In Figure 19-33, determine In for the circuit as "seen" by the load resistor. The beige area identifies the portion of the circuit to be nortonized.

Question

In Figure 19-33, determine [latex]I_{n}[/latex] for the circuit as "seen" by the load resistor. The beige area identifies the portion of the circuit to be nortonized.

Accepted Answer

Short the terminals A and B, as shown in Figure 19-34,

[latex]I_{n}[/latex] is the current through the short and is calculated as follows. First, the total impedance viewed from the source is

[latex]Z= X_{C1}+ \frac{RX_{C2}}{R+ X_{C2}}= 50 \angle -90°\Omega + \frac{(56\angle 0°\Omega )(100\angle -90°\Omega )}{56 \ \Omega - j100 \ \Omega } [/latex]

[latex] \ \ \ =50 \angle -90°\Omega + 48.9 \angle -29.3°\Omega[/latex]

[latex] \ \ \ =-j50 \ \Omega + 42.6 \ \Omega - j23.9 \ \Omega = 42.6 \ \Omega - j73.9 \ \Omega [/latex]

Converting to polar form yields

[latex]Z=85.3 \angle -60.0°\Omega[/latex]

Next, the total current from the source is

[latex]I_{s}= \frac{V_{s}}{Z}= \frac{6\angle 0° V}{85.3\angle -60.0° \Omega} = 70.3\angle 60.0°mA[/latex]

Finally, apply the current-divider formula to get [latex] I_{n}[/latex] (the current through the short between terminals A and B).

[latex]I_{n}= \left( \frac{R}{R+ X_{C2}}\right) I_{s}= \left( \frac{56\angle 0°\Omega }{56 \ \Omega - j100 \ \Omega }\right)70.3\angle 60.0°mA =34.4\angle 121°mA [/latex]

This is the value for the equivalent Norton current source.

Question 19.13: In Figure 19-33, determine In for the circuit as "seen" by t...

Related Answered Questions