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LEARNING OUTCOMES

LEARNING OUTCOMES. The end of this lesson, students should be able to : Explain Norton’s Theorem. Outline the procedure adopted when using Norton’s Theorem Determine the short circuit current ISC and the equivalent Resistance R N (parallel) Construct circuit base on Norton’s Theorem.

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LEARNING OUTCOMES

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  1. LEARNING OUTCOMES • The end of this lesson, students should be able to : • Explain Norton’s Theorem. • Outline the procedure adopted when using Norton’s Theorem • Determine the short circuit current ISC and the equivalent Resistance RN (parallel) • Construct circuit base on Norton’s Theorem. • Solve linear circuits problem using Norton’s Theorem.

  2. Nortons Theorem states that "Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor. As far as the load resistance, RL is NORTON’S THEOREM

  3. Nortons Analysis • The basic procedure for solving Nortons Analysis equations is as follows 4 steps: 1. Remove the load resistor RL or component concerned and short circuit the terminals. Find IN on the output terminals A and B. 2. Short-circuit the two voltage sources and open circuit terminals A and B, the two resistors are now connected together in parallel. Find RN

  4. 3.Having found both the short circuit current, IN and equivalent internal resistance, RN. Draw the Nortons equivalent. 4. Replace the original load resistor (remove resistor) connected across terminals A and B and calculate current, I

  5. Example 1 i. I Refer to the circuit above determine the current I flows in the 40Ω resistance by using Norton’s Theorem.

  6. Following the procedures; Step 1 & 2. Remove the load resistor RL or component concerned. Short circuit terminals A and B to give us the following circuit. I1 I2 IN

  7. When the terminals A and B are shorted • together the two resistors are connected in • parallel across their two voltage sources and • the currents flowing through each resistor as • well as the total short circuit current can now • be calculated as: • Determine the short-circuit current,IN flowing in • the branch Therefore IN= 2 Amps

  8. Step 3. Short-circuit the two voltage sources and open circuit terminals A and B, the two resistors are now connected together in parallel. Determine the value of the internal resistor RN is found by calculating the total resistance at the terminals A and B giving us the following circuit RN

  9. Find the Equivalent Resistance (RN) Therefore RN = 6.67 Ω

  10. Step 4. Having found both the short circuit current, IN and equivalent internal resistance, RN. Draw the following Nortons equivalent circuit below. • Nortons equivalent circuit.

  11. Step 4. Replace the original 40Ω load resistor connected across terminals A and B as shown below. Calculate current, I ; I Norton Equivalent circuit

  12. By using current divider I = RN X IN RN + RL = 6.67Ω x 2 A 6.67Ω+40Ω = 0.286 Amps I = current flowing through 40 is 0.286 Amps

  13. RELATION THEVENIN THEOREM & NORTON THEOREM • To convert Thevenin Equivalent circuit to Norton Equivalent circuit • Norton resistance RN = RTH • Norton current , IN = VTH /RTH • 2. To convert Norton Equivalent circuit to Thevenin Equivalent circuit • Thevenin resistance, RTH = RN • Thevenin Voltage, VTH = IN X RN

  14. I2 I3 15 10 I1 20 15V 20V Exercises • Refer to the circuit below, calculate I3 by using Norton’s Theorem Answers ; IN = 3 A RN = 6 ohm I3= 0.692 A

  15. 5.3k I1 RL =3k Exercise :Refer to the circuit, convert Thevenin Equivalent to Norton Equivalent circuit 12V

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