✔ 最佳答案
圖片參考:
http://imgcld.yimg.com/8/n/HA00639401/o/701008260180513873381990.jpg
1.
1/(1/R1 + 1/R2)
= 1/ (1/1000+1/680)
= 1/ (0.001+0.00147)
= 1/0.00247
= 405Ω
2.
I = V / R
= 10 / 405
I = 0.0247A
I = 24.7mA
圖片參考:
http://imgcld.yimg.com/8/n/HA00639401/o/701008260180513873382001.jpg
1.
Ra.
= R2 + R4
= 680Ω + 330Ω = 1010Ω
Rb.
= R3 + Ra
= 1/(1/1000Ω + 1/1010Ω)
= 1/(0.001 + 0.00099)
= 1/0.00199
= 502Ω
Rt
= R1 + Rb
= 470Ω + 502Ω
= 972Ω
I1 = V/R
= 10V/972Ω
= 0.0103A
V1 = I1 x R1
= 0.0103 * 470
= 4.83V
V2 = I1 x Rb
= 0.0103 * 502
= 5.17V
I2 = V2/R3
= 5.17/1000
= 0.00517A
I3 & I4 = V2/Ra
= 5.17 / 1010
= 0.00512A
V3
= I3 x R2
= 0.00512 * 680
= 3.48V
V4
= I4 x R4
= 0.00512 * 330
= 1.69V
如果全部電阻都短路, 假設電源10V有無限大既功率供應, 其電流就無限大, 當然導線本身都會有電阻, 無限大電流就不成立。