Answer to Question #261198 in Math for Henny

Question

Norton’s Circuit

 

Norton’s theorem states that it is possible to simplify any circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance. Using the same ‘individual’ circuit parameters as in task 1) and working to an accuracy of three decimal places and correctly rounding up/down at each calculation you are requested to;

a) Calculate the short circuit current INORT

 

b) Calculate the open circuit resistance RNORT

 

c) Draw Norton’s equivalent circuit

 

d) Calculate the current flowing through the load resistor RL

 

e) Calculate the voltage across the load resistor RL

 

f) Calculate the power dissipated by the load resistor RL

 

g) Prove that your answers for steps d), e) and f) are correct by simulating your Norton’s equivalent circuit using the Proteus simulation software and inserting appropriate meters.

(It is important that you import your Proteus circuits into WORD as part of your submission)