電路分析基礎(chǔ)第1章 答案

1、一、填空題第1章答案1-1 ,電流1-1 ,電流1-2 .正電荷1-3.電壓收），負載1-8.產(chǎn)生（或發(fā)出）,1-4.關(guān)聯(lián)參考15非關(guān)聯(lián)參考16一致1-7.消耗（或吸電源1-9.歐姆，基爾霍夫，KCL,支路電流，KVL ,元件上電壓1-10 .代數(shù)和為零1-11 .電壓二、選擇題1-4. A 1-5. B 1-6C1-7. C, B, A1-4. A 1-5. B 1-6C1-7. C, B, A1-4. A 1-5. B 1-6C1-7. C, B, A1-1. B 1-2. B 1-3. B1-8. C 1-9. B 1-10. B1-4. A 1-5. B 1-6C1-7. C, B,

2、A三、計算題1-1解：頻率為108MHz周期信號的波長為一=上嗎=278m/108xl06幾何尺寸d2.78機的收音機電路應(yīng)視為集中參數(shù)電路。說明：現(xiàn)在大多數(shù)收音機是超外差收音機，其工作原理是先將從天線接收到的高頻信號變換 為中頻信號后再加以放大、然后再進行檢波和低頻放大，最后在揚聲器中發(fā)出聲音。這種收 音機的高頻電路局部的幾何尺寸遠比收音機的幾何尺寸小。1-2解：二端元件的吸收功率為葉以其中任兩個量可以求得第三個量。A. p吸=z = 5xl(T3xi=5xl(r3w = 5mWC.E.u=E =A. p吸=z = 5xl(T3xi=5xl(r3w = 5mWC.E.u=E =- = 2xl

3、O3V = 2kVi 1x10.p 10 x10-3i = = 1x10 A = 1mAu 10一 . p sin 2t2 sin Z cos Z 八 AG. i = = 2 cos r Au sin t sin tB. p吸=-ui = 一5 x 10-3 * i x 10-3 = _5 x io w = -5gWp2D.=2Vi1口.p10 x10-3F. i = 1x10 A = 1mAu 10H. p 吸=-ui = -2 x 2e- = -4e-/W1-3解：局部支路電流，可以根據(jù)節(jié)點和封閉面KCL求得另外一些支路電流。根據(jù)封閉面的KCL求得以=-i-i-i =24A 1A 5A =

4、 30A根據(jù)節(jié)點a 的KCL求得。=-24A-(-5)A = -19A根據(jù)圭寸閉面2的KCL求得 i6 = i. + i4 -z7 = 1A + 5A (5A) = 11A根據(jù)封閉 jSB的KCL求得 4 = 3 ，4 一Go = -1A 5A (3A) = 3A根據(jù)節(jié)點h 的KCL求得 右=-i4 -ii0= -5A-(-3)A = -2A根據(jù)節(jié)點e的KCL求得i“=T；=-24A根據(jù)節(jié)點d的KCL求得il2 =-i4 =5A1-4解：局部支路電壓，可以根據(jù)回路和閉合節(jié)點序列的KVL方程，求得另外一些支 路電壓?！? = % “6 = 3 V 2 V = 5V 10 u 2 + 6 12 +

5、 “4 = 5 V + 2 V 8 V 3 V 14 Vu 2 + 7 + = 5 V + 2 V (3 V) +1OV - 1OV假設(shè)要求得電壓的，8, 9尚需要知道其中任意一個電壓1-9解：根據(jù)KCL, KVL求出各支路電壓和支路電流，然后計算各二端元件的吸收功率。i3 = -i,-i2 = -10A i6 = i2-z4 =8A-6A = 2Ag =i +4 =4A1-10解：根據(jù)KCL, KVL求出各支路電壓和支路電流，然后計算各二端元件的吸收功率。Pi =10 x2 = 20WPi =10 x2 = 20W3 =詠=-(-5) x (-3) = -15WP2 212 = (10 +

6、5) x 2 = 30W 4 - 44 = 5 x (-2 + 3) = 5W1-111-11解：l.u = Ri = (20)1 x IO_31-11解：l.u = Ri = (20)1 x IO_3 V = 2 x m v = 2mVa . u 5V 一3. i = =； = 5mAR 1x103Q5. w = (3Q) x 5e_2z =15e-2zV2. u = -Ri = -(5Q) x 1A = -5V4. 7? = - = -= -2Qi 1A 八 8cosfV 口6. R = 4Q2cos/A1-121-12解:。x2A = 6V(c)u= 10W 二1-12解:。x2A =

