计算图中A点电压和流过r2电流，要过程
\u5982\u56fe\u6240\u793a,R1=20\u03a9,\u7535\u8def\u4e2d\u603b\u7535\u963b\u4e3a12\u03a9,\u7535\u6d41\u8868\u793a\u6570\u4e3a0.3A,\u8bf7\u7b97(1)\u7535\u6e90\u7535\u538b(2)\u901a\u8fc7R2\u7684\u7535\u6d41\uff083\uff09\u7535\u963bR2\u7684\u963bR1=20\u6b27\u59c6 R\u603b=12\u6b27\u59c6
\u6839\u636e\u5e76\u8054\u7535\u8def\u603b\u7535\u963b\u8ba1\u7b97\u516c\u5f0f 1/R\u603b=1/R1 +1/R2 \u53ef\u4ee5\u89e3\u51faR2=30\u6b27
\u53c8\u7535\u6d41\u8868\u6d4bR1\u7535\u6d41\uff0c\u56e0\u6b64R1\u7684\u7535\u538bU1=0.3Ax20\u6b27=6V\uff0c\u800cR1\u5e76\u8054\u5728\u7535\u6e90\u4e24\u7aef\uff0c\u56e0\u6b64\u5176\u7535\u538b\u4e0e\u7535\u6e90\u7535\u538b\u76f8\u7b49\uff0c\u56e0\u6b64\u7535\u6e90\u7535\u538bU=6V
\u7531\u4e8e\u7535\u6e90\u7535\u538b\u4e0eR2\u4e24\u7aef\u7535\u538b\u76f8\u7b49\uff0c\u901a\u8fc7R2\u7684\u7535\u6d41=U2/R2=6V/30\u6b27=0.2A
\u8fd9\u4e48\u7b80\u5355\u7684\u9898\u76ee\u5c45\u7136\u6ca1\u6709\u4eba\u56de\u7b54\uff0c\u8fd8\u4ee5\u4e3a\u4f1a\u6709\u4eba\u62a2\u7b54\u2026\u2026
\u4e00\u3001\u65ad\u5f00\u5f00\u5173S\uff0c\u4e24\u4e2a\u7535\u963b\u5f62\u6210\u4e32\u8054\u7535\u8def\uff0c\u603b\u7535\u963b\u7b49\u4e8e\u4e24\u4e2a\u7535\u963b\u4e4b\u548c\uff0c\u5373\uff1a
R=R1+R2=2R2=2\u00d730=60\uff08\u6b27\u59c6\uff09\uff1b
\u6839\u636e\u7535\u8def\u5206\u538b\u539f\u7406\uff0c\u7535\u963bR2\u5206\u5f97\u4e00\u534a\u7684\u7535\u538b\uff0c\u5373\uff1a
U2=U/(R1+R2)=U\u00d7R2/2R2=6\u00d730/(2\u00d730)=3\uff08\u4f0f\uff09
\u4e8c\u3001\u5f00\u5173S\u5408\u4e0a\u540e\uff0cR2\u88ab\u65c1\u8def\uff08\u77ed\u8def\uff09\uff0c\u7535\u8def\u4e0a\u53ea\u6709\u4e00\u4e2a\u7535\u963bR1\uff0c\u6240\u4ee5\uff0c\u7535\u6d41\u8868\u6307\u793a\u7535\u6d41\u4e3a\uff1a
I=U/R1=6/30=0.2\uff08\u5b89\uff09
\u56e0\u4e3a\u7535\u963bR2\u4e24\u7aef\u88ab\u77ed\u63a5\uff0c\u6240\u4ee5R2\u4e0a\u7684\u7535\u538b\u7b49\u4e8e\u96f6\u3002
\u5509\uff0c\u8fd8\u662f\u88ab\u62a2\u7b54\u4e86\u2026\u2026
Va=5x1.6k/(1k+1.6k)=3.08v。R2电流=Va/R2=0.38mA
5) 点1=3.3v，点2,3=3.3v，点4=0v。。
绛旓細灏R2鏂紑锛屾湁锛氬紑璺鐢靛帇Uoc锛10锛10x4锛濓紞30V锛岀瓑鏁堢數闃籖eq锛4娆э紝鏁匯2鐨鐢垫祦I2锛濓紞30/(4锛1)锛濓紞6A銆
绛旓細鍏堢湅鐢佃矾锛屽疄闄呯數璺槸R1+R2*R3/(R2+R3)+R4鐨勪覆鑱旂數璺紝B鐐圭數鍘嬪紡R4鐨勭數鍘嬶紝A鐐圭數鍘鏄疪2*R3/(R2+R3)+R4鐨勭數鍘 鐢辩數璺垎鍘嬪師鐞嗭紝B鐐圭數鍘嬩负V*R4/[R1+R2*R3/(R2+R3)+R4]=5*3.9/[10+27*47/(27+47)+3.9]A鐐圭數鍘嬩负V*[R2*R3/(R2+R3)+R4]/[R1+R2*R3/(R2+R3)+R4]=5*[3...
