Electric Circuits Assessment Problems Chapter 6 Inductance, Capacitance, and Mutual Inductance

Ppt下載 140.125.35.23

6.1 The current source in the circuit shown generates the current pulse 𝑖 𝑔 𝑡 =0, 𝑡<0 𝑖 𝑔 𝑡 =8 𝑒 −300𝑡 −8 𝑒 −1200𝑡 𝐴, 𝑡≥0 Find (a) 𝑣(0); (b) the instant of time, greater than zero, when the voltage v passes through zero; (c) the expression for the power delivered to the inductor; (p.204)

(d). the instant when the power delivered to the (d) the instant when the power delivered to the inductor is maximum; (e) the maximum power; (f) the instant of time when the stored energy is maximum; (g) the maximum energy stored in the inductor.

6.1 Ans 𝑖 𝑔 𝑡 =8 𝑒 −300𝑡 −8 𝑒 −1200𝑡 𝐴 , 𝑡≥0 𝑣=𝐿 𝑑 𝑖 𝑔 𝑑𝑡 𝑖 𝑔 𝑡 =8 𝑒 −300𝑡 −8 𝑒 −1200𝑡 𝐴 , 𝑡≥0 𝑣=𝐿 𝑑 𝑖 𝑔 𝑑𝑡 =4𝑚 [ −300∙8 𝑒 −300𝑡 −(−1200∙8 𝑒 −1200𝑡 )] = −9.6 𝑒 −300𝑡 + 38.4 𝑒 −1200𝑡 𝑉 𝑣 0 + =−9.6+38.4 =28.8𝑉

6.1 Ans (b) −9.6 𝑒 −300𝑡 + 38.4 𝑒 −1200𝑡 =0 38.4 𝑒 −1200𝑡 =9.6 𝑒 −300𝑡 38.4 9.6 = 𝑒 −300𝑡 𝑒 −1200𝑡 𝑒 900𝑡 =4 ln 𝑒 900𝑡 = ln 4 𝑡= ln 4 900 =1.54𝑚𝑠

6.1 Ans (c) 𝑣 𝑡 =−9.6 𝑒 −300𝑡 + 38.4 𝑒 −1200𝑡 𝑖 𝑔 𝑡 =8 𝑒 −300𝑡 −8 𝑒 −1200𝑡 𝑝=𝑣𝑖 = −9.6 𝑒 −300𝑡 ∙38.4 𝑒 −1200𝑡 ∙ 8 𝑒 −300𝑡 −8 𝑒 −1200𝑡 = −76.8 𝑒 −600𝑡 +76.8 𝑒 −1500𝑡 +307.2 𝑒 −1500𝑡 −307.2 𝑒 −2400𝑡 =384 𝑒 −1500𝑡 −76.8 𝑒 −600𝑡 −307.2 𝑒 −2400𝑡

6.1 Ans (d) 𝑑 𝑑𝑡 384 𝑒 −1500𝑡 −76.8 𝑒 −600𝑡 −307.2 𝑒 −2400𝑡 =0 2𝑒 1800𝑡 −25 𝑒 900𝑡 +32=0 令𝑥= 𝑒 900𝑡 , 2 𝑥 2 −25𝑥+32=0 𝑥1=1.44766, 𝑥2=11.05234 𝑡1= ln⁡(1.44766) 900 =411.054𝑢𝑠 →𝑝 𝑡1 =32.719W 𝑡2= ln⁡(11.05234) 900 =2.6696𝑚𝑠 →𝑝 𝑡2 =−8.983W

6.1 Ans (e) 𝑡 𝑝𝑚𝑎𝑥 =2.6696𝑚𝑠 𝑝 𝑚𝑎𝑥 =𝑝 𝑡 𝑝𝑚𝑎𝑥 𝑝 𝑚𝑎𝑥 =𝑝 𝑡 𝑝𝑚𝑎𝑥 =384 𝑒 −1500∙411.054𝑢 −76.8 𝑒 −600∙411.054𝑢 −307.2 𝑒 −2400∙411.054𝑢 =32.72𝑊

