✔ 最佳答案
(i) Consider the loop abcd, we have two 1-ohm resistors in series and then parallel with another 1-ohm resistor.
Equivalent resistance = 1x2/(1+2) ohms = 2/3 ohm = 0.6667 ohm
Hence, in loop ecdf, equivalent resistance = (1 + 0.6667 + 1) ohms = 2.6667 ohms
Final equivalent resistance at ef = 2.6667 x 1/(2.6667+1) ohm = 0.727 ohm
(ii) The circuit is similar to the one in (i)
Equivalent resistance between cd = 1x2/(1+2) ohm = 0.6667 ohms
Equivalent resistance between dcef = (0.6667 + 1 + 1) ohms = 2.6667 ohms
Total equivalent resistance between df = 1 x 2.6667/(1+2.6667) ohm = 0.727 ohm