By Allan R. Hambley
<P style="MARGIN: 0px"> For undergraduate introductory or survey classes in electric engineering.
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<P style="MARGIN: 0px"> ELECTRICAL ENGINEERING: ideas AND functions, 5/e is helping scholars research electrical-engineering basics with minimum frustration. Its ambitions are to offer easy strategies in a common atmosphere, to teach scholars how the foundations of electric engineering observe to express difficulties of their personal fields, and to augment the final studying technique. Circuit research, electronic platforms, electronics, and electromechanics are coated. a large choice of pedagogical positive factors stimulate pupil curiosity and engender information of the material’s relevance to their selected profession.
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Additional info for Electrical Engineering: Principles and Applications 5th - Solutions
1 / a hundred = 1 ms (b) in advance of the change opens, the circuit is in dc regular nation with an inductor present of Vs / R1 = 1. five A. This present keeps to movement in the inductor instantly after the change opens so we now have i (0+) = 1. five A . This present needs to circulate (upward) via R2 so the preliminary worth of the voltage is v (0 +) = −R2i (0 +) = −150 V. (c) We see that the preliminary value of v(t) is ten occasions greater than the resource voltage. (d) The voltage is given by way of v (t ) = − Vs L exp( −t / τ ) = −150 exp( −1000t ) R1τ allow us to denote the time at which the voltage reaches 1/2 its preliminary value as tH. Then we've zero. five = exp( −1000tH ) fixing and substituting values we receive tH = −10 −3 ln(0. five) = 10 −3 ln(2) = zero. 6931 ms 122 E4. five First we write a KCL equation for t ≥ zero. t v (t ) 1 + ∫ v (x )dx + zero = 2 R L zero Taking the spinoff of every time period of this equation with appreciate to time and multiplying each one time period by means of R, we receive: dv (t ) R + v (t ) = zero dt L the answer to this equation is of the shape: v (t ) = okay exp( −t / τ ) during which τ = L / R = zero. 2 s is the time consistent and ok is a continuing that has to be selected to slot the preliminary stipulations within the circuit. because the preliminary (t = 0+) inductor present is distinct to be 0, the preliminary present within the resistor needs to be 2 A and the preliminary voltage is 20 V: v (0+) = 20 = okay hence, we've v (t ) = 20 exp( −t / τ ) 1 iR = v / R = 2 exp( −t / τ ) t t 1 iL (t ) = ∫ v (x )dx = [− 20 τ exp( −x / τ )] = 2 − 2 exp( −t / τ ) L0 2 zero E4. 6 sooner than t = zero, the circuit is in DC regular kingdom and the similar circuit is hence we've i(0-) = 1 A. but the present during the inductor can't swap right away so we even have i(0+) = 1 A. With the change open, we will write the KVL equation: di (t ) + 200i (t ) = a hundred dt the answer to this equation is of the shape i (t ) = okay 1 + ok 2 exp( −t / τ ) within which the time consistent is τ = 1 / 2 hundred = five ms. In regular kingdom with the change open, we've i (∞) = ok 1 = a hundred / 2 hundred = zero. five A. Then utilizing the preliminary 123 current, we have now i (0+) = 1 = ok 1 + okay 2 , from which we confirm that okay 2 = zero. five. therefore we've i (t ) = 1. zero A for t < zero = zero. five + zero. five exp( −t / τ ) for t > zero. v (t ) = L di (t ) dt = zero V for t < zero = −100 exp( −t / τ ) for t > zero. E4. 7 As in instance four. four, the KVL equation is t 1 Ri (t ) + ∫ i (x )dx + v C (0+) − 2 cos(200t ) = zero C zero Taking the spinoff and multiplying via C, we receive di (t ) RC + i (t ) + 400C sin(200t ) = zero dt Substituting values and rearranging the equation turns into di (t ) five × 10 −3 + i (t ) = −400 × 10 −6 sin(200t ) dt the actual answer is of the shape i p (t ) = A cos(200t ) + B sin(200t ) Substituting this into the differential equation and rearranging phrases ends up in five × 10 −3 [− 200A sin(200t ) + 200B cos(200t )] + A cos(200t ) + B sin(200t ) = −400 × 10 −6 sin(200t ) Equating the coefficients of the cos and sin phrases supplies the subsequent equations: − A + B = −400 × 10 −6 and B + A = zero from which we ascertain = 2 hundred × 10 −6 and B = −200 × 10 −6 .