物理的电路题如图,已知R1=R2=R3=R4=R,则AF两端的电阻为 给个有计算的解释，答案是D 大连理工大学的物理学专业怎么样？
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R2和R3和R4并联，再和R1串联 R总=R/3+R=4R/3绛旓細锛1锛夊綋寮鍏硈鏂紑鏃 鐢甸樆R1鍜孯2骞惰仈锛岀數闃籖3鍜孯4涓茶仈鍚庡啀涓嶳2骞惰仈锛屾墍浠34锛漅3+R4锛20锛堟濮嗭級R12锛漅1/2锛10/2锛5锛堟濮嗭級鍒欑瓑鏁堢數闃籖锛漅12*R34/锛圧12+R34锛夛紳20*5/锛20+5锛夛紳4锛堟濮嗭級锛2锛夊綋寮鍏硈闂悎鏃 鐢甸樆R1銆丷2鍜孯4骞惰仈锛岃孯3琚煭鎺 鎵浠ョ瓑鏁堢數闃籖锛漅1/3锛10/3锛3.3锛堟...
绛旓細灏哛1銆丷2鍜孯3鐨刌褰㈡帴娉曡浆鍖栦负鈻虫帴娉锛屽鍥锛氭牴鎹叕寮忥細R12=锛R1R2+R2R3+R3R1锛/R3锛岀敱浜R1=R2=R3=50惟锛屾墍浠12=3脳50=150惟銆傚悓鏍凤細R23=R31=150惟銆俁134=R13鈭4=60锛埼╋級銆俁235=60锛埼╋級銆傛墍浠ユ荤數闃讳负锛歊=锛圧134+R235锛夆垾R12=锛60+60锛夆垾150=200/3锛埼╋級銆鐢佃矾鎬荤數娴佷负锛欼...
绛旓細棣栧厛US1鍗曠嫭宸ヤ綔US2瑙嗕綔鐭矾I1涓2/1=2A I2锛孖3涓1A.鎺ョ潃US2鍗曠嫭宸ヤ綔US1瑙嗕綔鐭矾I2=2A,I1=I3=1A 缁间笂I1鎬讳负2+1=3A,I2鎬讳负1+2=3A锛 I3鎬讳负1+1=2A
绛旓細璁＄畻鏃跺彲浠ユ妸涓嬮潰鐨凴2鏂紑璁＄畻銆傜敱鐢甸樆涓插苟鑱鐢佃矾璁＄畻寰楁荤數闃讳负Rab=0.5惟銆傚彟涓绉嶈绠楁柟娉曪細鐢变簬鐢佃矾鐨勫绉版э紝涓嬮潰R2涓棤鐢垫祦銆傦紙鐢垫ˉ骞宠 锛夎a銆乥鍔犱笂鐢靛帇U锛屾荤數娴侊細Iab=Uab/(R1+R2)+Uab/R1+Uab/(R1+R2)Rab=Uab/Iab=1/(R1+R2)+1/R1+1/(R1+R2)=0.5惟 ...
绛旓細瑙ｏ細B銆丏杩炵嚎鏂 R1銆丷2鐢垫祦婧怚s1涓茶仈璇ユ敮璺數娴両s鍚戝悜宸 鏍规嵁KCLR4銆丷3鐢靛帇婧怳s涓茶仈鏀矾鐢垫祦锛欼s2-Is1=5-1=4锛圓锛夊悜鍚戝乏 UBC=-Is1脳R2=-1脳5=-2锛圴锛塙CD=锛圛s2-Is1锛壝桼3-Us=4脳2-24=-16锛圴锛夋墍锛歎oc=UBD=UBC+UCD=-2-16=-18锛圴锛夊啀鐢靛帇婧愮煭璺佺數娴佹簮璺細Req=RBD=R2+R...
绛旓細A銆佽R=R1=R2=R3锛屽凡鐭1涓ょ鐢靛帇涓3V锛孯3涓ょ鐢靛帇涓1V锛屾牴鎹濮嗗畾寰嬶紝娴佽繃R1鐨勭數娴佷负3R1=3R锛屾祦杩嘡3鐨勭數娴佷负1R3=1R锛涙晠娴佽繃鐢甸樆R2鐨勭數娴佸彲鑳戒负锛3R1-1R3=2R锛屼篃鍙兘涓猴細3R1+1R3=4R锛涜嫢娴佽繃R2鐨勭數娴佷负2R锛屽垯R2涓ょ鐢靛帇涓2V锛涜嫢娴佽繃R2鐨勭數娴佷负4R锛屽垯R2涓ょ鐢靛帇涓4V锛涙晠A姝ｇ‘锛...
绛旓細鏍规嵁鎴寸淮鍗楀畾鐞,鍘绘帀RL鍚庣殑寮璺數鍘婨=12+0.5X(R1//R3+R2)=16V 鐢甸樆R0=R1//R3+R2=4+4=8娆 閭ｄ箞鐢垫祦i=E/(R0+RL)=16/(8+4)=1.33A
绛旓細璁綬1銆丷2銆丷5鐢垫祦涓簒,y,z. 鍒欑敱鍩哄皵闇嶅か瀹氬緥R3銆丷4鐢垫祦涓簒-z,y+z 浠R1R3銆丷2R4鍥炶矾鐪婾AB鏈 xR1+(x-z)R3=yR2+(y+z)R4 浠嶳1R5鍜孯2涓や釜鍥炶矾鐪 xR1+zR5=yR2 浠ｅ叆鏁板间负 50x+15(x-z)=40y+25(y+z) (1)50x+10z=40y (2)鐢(2)寰 z=4y-5x (3)浠ｅ叆(1)鍖栫畝寰 y...
绛旓細瑙ｏ細璁R1銆丷2鍜孯3涓変釜鏀矾鐨勭數娴佸垎鍒负I1銆両2鍜孖3锛屾柟鍚戝垎鍒负鍚戝彸銆佸悜宸﹀拰鍚戜笅銆備娇鐢ㄥ彔鍔犲畾鐞嗘潵璁＄畻銆1銆両s=5A鍗曠嫭浣滅敤鏃讹紝Us=110V鐭矾銆傚垯锛欼1'=Is=5A銆俁2鈭3=5鈭10=10/3锛埼╋級锛屽叾涓ょ鐢靛帇锛歎=I1'脳锛圧2鈭3锛=5脳10/3=50/3锛圴锛夈傛墍浠ワ細I2'=-U/R2=-锛50/3锛/5=-10/...
绛旓細I11 = E1/R = E1/(R1+R2//R3) = 4/(10+10//20) = 0.24A R2//R3琛ㄧずR2鍜孯3骞惰仈 I21 = I11 * R3/(R2+R3) = 0.24*2/3 = 0.16A I31 = I11 * R2/(R2+R3) = 0.24*1/3 = 0.08A 2. 璁綞1涓洪浂锛屽垯E2浜х敓鐨勫悇鏀矾鐢垫祦涓 I12 = E2/r = E2/(R2+R1//R3) =...
