R15QR3R1aoaoR2R34R2R15Q10Qbo50bo(d)(c)R,R3415×10R34 = R3 + R4 = 102R234Q=62R2 + R3415+10R162aoao122RabR23462bobo(e)(f)Rab = R + R234 = 62+62=122R2(R3 + R4)15(5 +5)Rab2=12QR62-R2 + R3 + R415+5+5
5 5 15 10 6 6 12 R34 = R3 + R4 =10 = + = + = 6 15 10 15 10 2 34 2 34 234 R R R R R Rab = R1 + R234 = 6+ 6 =12 = + + + = + + + + = + 12 15 5 5 15(5 5) 6 ( ) 2 3 4 2 3 4 ab 1 R R R R R R R R
dc0QR3R1ao5215Q52R2R4bo(a)显然，cd两点间的等效电阻为R3(R2 +R4)5(15 + 5)RedQ= 4Q5+15+5R3 +R2 + R4
显然，cd两点间的等效电阻为 = + + + = + + + = 4 5 15 5 ( ) 5(15 5) 3 2 4 3 2 4 cd R R R R R R R 15 5 5