Electric Circuit Problem: finding open circuit voltage/ short circuit current?

Question

a) What is the open circuit voltage and the short circuit current for this circuit.

b) calculate Equivalent Resistor directly

c) What would be the values for Norton Equivalent of the circuit ?

Answer 1

use superposition method....

first open current source ....

voltage 250 divided between 25 and 100

25 x 100 / ( 100 + 25 ) = 200 v ....across ab

now short ab.....5II 100 >> 4.76

voltage 250 divided between 25 and 4.76

i = 250 / ( 25 + 4.76 ) = 8.4 A ...this current divided between 100 and 5 ........8.4 x 4.76 / 5 = 7.99 A ...short circuit ab



now short the voltage source....current 8A divided between 100 and 25.......100 II 25 = 20 ..8 x 20 = 160 v. voc on ab

now short ab ....25 II 100 II 5 >> 4. ...8 x 4 / 5 = 6.4 A ...sc current on ab

now do algebric sum..they are on opposite directions...

on first step ...a + b- 200 v current downward

on second step b+ a- 160 v current upward..so

voltage is 40 v a+ b- current 1.59 A downwarn



norton equivalent >>>current source 1.6 A in parallel 40 / 1.6 = 25 ohms resistance

Answer 2

Simplify. Remove the 125Ω resistor next to the voltage source. It loads the source, but has no impact on the questions. Remove the 10Ω resistor in series with the 8A current sink because a current sink has infinite impedance and 10Ω will NOT affect that, either.



Now convert the 250V source in series with the 25Ω into the Norton equivalent of a 10A source and a 25Ω in parallel with it. Now you can see 10A flowing into the V₂ node and 8A flowing out through the 8A current sink, leaving a net 2A into the node. Also, the 25Ω and 100Ω are in parallel, so convert that to a 20Ω resistor. Convert this 2A current source + 20Ω resistor in parallel to it into the Thévenin equivalent of 40V with a series 20Ω resistor and combine the 20Ω with the 5Ω to get 25Ω as the final Thévenin equivalent of the circuit.



To convert to the Norton of that, you have 40V ⁄ 25Ω = 1.6A. So that would be 1.6A in parallel with 25Ω resistor.

Answer 3

The 125 ohm can be safely removed (ignored) for these calculations at a/b. The only thing that 125 ohm resistor affects, is how much current the 250 Volt supplies. The 125 ohm resistor will always have 250 V across it, and does not affect the rest of the circuit.