Question

Два проводника сопротивление которых 20 Ом и 30 ОМ, соединяют последовательно и подключают к источнику тока. В первом выделилось 60 дж теплоты. Какое количество теплоты выделилось во втором проводнике за это же время? ОФОРМИТЕ ВСЁ КАК ПОЛОЖЕНО С ДАНО СИСТЕМОЙ СИ, ФОРМУЛАМИ И РЕШЕНИЯМИ.

Answer 1

Решение в приложении.

Answer 2

Problem Analysis

We are given two resistors with resistances of 20 Ω and 30 Ω, connected in series and connected to a current source. The first resistor dissipates 60 J of heat. We need to find the amount of heat dissipated in the second resistor during the same time.

Solution

When resistors are connected in series, the total resistance is the sum of the individual resistances. In this case, the total resistance is 20 Ω + 30 Ω = 50 Ω.

The amount of heat dissipated in a resistor can be calculated using the formula:

Q = I^2 * R * t

Where: - Q is the heat dissipated (in joules) - I is the current flowing through the resistor (in amperes) - R is the resistance of the resistor (in ohms) - t is the time (in seconds)

Since the resistors are connected in series, the current flowing through both resistors is the same. Let's assume the current flowing through the circuit is I.

For the first resistor with a resistance of 20 Ω, we are given that it dissipates 60 J of heat. Using the formula, we can write:

60 = I^2 * 20 * t

For the second resistor with a resistance of 30 Ω, we need to find the amount of heat dissipated. Using the same formula, we can write:

Q2 = I^2 * 30 * t

To find Q2, we need to find the value of I. We can rearrange the equation for the first resistor to solve for I:

I^2 = 60 / (20 * t)

Taking the square root of both sides, we get:

I = sqrt(60 / (20 * t))

Substituting this value of I into the equation for Q2, we get:

Q2 = (sqrt(60 / (20 * t)))^2 * 30 * t

Simplifying, we get:

Q2 = 3 * 30 = 90 J

Therefore, the second resistor dissipates 90 J of heat during the same time.

Answer

The amount of heat dissipated in the second resistor during the same time is 90 J.