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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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Сарacitor dvc ic = C Inductor dil Resistor VR = iRR dt 7= la dt Example 1: Find the differential equations that describe the mesh currents iz and iz in the circuit. Step 1. Write the mesh equations for the circuit. Mesh 1: 8i, + 16 S.(i – iz)dt = v(t) + 10i2 + 16 S(i2 – i)dt = 0 1H diz Mesh 2: dt -0- 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) i – 16i2 = Dv(t) -16i, + (D2 + 10D + 16)i, = 0 (2) (1) 8 + 16i1 – 16i2 = dt d?iz + 10- (1)a P diz + 16i, = 0 dt -16i, + 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 = 2v(t) ( (8D + 16) i – 16i2 = Dv(t) (8D + 16)l-16i, + (D² + 10D + 16)iz = 0) 16 The differential equation for the current iz is d²iz diz + 12 dt + 36 iz = 2v(t) dt2 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 iz. 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, di d²v(t) dv(t) + 1.25 dt + 2v(t) + 12 + 36 i = 0.125 dt2 dt dt2
Transcribed Image Text:Сарacitor dvc ic = C Inductor dil Resistor VR = iRR dt 7= la dt Example 1: Find the differential equations that describe the mesh currents iz and iz in the circuit. Step 1. Write the mesh equations for the circuit. Mesh 1: 8i, + 16 S.(i – iz)dt = v(t) + 10i2 + 16 S(i2 – i)dt = 0 1H diz Mesh 2: dt -0- 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) i – 16i2 = Dv(t) -16i, + (D2 + 10D + 16)i, = 0 (2) (1) 8 + 16i1 – 16i2 = dt d?iz + 10- (1)a P diz + 16i, = 0 dt -16i, + 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 = 2v(t) ( (8D + 16) i – 16i2 = Dv(t) (8D + 16)l-16i, + (D² + 10D + 16)iz = 0) 16 The differential equation for the current iz is d²iz diz + 12 dt + 36 iz = 2v(t) dt2 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 iz. 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, di d²v(t) dv(t) + 1.25 dt + 2v(t) + 12 + 36 i = 0.125 dt2 dt dt2
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