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有一題物理問題唔明白

發問:

A 11.2 {\rm V} battery (assume the internal resistance = 0) is connected to two resistors in series. A voltmeter whose internal resistance is 12 {\rm k}\Omega measures 5.7 {\rm V} and 2.6 {\rm V}, respectively, when connected across each of the resistors.What is the... 顯示更多 A 11.2 {\rm V} battery (assume the internal resistance = 0) is connected to two resistors in series. A voltmeter whose internal resistance is 12 {\rm k}\Omega measures 5.7 {\rm V} and 2.6 {\rm V}, respectively, when connected across each of the resistors. What is the resistance of each resistor? Express your answers using two significant figures. Enter your answers numerically separated by a comma. R_{5.7 {\rm V}}, R_{2.6 {\rm V}} =?

最佳解答:

Let the resistances be R1 and R2 and when the voltmeter connected, their measured voltages are 5.7 V and 2.6 V resp. So considering the condition when the voltmeter is connected across R1: voltage across R2 = 11.2 - 5.7 = 5.5 V Equivalent reisstance of voltage-R1 = R1(12k)/(R1 + 12k) So R1(12k)/[R2(R1 + 12k)] = 57/55 R1R2 + R2(12k) = 55R1(12k)/57 R1R2 = 55R1(12k)/57 - R2(12k) ... (1) Considering the condition when the voltmeter is connected across R2: voltage across R1 = 11.2 - 2.6 = 8.6 V Equivalent reisstance of voltage-R2 = R2(12k)/(R2 + 12k) So R2(12k)/[R1(R2 + 12k)] = 13/43 R1R2 + R1(12k) = 43R2(12k)/13 R1R2 = 43R2(12k)/13 - R1(12k) ... (2) Combining: 55R1(12k)/57 - R2(12k) = 43R2(12k)/13 - R1(12k) 23579R1 = 51692R2 R1 = 2.192R2 Sub into (1): 2.192R2^2 = 13385R2 R2 = 6105 Ω R1 = 13385 Ω

其他解答:

R_{5.7 <units>{\rm V}</units>}, R_{2.6 <units>{\rm V}</units>} =? 唔明呢個系咩意思~

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