Thévenin and Norton Equivalent Circuits: Finding VTH and RTH - Prof. Marie-Christine Brune, Exams of Electrical and Electronics Engineering

Download Thévenin and Norton Equivalent Circuits: Finding VTH and RTH - Prof. Marie-Christine Brune and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 THEVENIN AND NORTON CIRCUITS I(A) I V + _? V (volts) I V +? I(A) V_ (volts) M.-C. BRUNET ECE 110 UIUC 22.1 Are simple forms of linear circuits with ideal sources THEVENIN AND NORTON CIRCUITS and resistances. +_VTH RTH is called a Thévenin circuit I R is called a Norton circuitN N Related important topics: IV characteristics Equivalent circuits Circuits are said to be equivalent if they have the same IV characteristic. Note: Thévenin & Norton circuits have linear IV characteristics…. M.-C. BRUNET ECE 110 UIUC 22.2 ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 Example 1: Give the IV equations for the circuits I I(a) (b) 2Ω V + _ 5Ω V + _ I = I = (c) I 10Ω V + 10Ω 2Ω _ I = Comments? (d) I10Ω V+ _ 10Ω 2Ω M.-C. BRUNET ECE 110 UIUC 22.3 I = Example 2: Give the IV equation and draw the IV graph for the following circuits (a) I 20V V + _+_ 2Ω I = = I = x V + For V = 0 I = . . . . (this is a short-circuit connection) I For I = 0 V = . . . . (this is an open-circuit connection) V M.-C. BRUNET ECE 110 UIUC 22.4 ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 First Summary I IRTH V + _C I = aV + b I = 1RTH _ V + VTHRTH V + _+_VTH Identical IVs VTH RTH I i for a short circuit VTH v for an open-circuit i I=0 V +V= 0_CKT CKT v + - M.-C. BRUNET ECE 110 UIUC 22.9 “SECOND” WAY “compute” v open-circuit; VTH = v “compute” i short-circuit; RTH = v/i Back to Example 2: (part(b)) Find the equivalent Thévenin circuit I3Ω 30V V + _+_ 6Ω Let us use the “second” way! (a) open-circuit connection I = 0 30V +_ 3Ω 6Ω v =?+_ VTherefore TH = (b) short-circuit connection i = ?3Ω 30V +_ 6Ω +V= 0_ Therefore RTH = E i l M.-C. BRUNET ECE 110 UIUC 22.10 qu va ent Thévenin circuit +_ ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 Example 5: Find the equivalent Thévenin circuit + _ 10V 10A 5Ω 10V (a) open-circuit connection I 0 +_ 10A 5Ω = v =?+_ (b) short-circuit connection +_ 10A 5Ω 10V i = ? V=0 + _ Equivalent Thévenin Circuit (solution!) 5Ω M.-C. BRUNET ECE 110 UIUC 22.11 +_40V Norton Equivalent Circuit (Very similar to Thévenin Analysis!!) I I V + _C IN RN V+_ I = aV + b I = = I = V + Identical IVs (a) open-circuit connection Norton Circuit Special Connections: IN RN I = 0 ?+ I (b) short-circuit connection: v =_ i = + M.-C. BRUNET ECE 110 UIUC 22.12 v=0_IN RN ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 Example 4: (continued) I +C I 5 V_ 15 V Thévenin equivalent (done previously) Norton equivalent ? +_ Remarks: IV Equations: Thévenin: I = V M.-C. BRUNET ECE 110 UIUC 22.13 Norton: I = V open-circuit (i=0) v =(Thévenin) =(Norton) RN = i v (Thévenin) short-circuit (v=0) i = (Thévenin)=(Norton) R TH = i v Source Conversion Theorems (Norton) Thévenin Given Norton? +_VTH RTH RN =IN = Norton Given Thévenin? RN VTH RTH = IN M.-C. BRUNET ECE 110 UIUC 22.14 +_ = ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 Example 7: Find RTH (=RN) using the ‘fast’ algorithm. (a) 7V 5Ω 6Ω+_ 2Ω (b) 8V 10Ω 5Ω+_ 2A (c) 2Ω 5A 10Ω +_ 6V 10Ω M.-C. BRUNET ECE 110 UIUC 22.19 SUMMARY(FINAL) 2 3 Thévenin Norton 12 3 v+C Open-circuit C Short-circuit i vi= use , or RTH _ compute v VTH = v compute i IN = i 1 v i=RN Use ‘fast’ algorithm produces RTH Use ‘fast’ algorithm produces RN Conversion Theorems =VTHRTH VTH =RN NI.IN M.-C. BRUNET ECE 110 UIUC 22.20 =RTH = RNRTHRN ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 Back to Example 2: use 2 I3Ω 30V V + _+_ 6Ω Recall v (open-circuit) = (=VTH) Fast algorithm: 3Ω 30V +_ 6Ω RTH = Remark: It is a good idea to check your results! i (short-circuit) found using 1 = d VTH / RTH M.-C. BRUNET ECE 110 UIUC 22.21 an = Example 8: Find the Thévenin and Norton circuits for: (a) 10V 2Ω 4Ω +_ 3Ω 2A VTH RTH = RNIN M.-C. BRUNET ECE 110 UIUC 22.22 (a) ECE 110 M.-C. Brunet Thévenin & Norton Equivalent Circuits Handout 22 Example 8: (cont’d) Find the Thévenin and Norton circuits for: (b) 6A 3Ω 10Ω + - 5V 5Ω VTH RTH = RNIN M.-C. BRUNET ECE 110 UIUC 22.23 (b) 1 Find the equivalent circuits for: (a) + 4Ω 24 (b) 5ΩI_V Norton? N Thévenin? 2 Find the Thévenin & Norton equivalent circuits: 24Ω (a) 60V 24Ω+_ 10Ω (b) -+ 10Ω 10Ω 5V 7V +_ 3A (c) 2Ω 5Ω + IN = ? Solutions 6A 4Ω 1 (a) (b) +_ 5Ω 5IN 10V _ +_ 10V M.-C. BRUNET ECE 110 UIUC 22.24 2 (a) VTH = 30V RTH = RN = 12Ω ΙΝ = 2.5Α (b) VTH = 1V RTH = RN = 5Ω ΙΝ = 0.2Α (c) ΙΝ = 7Α ( if you find 2A it’s wrong!)