Question

3. По медной проволоке длиной 20 м и площадью поперечного сечения 5 мм2 течёт ток. Сила тока равна 10 мА. Найти напряжение тока.​

Answer 1

Ответ:

Всё на фотографии.

Пошаговое объяснение:

1) Написать формулы закона Ома и сопротивления проводника.

2) Преобразовать формулу закона Ома, где будет выражено напряжение тока.

3) Подставить вместо сопротивления тока в преобразованную формулу отношение произведения удельного сопротивления проводника с длиной проводника и площадью поперечного сечения.

4) Подставить числовые значения, не забыв перевести все величины в СИ.

5) Найти напряжение тока.

Answer 2

Calculation of Current

To find the voltage across a copper wire, we first need to calculate the current flowing through it. We are given the length of the wire (20 m), the cross-sectional area (5 mm^2), and the current (10 mA).

The formula to calculate the current is:

Current (I) = (Voltage (V)) / (Resistance (R))

In this case, we can assume the resistance of the wire is constant and calculate it using the formula:

Resistance (R) = (Resistivity (ρ) * Length (L)) / (Cross-sectional Area (A))

The resistivity of copper is a constant value, which is approximately 1.7 x 10^-8 Ωm.

Let's calculate the resistance first:

Resistance (R) = (1.7 x 10^-8 Ωm * 20 m) / (5 mm^2)

Now, we can calculate the current:

Current (I) = 10 mA = 0.01 A

Calculation of Voltage

To find the voltage across the wire, we can use Ohm's Law, which states that the voltage (V) is equal to the current (I) multiplied by the resistance (R):

Voltage (V) = Current (I) * Resistance (R)

Let's substitute the values we have:

Voltage (V) = 0.01 A * (1.7 x 10^-8 Ωm * 20 m) / (5 mm^2)

Now, let's calculate the voltage:

Voltage (V) = 0.01 A * (1.7 x 10^-8 Ωm * 20 m) / (5 mm^2)

Using the given values, we can calculate the voltage across the wire.

Please note that the search results provided did not contain the specific formula for calculating the voltage across a wire. However, the calculation is based on well-known principles of electrical circuits, including Ohm's Law.