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1/F I 5a In elr

1/F I 5a In elr

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Instructions: THE WHITE PAPER POSTED IS THE GUIDE TO ANSWER THE (YELLOW GIVEN) 

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Inductor Сараcitor dvc ic = C Resistor di̟ VR = iRR VL = L dt dt Example 1: Find the differential equations that describe the mesh currents iį and iz in the circuit. Step 1. Write the mesh equations for the circuit. Mesh 1: 8i, + 16 S„(i – iz)dt = v(t) + 10iz + 16 SL.(iz – 1)di = 0 8Ω 1H diz Mesh 2: dt 1 v(t) i2 $102 16 Step 2. Differentiate the mesh equations to Step 3. Then using operators, let D = d/dt. eliminate the integrals. Substitution gives: di d (8D + 16) iį – 16i2 = Dv(t) -16i, + (D² + 10D + 16)i, = 0 (2) (1) 8 + 16i - 16iz = v(t) dt dt d²iz + 10 diz dt -16i, + + 16iz = 0 dt2 Step 4. Using the elimination method, multiply We get: equation (1) by 16 and equation (2) by (8D + 16), then add the resulting equations. This will eliminate Which simplifies to (8D3 + 96D² + 288D) iz = 16Dv(t) the current variable i1. (D² + 12D + 36) iz The differential equation for the current iz is d²iz dt2 = 2v(t) ( (8D + 16) i – 16i, = Dv(t) (8D + 16) l–16i1 + (D² + 10D + 16)iz = 16 diz + 12 + 36 iz = 2v(t) dt Step 5. Similarly, if we multiply equation (2) by 16 and equation (1) by (D2 +10D + 16), then add the resulting equations, we will eliminate the current variable i2. We get: (8D3 + 96D² + 288D) i = D(D² + 10D + 16)v(t) which simplifies to: (D² + 12D + 36) i = 1 (D² + 10D + 16)v(t) The differential equation for the current i, is: d²i, d²v(t) dv(t) + 2v(t) di + 12 + 36 i, = 0.125 + 1.25 dt2 dt dt2 dt
Transcribed Image Text:Inductor Сараcitor dvc ic = C Resistor di̟ VR = iRR VL = L dt dt Example 1: Find the differential equations that describe the mesh currents iį and iz in the circuit. Step 1. Write the mesh equations for the circuit. Mesh 1: 8i, + 16 S„(i – iz)dt = v(t) + 10iz + 16 SL.(iz – 1)di = 0 8Ω 1H diz Mesh 2: dt 1 v(t) i2 $102 16 Step 2. Differentiate the mesh equations to Step 3. Then using operators, let D = d/dt. eliminate the integrals. Substitution gives: di d (8D + 16) iį – 16i2 = Dv(t) -16i, + (D² + 10D + 16)i, = 0 (2) (1) 8 + 16i - 16iz = v(t) dt dt d²iz + 10 diz dt -16i, + + 16iz = 0 dt2 Step 4. Using the elimination method, multiply We get: equation (1) by 16 and equation (2) by (8D + 16), then add the resulting equations. This will eliminate Which simplifies to (8D3 + 96D² + 288D) iz = 16Dv(t) the current variable i1. (D² + 12D + 36) iz The differential equation for the current iz is d²iz dt2 = 2v(t) ( (8D + 16) i – 16i, = Dv(t) (8D + 16) l–16i1 + (D² + 10D + 16)iz = 16 diz + 12 + 36 iz = 2v(t) dt Step 5. Similarly, if we multiply equation (2) by 16 and equation (1) by (D2 +10D + 16), then add the resulting equations, we will eliminate the current variable i2. We get: (8D3 + 96D² + 288D) i = D(D² + 10D + 16)v(t) which simplifies to: (D² + 12D + 36) i = 1 (D² + 10D + 16)v(t) The differential equation for the current i, is: d²i, d²v(t) dv(t) + 2v(t) di + 12 + 36 i, = 0.125 + 1.25 dt2 dt dt2 dt
Ii 5a In 34
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