绛旓細浜屾瀬绠′腑鎵嶄細鏈鐢垫祦娴佽繃銆傚湪鍥綽涓紝 鍏堝亣璁句簩鏋佺D瀵奸氾紝 R2锛孯3鐨勫苟鑱旂數闃讳负 5/3 ,閭ｄ箞姹傚緱A鐐圭數浣嶄负 10x(5/3)/(3+5/3)=3.57V 銆 3,57V澶т簬浜屾瀬绠＄殑瀵奸氱數鍘嬶紝 鎵浠ュ亣璁炬纭紝 姝ゆ椂浜屾瀬绠″閫氱殑锛 A鐐圭數鍘涓3.57V,浜屾瀬绠涓數娴涓3.57V/2000惟=1.785mA ...
绛旓細璁＄畻鏂规硶濡備笅
绛旓細璁惧悇鐢甸樆鐢垫祦鍙傝冩柟鍚戝鍥撅紝搴旂敤鑺傜偣鐢靛帇娉曪紝B涓哄弬鑰冪偣锛A鐐鏂圭▼: (1/2 + 1/40 + 1/8)VA=16/2 - 8/40锛26VA=312锛孷A=12v锛宨1=(VA-E1)/R1=-2A锛宨2=(VA+E2)/R2=0.5A锛宨3=VA/R3=1.5A銆
绛旓細AB涓ょ偣闂鐢垫祦鐨勫ぇ灏忥紝涓庣數鍘琛ㄧ殑鐢甸樆RV1銆丷V2銆丷1銆R2鐨勫ぇ灏忔湁鍏炽傝瑙ｅ涓嬨
绛旓細鈶″綋鐢垫祦鍙嶅悜鏃鐢垫祦娴佽繃鐢甸樆涓嶆祦杩囦簩鏋佺锛屾祦杩嘡1鍜孯2涓や釜鐢甸樆锛屽苟涓斾袱鑰呬覆鑱斻傜數璺浘濡傚浘 杩欐椂鐢垫簮20V鐨鐢靛帇鍔犲湪涓や釜鐢甸樆涓ょ銆備袱涓數闃婚樆鍊肩浉鍚岋紝鍒嗗帇鐩稿悓锛屽垎鍒负10V.鍗宠繖鏃跺彲浠ョ湅鎴1/2T鍒癟鍐咃紝U1锛10V.U2锛10V.杩欐牱缁撳悎鍓嶅崐涓懆鏈熷唴鐨勭數鍘嬫暟鍊硷紝鍒╃敤鐢垫祦鐨勭儹鏁堝簲绛夋晥鐢靛帇鐨勬眰娉曪紝鍙互鐭ラ亾 璁＄畻...
绛旓細鐢遍鍙煡锛孷S=6V锛1锛塕1=4V/0.4A=10惟锛孯2=(6V-4V)/0.4A=5惟锛2锛夊苟鑱旂數闃璁＄畻鍏紡 R骞=R1*R2/(R1+R2)锛屾湰棰樹笉闇鐢ㄦ鍏紡涔熻兘姹傦紝鍙槸甯ゼ涓婄殑澶嶄範涓涓嬨傛祦杩嘡1鐨鐢垫祦I1=6V/10惟=0.6A锛娴佽繃R2鐨勭數娴両2=6V/5惟=1.2A锛屾墍浠=0.6A+1.2A=1.8A銆
绛旓細閫氳繃r1鐢垫祦涓篿1锛岄氳繃r2鐢垫祦涓篿2锛岄偅涔堥氳繃r3鍜宺4鐨勪负i1+i2 鍙互鍒楀嚭涓や釜鏂圭▼ 12-r1*i1-锛坮3+r4锛*锛坕1+i2锛=0 6-r2*i2-锛坮3+r4锛*锛坕1+i2锛=0 i1=3A鏂瑰悜浠庡乏寰鍙 i2=-0.75A 璇存槑鏂瑰悜浠庝笂寰涓0.75A
绛旓細U2K=1.4mA*2K=2.8V UR1=1.4mA*4.2K=5.9V;UR2=1.4mA*4.2K=5.9V锛沀AB=U2K-6V+UR2=1.4V-6V+5.9V=2.7V銆侫B鐐逛箣闂达紝娴婻1鍜1鍗冩涔嬮棿鐢靛帇锛岃窡娴R2鍜2鍗冩涔嬮棿鐨勭數鍘嬫槸涓鑷寸殑锛侊紒鍚屾牱锛孶AB=10V-U1K-UR1=10V-1.4V-5.9V=2.7V銆傚鏋滄眰UAB鐢靛帇锛屾病娉曠畝鍖栵紝濡傛灉姹傛鐢垫祦鐨...