7、6V(c)u= 10W 二=5V2A10V j=5A2Q = 一& = (5Q)x3A = 15V(d) u = 4PR = V20W x 5Q =10VP 10W(g)i =5Au 2V20V “=4A5Q理=2人(h) i=5Q1-13 解:()R = y = 12K = 5Q,P=ui = 0Vx2A = 2QWi 2A(c) u = i/ = 2Ax5Q = lOV,(d) i =uR(c) u = i/ = 2Ax5Q = lOV,(d) i =uRP=-ui = -l2Vx3A = -36WP=i?R = 20Wp=土 二 5WRP 10W(e)i = = b = 2A,u 5VP

8、(7) = 二iP(7) = 二i20W2A= 1()匕u = 4pR = j20Wx5Q = 10Vn P 20W sR=f =7 = 5Qi2 (2A)2d /(5V)2 nosP40卬1-15 解:u4 =2QxlA = 2Vz4 =1Ai. = -z4u4 =2QxlA = 2Vz4 =1Ai. = -z4 = -1A% = 4。x (-1A) = -4V6Vu2 = % + U4 = (4 V) + 2V = 6Vz9 = = 2Ai = i2 + i3 =1A-(-2A)=3A% =10Qx3A = 30V = +%= 30V + 6V = 36 Vz. = 3AI411-16 解

9、:u4 = 2Q x (2 A) = 4V h = 74 = 2 Ai4= -2Au3 = 4。x 2A = 8Vi(2 = % + % = 8 V 4V = - 12Vj = i9 i3 = 4A 2A = -6A u= % + u2 = -60 V -12V = -72 V-12V3Q= -4Aw, =10Qx(-6A) = -60VZi = 6AA = =2VxlA = 2WA =%乙=-4V x (-1A) = 4WP2 =u2i2 =6Vx2A = 12W片=* =30Vx 3A = 90WP = -uix = 36V x (-3A) = -108WP4 = -4V x (-2A)

10、= 8W=W4z4=8Vx2A = 16WP1=u2i2=-12Vx (-4A) = 48W= -60V x (6A) = 360WP = -uix = -(-72 V) x (-6A) = -432W說明：題圖12電路是只包含一個獨立電壓源的線性電阻電路，當電流i4數(shù)值增加一 倍時，即乙=2A,電路中各電壓以及各電流均增加一倍，而各二端元件的功率增加到原來 的4倍。當電流由乙=2A變化為，4=-2A時，各電壓電流的符號改變，而功率符號不改 變。1-19 解：P. =uJ. =u.i? =10Vx3A = 30WPs2 = -us2is2 = -(-5i + 乂2 = -(-5 x (-3A)

11、 + 10V)x3A = -75W1-20 解:(c)uc =5QxlA=5V aCit =5Qx(lA) = 5V-6V CA=2A-3Qwdc=-10QxlA = -10Vwdc =-10Qx(-lA) = 10Vii =5Qx2A = 10V dv4b =(5Q + 10Q-3Q)xlA = 12V% = (5Q +10。- 3Q) x (-1A) = -12V% =-10Qx2A = -20V% = (5Q +1 ()Q - 3Q) x 2 A = 24V1-21解:(a)z = IOA-2A=8A(b)1-23(c)(e)解: = -2x5V 20V 2x5V = -40Vw = (

12、2Q)xOA-2V = -2V(d)(f)w = (OQ)xlOA = OV. /a 10V 一i = 6A H= 8A5。IVz = lA + = 2AIQ先計算電流= -2xlO_3A = -2mA(25 + 10) V(1+5 + 1.5)x10先計算電流= -2xlO_3A = -2mA一25 V電壓源發(fā)出的功率為 p = (-25V) x (-2mA) = 50mW 10V電壓源發(fā)出的功率為 p = (10 V) x (-2mA) = -20m W1-24解:彳=5A 1A = 6A4d =8V + 3Qx(6A) = 10Vp= u, x5A = 10Vx5A=50W ，Clcl，2 = 5 A 1A 3 A (6 A) = -3A wab=-12V + 4Qx (-3A) = -24Vp - u, x 1A = 24W *Dd4 =2A 3A + 1A = 4A wbc = 10 V + 2Q x (-4 A) = 2 V1-251-25p = cb x 2A =