• 8 - j8 = 11.31 \angle -45^\circ
• 8 - j4 = 8.94 \angle -26.5^\circ
• I_N = (20 \angle -23.13^\circ) \times \frac{11.31 \angle -45^\circ}{8.94 \angle -26.5^\circ}
• I_N = 25.3 \angle -41.63^\circ A

Drawing the Norton Equivalent Circuit

• The Norton equivalent circuit consists of a current source (IN) in parallel with the Norton impedance (ZN) connected between points A and B.
• The direction of the current source is the same as the original voltage source (from negative to positive).
• The load (originally connected between A and B) is placed back in the circuit, connected to the Norton equivalent.