Question

ПОМОГИТЕ !!!!!!! В лекции "Мультиметр" при измерении тока в режиме 200 мA мы получили ток (19,3±0,2) мА, а при измерении тока в режиме 20 мA получилось (18,75±0,05) мА, то есть диапазоны не пересекаются. Это объясняется тем, что сопротивление мультиметра неодинаково в различных режимах измерения тока: в них используются различные шунты. Пусть сопротивление мультиметра в режиме 20 мA равно R1, а в режиме 200 мА R2. Определите отношение R1/R2. Ответ округлите до целого числа.

Answer 1

Ответ:

10

Объяснение:

ПОДОШЛО В СИРИУСЕ

Answer 2

Understanding the Problem

In the lecture on the multimeter, the current was measured in two different modes: 200 mA and 20 mA. The measured currents were (19.3 ± 0.2) mA and (18.75 ± 0.05) mA, respectively. The ranges do not overlap, indicating that the resistance of the multimeter is different in these two modes due to the use of different shunts. We need to determine the ratio of the resistance of the multimeter in the 20 mA mode (R1) to the resistance in the 200 mA mode (R2), rounding the answer to the nearest whole number.

Solution

To find the ratio of R1 to R2, we can use the formula:

R1/R2 = (I2 ± ΔI2)/(I1 ± ΔI1)

where I1 and I2 are the measured currents in the 200 mA and 20 mA modes, and ΔI1 and ΔI2 are the corresponding uncertainties.

Given: - I1 = 19.3 mA - ΔI1 = 0.2 mA - I2 = 18.75 mA - ΔI2 = 0.05 mA

Substituting the values into the formula, we get:

R1/R2 = (18.75 ± 0.05)/(19.3 ± 0.2)

To simplify the calculation, we can ignore the uncertainties and calculate the ratio using the central values:

R1/R2 = 18.75/19.3

Calculating the ratio, we find:

R1/R2 ≈ 0.9705

Rounding this to the nearest whole number, we get:

R1/R2 ≈ 1

Therefore, the ratio of R1 to R2 is approximately 1.

Please note that the above calculation assumes that the uncertainties in the measured currents are independent and do not affect the ratio significantly.