中學Physics物理題目~~~~~~

2. Uquivalent resistance of whole circuit = 12 ohms
Hence, main current = 12/12 A = 1 A
P.d. across each of the parallel branch = 12 x 6/12 v = 6 v
hence, p.d. across 5-ohm resistor = 6 x 5/10 v = 3 v
Current i1 = 3/5 A = 0.6 A

Current through the right-most 5-ohm resistor = (1 - 0.6) A = 0.4 A
Current i2 = 0.4/2 A = 0.2 A

3. Current delivered by the voltage source
= 2/(1+1) A = 1 A
Current delivered by the current source
= 3/2 A = 1.5 A
Hence, current i = (1.5 - 1) A = 0.5 A (in direction opposite to that indicated)

Power dissipated at the bottom 1-ohm resistor = 0.5^2 x 1 w = 0.25 w
Current through the other 1-ohm resistor = (3 - 0.5) A = 2.5 A
Power dissipated = 2.5^2 x 1 w = 6.25 w

The voltage source is being charged at a power of (0.5 x 2) w = 1 w
.

1.　　A and B

　　path CDB = R//3R = 0.75R
　　path ACB = R + 0.75R = 1.75R

　　path AB = 2R//1.75R = 0.933R

　　　A and C

　　path BDC = path CDB above = 0.75R
　　path ABC = 2R + 0.75 R = 2.75R

　　path AC = R//2.75R = 0.733R

2014-03-09 23:21:07 補充：
　　　A and D

　　path AB = 2R
　　path BD = 2R
　　path AC = R
　　path CD = R

　　B and C are at the same potential. No current flow through R between B and C.
　　
　　path AD = 2R//R + 2R//R = 1.33R