6.1 Ans (f) 𝑊 is max when 𝑖 is max, 𝑖 is max when 𝑑𝑖 𝑑𝑡 is zero. When 𝑑𝑖 𝑑𝑡 =0,v=0,therefore t=1.54ms (g) 𝑖 𝑚𝑎𝑥 =8 𝑒 −300∙1.54𝑚 −8 𝑒 −1200∙1.54𝑚 =3.78𝐴 𝑊 𝑚𝑎𝑥 = 1 2 𝐿 𝑖 𝑚𝑎𝑥 2 = 1 2 ∙4𝑚∙ 3.78 2 =28.6𝑚𝐽

6. 2 The voltage at the terminals of the 0 6.2 The voltage at the terminals of the 0.6𝑢F capacitor shown in the figure is 0 for t<0 and 40 𝑒 −15000𝑡 sin30000t V for t>0. Find (a) 𝑖 0 ; (b) the power delivered to the capacitor at 𝑡= 𝜋 80 ms (c) the energy stored in the capacitor at 𝑡= 𝜋 80 ms 0.6𝑢𝐹 𝑣 𝑖 − + (p.208)

6.2 Ans 𝑖=C 𝑑𝑣 𝑑𝑡 =24𝑢 𝑑 𝑑𝑡 𝑒 −15000𝑡 𝑠𝑖𝑛30000𝑡 =24𝑢 −15000 𝑒 −15000𝑡 𝑠𝑖𝑛30000𝑡+30000 𝑒 −15000𝑡 𝑐𝑜𝑠30000𝑡 =[0.72𝑐𝑜𝑠30000𝑡−0.36𝑠𝑖𝑛30000𝑡] 𝑒 −15000𝑡 𝑖 0 =0.72 𝐴

6.2 Ans (b) 𝑖 𝜋 80 = [0.72𝑐𝑜𝑠30000 𝜋 80 −0.36𝑠𝑖𝑛30000 𝜋 80 ] 𝑒 −15000 𝜋 80 =−31.66𝑚𝐴 𝑣 𝜋 80 = 40 𝑒 −15000 𝜋 80 sin30000 𝜋 80 =20.505 𝑉 𝑝=𝑖𝑣=−649.23𝑚𝑊

6.2 Ans (c) 𝑊 𝜋 80 = 1 2 𝐶 𝑣 𝜋 80 2 = 1 2 ∙0.6𝑢∙ 20.505 2 =126.137𝑢𝐽

6. 3 The current in the capacitor of assessment problem 6 6.3 The current in the capacitor of assessment problem 6.2 is 0 for t<0 and 3 cos 50000𝑡 𝐴 for t≥0. Find (a) 𝑣(𝑡) (b) the maximum power delivered to the capacitor at any one instant of time (c) the maximum energy stored in the capacitor at any one instant of time. (p.208)

6.3 Ans 𝑣= 1 𝐶 0 𝑡 𝑖 𝑑𝑥 = 1 0.6𝑢 0 𝑡 3 cos 50000𝑥 𝑑𝑥 =100𝑠𝑖𝑛50000𝑡 𝑉, 𝑡≥0

6.3 Ans (b) 𝑝 𝑡 =𝑣𝑖= 100𝑠𝑖𝑛50000𝑡 ∙3 cos 50000𝑡 =150𝑠𝑖𝑛100000t W 𝑝 𝑚𝑎𝑥 =150𝑊 (c) 𝑤 𝑚𝑎𝑥 = 1 2 𝐶 𝑣 𝑚𝑎𝑥 2 = 1 2 ∙6𝑢∙ 100 2 =3𝑚𝐽

6.3 Ans (c) 𝑊 𝜋 80 = 1 2 𝐶 𝑣 𝜋 80 2 = 1 2 ∙0.6𝑢∙ 20.505 2 =126.137𝑢𝐽