kfsdgte的部落格

標題:

F5 Phy resistance

發問:

A resistor is to have a constant resistance of R = 30。Ω, independent of temperature. For this, an aluminium resistor with resistance R01 at 0℃ is connected in series with a carbon resistor with R02 at 0℃. Evaluate R01 and R02. Given that linear temperature coefficient α1= 3.9×10^-3 for aluminium and α2 = -0.5×10^-3... 顯示更多 A resistor is to have a constant resistance of R = 30。Ω, independent of temperature. For this, an aluminium resistor with resistance R01 at 0℃ is connected in series with a carbon resistor with R02 at 0℃. Evaluate R01 and R02. Given that linear temperature coefficient α1= 3.9×10^-3 for aluminium and α2 = -0.5×10^-3 for carbon. Could you please tell me the method to do it? Thanks.

• 想買CPU同底板
• equilibrium@1@
• 想問下大陸既地方 (熟識大陸地方既人請進,10點)
• 點解要扼綜援-@1@
• JR East Pass - 預訂 and 補錢問題
• 中五M1 概率
• 解析幾何之對稱問題@1@
• 百家姓的來源
• nokia有咩新手機@1@
• 耶和華吩咐亞當管理海魚跟不吃魚翅運動的精神相符合否 -

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

Since you did not give a unit to the temperature coefficients, I suppose they are in unit of C^-1 It is required that: R01 + R02 = 30 ohms --------------------- (1) Suppose the temperature rises by T degrees C change of resistance of R01 = (R01) x (3.9x10^-3) x T change of resistance of R02 = (R02) x (-0.5x10^-3) x T For the total resistance to be kept at 30 ohms, we need to have, change of resistance of R01 + change of resistance of R02 = 0 i.e. [(R01) x (3.9x10^-3) x T] + [ (R02) x (-0.5x10^-3) x T] = 0 [(R01) x (3.9x10^-3) = (R02) x (0.5x10^-3) (R02) = (R01) x [3.9x10^-3/0.5x10^-3] = 7.8(R01) Substitute into equation (1) R01 + 7.8(R01) = 30 hence, R01 = 3.409 ohms and R02 = 7.8 x 3.409 ohms = 26.59 ohms

其他解答:0D7DAC4E6DF60DFA