Question

Задания из задачника№17.20 №17.23 №17.20На лампе для карманного фонарика написано: 4 B 1Вт, а на лампе в прихожей - 220 B 40Вт. Какая из ламп рассчитана на большую силу тока? Во сколько раз? Каково сопротивление нитей накала ламп в рабочем состоянии?? №17.23В проводнике за 10мин при силе тока 5 А выделяется количество теплоты 5 кДж. Каково сопротивление проводника?

Answer 1

Дано:I=5At=600cQ=5000ДжНайти:
R-?Решение:Q=I^2Rt =>R=Q/I^2t=5000Дж/25А * 600с=0,33Ом 

Answer 2

Task 17.20:

The question asks which of the two lamps is designed for a higher current and by how many times. It also asks for the resistance of the filament in the lamps when they are in the working state.

To determine which lamp is designed for a higher current, we need to compare the power ratings of the two lamps. The power rating of a lamp is given by the formula P = V * I, where P is power, V is voltage, and I is current.

The power rating of the pocket flashlight lamp is given as 1W at 4V. Therefore, the current can be calculated as I = P / V = 1W / 4V = 0.25A.

The power rating of the hallway lamp is given as 40W at 220V. Therefore, the current can be calculated as I = P / V = 40W / 220V = 0.182A.

Comparing the two currents, we can see that the pocket flashlight lamp is designed for a higher current of 0.25A, while the hallway lamp is designed for a lower current of 0.182A.

To calculate the ratio of the currents, we can divide the higher current by the lower current: 0.25A / 0.182A ≈ 1.37.

Therefore, the pocket flashlight lamp is designed for a higher current by approximately 1.37 times.

Now let's calculate the resistance of the filament in the lamps when they are in the working state. The resistance of a filament can be calculated using Ohm's Law: R = V / I, where R is resistance, V is voltage, and I is current.

For the pocket flashlight lamp, the voltage is given as 4V and the current is 0.25A. Therefore, the resistance can be calculated as R = 4V / 0.25A = 16Ω.

For the hallway lamp, the voltage is given as 220V and the current is 0.182A. Therefore, the resistance can be calculated as R = 220V / 0.182A ≈ 1209.89Ω.

Therefore, the resistance of the filament in the pocket flashlight lamp is approximately 16Ω, while the resistance of the filament in the hallway lamp is approximately 1209.89Ω.

Task 17.23:

The question asks for the resistance of a conductor given the amount of heat generated and the current flowing through it.

To calculate the resistance of the conductor, we can use the formula Q = I^2 * R * t, where Q is the amount of heat generated, I is the current, R is the resistance, and t is the time.

In this case, the amount of heat generated is given as 5 kJ (kilojoules), the current is given as 5A, and the time is given as 10 minutes, which is equivalent to 600 seconds.

Substituting the values into the formula, we have 5 kJ = (5A)^2 * R * 600s.

Simplifying the equation, we have 5 kJ = 25A^2 * R * 600s.

To solve for the resistance R, we can rearrange the equation as R = (5 kJ) / (25A^2 * 600s).

Converting 5 kJ to joules, we have 5 kJ = 5000 J.

Substituting the values into the equation, we have R = (5000 J) / (25A^2 * 600s).

Simplifying further, we have R = 0.0333 Ω.

Therefore, the resistance of the conductor is approximately 0.0333 Ω.

Please note that the calculations provided are based on the information given in the question. If there are any additional factors or assumptions that need to be considered, please let me know.