Question

Свинцовая проволочка плавкого предохранителя имеет сечение S = 0,50 мм2 и длину l = 2,0 см. При каком напряжении сила тока в проволочке может принять предельно допустимое значение Iпред = 10 А? Решите подробно плиз)

Answer 1

I=U/RU=I*RR=p l/sU=I*p*l/sU-напряжениеI-сила токаp-удельное сопротивление свинцаl-длиннаs-поперечное сечение

Answer 2

Calculation of the Maximum Current in the Lead Wire

To determine the maximum current that the lead wire can carry, we need to consider the power dissipation and the resistance of the wire. The power dissipated in the wire can be calculated using the formula:

P = I^2 * R

Where: - P is the power dissipated in watts (W) - I is the current flowing through the wire in amperes (A) - R is the resistance of the wire in ohms (Ω)

The resistance of the wire can be calculated using the formula:

R = ρ * (l / S)

Where: - ρ is the resistivity of the material in ohm-meters (Ω·m) - l is the length of the wire in meters (m) - S is the cross-sectional area of the wire in square meters (m^2)

Given: - The cross-sectional area of the wire, S, is 0.50 mm^2 (0.50 * 10^-6 m^2) - The length of the wire, l, is 2.0 cm (0.02 m) - The maximum allowable current, Iпред, is 10 A

We can now calculate the maximum current that the wire can carry.

Calculation Steps:

1. Convert the cross-sectional area from mm^2 to m^2: - S = 0.50 * 10^-6 m^2

2. Calculate the resistance of the wire: - Use the resistivity of lead, which is approximately 0.22 * 10^-6 Ω·m. - R = ρ * (l / S) = (0.22 * 10^-6 Ω·m) * (0.02 m / 0.50 * 10^-6 m^2)

3. Substitute the resistance value into the power dissipation formula: - P = I^2 * R = (10 A)^2 * R

4. Solve for the maximum current, I: - I = sqrt(P / R)

Let's calculate the maximum current in the wire.

Calculation:

1. Convert the cross-sectional area from mm^2 to m^2: - S = 0.50 * 10^-6 m^2

2. Calculate the resistance of the wire: - R = (0.22 * 10^-6 Ω·m) * (0.02 m / 0.50 * 10^-6 m^2) = 0.0088 Ω

3. Substitute the resistance value into the power dissipation formula: - P = (10 A)^2 * 0.0088 Ω = 0.88 W

4. Solve for the maximum current, I: - I = sqrt(0.88 W / 0.0088 Ω) = 10 A

Therefore, the maximum current that the lead wire can carry while maintaining the allowable power dissipation is 10 A.

Please note that this calculation assumes ideal conditions and does not take into account factors such as temperature rise, wire heating, or safety margins. It is always recommended to consult relevant standards and guidelines for specific applications.