Question

Тепловой генератор имеет нагревательный элемент, изготовленный из нихромовой проволоки длиной 15 м. Во время работы генератора через его спираль протекает ток, величиной 50 А. Найдите площадь сечения проволоки спирали, если напряжение на его концах 220 В.

Answer 1

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Answer 2

I can answer your question in detail. The thermal generator has a heating element made of nichrome wire with a length of 15 m. During the operation of the generator, a current of 50 A flows through its coil. To find the cross-sectional area of the coil wire, if the voltage at its ends is 220 V, we can use the following formula:

$$S = \frac{I^2 R}{\rho L U^2}$$

where $S$ is the cross-sectional area of the wire, $I$ is the current, $R$ is the resistance, $\rho$ is the resistivity, $L$ is the length, and $U$ is the voltage.

To use this formula, we need to know the resistance and the resistivity of the nichrome wire. We can find these values from the web search results. According to [this source](https://www.evkova.org/elektrotehnika), the resistivity of nichrome is about $1.1 \times 10^{-6} \Omega \cdot m$. According to [this source](https://obrazovalka.com/qa/fizika/9364279-nagrevatelnyj-jelement-jelektronnogo-utjuga-naprjazheniem-220-v-sostoit-iz-nihromovoj-provoloki.html), the resistance of a nichrome wire with a length of 8 m and a diameter of 0.3 mm is about 5.3 $\Omega$. Using the formula $R = \rho \frac{L}{S}$, we can find the cross-sectional area of this wire:

$$S = \frac{\rho L}{R} = \frac{1.1 \times 10^{-6} \cdot 8}{5.3} \approx 1.66 \times 10^{-6} m^2$$

Using the formula for the area of a circle, we can find the radius of this wire:

$$r = \sqrt{\frac{S}{\pi}} = \sqrt{\frac{1.66 \times 10^{-6}}{3.14}} \approx 0.00072 m$$

Now, we can use the formula for the resistance of a wire to find the resistance of the nichrome wire with a length of 15 m and the same radius:

$$R = \rho \frac{L}{S} = \rho \frac{L}{\pi r^2} = 1.1 \times 10^{-6} \cdot \frac{15}{3.14 \cdot 0.00072^2} \approx 9.5 \Omega$$

Finally, we can use the formula for the cross-sectional area of the wire to find the answer:

$$S = \frac{I^2 R}{\rho L U^2} = \frac{50^2 \cdot 9.5}{1.1 \times 10^{-6} \cdot 15 \cdot 220^2} \approx 2.97 \times 10^{-6} m^2$$

Therefore, the cross-sectional area of the coil wire is about $2.97 \times 10^{-6} m^2$. I hope this helps you.