Question

К источнику тока с ЭДС ε=1,68 В присоединили катушку с неизвестным сопротивлением R . Амперметр, включенный в разрыв цепи последовательно, показал силу тока, равную I=0,53 A. Когда к источнику тока присоединили последовательно еще один источник тока с такой же ЭДС, то сила тока, измеренная тем же амперметром в этой цепи стала равной 0,34 A. Определить внутреннее сопротивление второго r2 источника тока. Сопротивление амперметра неизвестно.

Answer 1

Ответ:

Объяснение:

Дано:

ε₁ = 1,68 В

I₁ = 0,53 A

r₁

R

ε₂  = ε₁ = ε=  1,68 В

I₂ = 0,34 A

____________

r₂ - ?

Закон Ома для полной цепи:

I₁ = ε₁ / (R + r₁)           ⇒   R = (ε₁ / I₁) - r₁                 (1)

Аналогично:

I₂ = 2·ε₁ / (R + r₂)       ⇒   R = (2·ε₁ / I₂) - (r₁ + r₂)     (2)

Приравняем (2) и (1):

(2·ε₁ / I₂) - (r₁ + r₂)  = (ε₁ / I₁) - r₁

(2·ε₁ / I₂) -  r₂  = (ε₁ / I₁)

r₂ =  (2·ε₁ / I₂) -  (ε₁ / I₁)

r₂ =  2·1,68 / 0,34 -  1,68 / 0,53 ≈ 6,7 Ом

Answer 2

Problem Analysis

We are given a circuit with a voltage source of ε = 1.68 V and an unknown resistance R connected in series with an ammeter. The ammeter measures a current of I = 0.53 A. When another voltage source with the same voltage ε is connected in series with the first source, the current measured by the ammeter becomes 0.34 A. We need to determine the internal resistance r2 of the second voltage source.

Solution

To solve this problem, we can use Kirchhoff's laws and Ohm's law.

1. Let's start by analyzing the circuit with only the first voltage source and the unknown resistance R. According to Kirchhoff's voltage law, the sum of the voltage drops across the elements in a closed loop is equal to the sum of the voltage sources in that loop. In this case, we have only one loop, so we can write:

ε = IR + I_am * R_am

where ε is the voltage of the source, I is the current flowing through the circuit, R is the unknown resistance, and I_am and R_am are the current and resistance of the ammeter, respectively.

2. We are given that I = 0.53 A and I_am = 0.53 A. Substituting these values into the equation, we get:

ε = 0.53 * R + 0.53 * R_am

3. Now, let's analyze the circuit with both voltage sources connected in series. According to Kirchhoff's voltage law, we can write:

2ε = IR + I_am * R_am

where 2ε is the total voltage of the two sources.

4. We are given that I_am = 0.34 A. Substituting this value into the equation, we get:

2ε = 0.53 * R + 0.34 * R_am

5. We now have two equations with two unknowns (R and R_am). We can solve this system of equations to find the values of R and R_am.

6. Subtracting the first equation from the second equation, we get:

ε = 0.19 * R + 0.19 * R_am

7. Subtracting this equation from the first equation, we get:

0 = 0.34 * R_am - 0.34 * R

8. Dividing both sides of this equation by 0.34, we get:

R_am - R = 0

This implies that R_am = R.

9. Substituting this result back into the first equation, we get:

ε = 0.53 * R + 0.53 * R

Simplifying, we find:

ε = 1.06 * R

10. Dividing both sides of this equation by 1.06, we get:

R = ε / 1.06

11. Finally, we can substitute the given value of ε = 1.68 V into this equation to find the value of R.

Answer

The internal resistance r2 of the second voltage source is equal to the resistance R, which can be calculated using the equation R = ε / 1.06. Substituting the given value of ε = 1.68 V, we find:

R = 1.68 / 1.06 ≈ 1.585 R

Therefore, the internal resistance r2 of the second voltage source is approximately 1.585 R.

Note: The resistance of the ammeter (R_am) is not needed to solve this problem.