-
Joule's law is defined as the power that occurs in a uniform conductor in the form of heat that is proportional to the product of the resistance of this conductor and the square of the current passing through it.
Specifically, it can be understood that the heat generated by the current passing through the conductor is proportional to the square of the current, or proportional to the resistance of the conductor.
After understanding the above concept, it is easy to explain:
1) The filament in the home circuit heats up and glows when energized, while the wire connected in series with the filament does not heat up significantly.
The reason is that the current flowing through the filament and the wire is the same, and the resistance of the filament is much greater than the resistance of the wire, so the heat generated on the two is also far greater than that of the latter, so the former is so hot that it glows, while the latter does not heat significantly.
2) If a high-power electric furnace is connected to the lighting circuit, the wire will be significantly heated, which may burn out its insulating skin and even cause disasters.
This is because, although the current flowing through the electric furnace and the wire is the same, generally speaking, the resistance of the electric furnace is much greater than the resistance of the wire, but compared with the "filament and wire", the degree of greater is much smaller, so I only say "much greater", not "far greater", therefore, the heat of the wire is more obvious, in serious cases, it is possible to burn the insulating skin of the wire, or even burn, thus leading to disaster.
The function of the fuse is to protect the electrical equipment in series in the circuit, but it has a certain range of protection, which is related to the power supply of the electrical appliance or the line, and for the fuse, it has an important parameter - rated current.
Although the current flowing through the fuse, wire, and electrical equipment is the same, the resistivity of the fuse is much greater than that of the wire, and the heat of the fuse is not large when the electricity is used under the specified current, so it will not break. However, if a short circuit occurs in the circuit, the current will increase dramatically to the point where the current value is much greater than the rated current of the fuse, and the heat generated on the fuse will also increase so much that the melting point of the fuse will be exceeded, and the fuse will be blown off.
-
Joule's law w=pt=i squared rt
1. The reason why the bulb is heated and glows after being energized, and the wire in series with the filament is not heated obviously, because the resistance of the two is very different, the wire resistance is very small, and the electric energy consumed is very small, so the heating is not obvious; When connected to a high-power electric furnace, the current increases significantly, and the electrical energy consumed by the wire is proportional to the square of the current, so the wire heats up significantly. When the current exceeds the maximum ampacity of the wire, the insulating skin melts and even burns.
2. When the circuit is short-circuited, the circuit load is only the resistance of the wire between the short-circuit point and the power supply, which is very small, so the current is very large. The melting point of the fuse material is very low, and the sudden increase in current causes the fuse to heat up and melt quickly, protecting the circuit.
-
Pure resistive electric alarm wide loop. Joule's law applies to purely resistive circuits as well as non-purely resistive circuits. Pure Resistance Circuits:
Circuits composed of electrical heating devices such as electric furnaces and soldering irons, and motors with incandescent lamps and rotors stuck are also pure resistive devices.
Joule's lawWhen an electric current passes through a conductor, it will produce heat, which is called the thermal effect of electric current, and an electric heater is a device that uses the thermal effect of electric power to heat up, electric stoves, electric soldering irons, electric irons, electric rice cookers, electric ovens, etc. are all common electric heaters. The main component of the electric heater is the heating element, which is made of resistance wire with high resistivity and high melting point wound on an insulating material.
Joule's law states that the heat generated by the current passing through the conductor is proportional to the resistance of the conductor, proportional to the square of the current passing through the conductor, and proportional to the time of energizing. This law was discovered by the British scientist Joule in 1841.
Joule's law is an experimental law that can be applied to any conductor in a wide range and can be used in all circuits. When encountering the thermal effect of electric current, such as calculating the heat emitted when an electric current passes through a circuit; Joule's law can be used when comparing how much heat is emitted by a certain section of a circuit or conductor, i.e., when considering the requirements for a circuit from the perspective of the thermal effect of the current.
-
Electric furnace wire and copper wire are connected in series, i electric furnace wire.
i Copper wire. The power-on time t is the same, q=i2
RT, R electric furnace wire.
r Copper wire. Heat generated by electric current: q electric furnace wire.
Q copper wire. Electric stove wire is hotter
So the answer is: electric stove
-
The fuse used in the home circuit should be connected in series in the circuit, because the fuse wire with the same length and cross-sectional area is larger than the resistance of the wire, and the melting point is lower than the wire, when the current in the circuit is too large, according to Joule's law, it can be known that the fuse generates more heat, and it can be blown when its temperature reaches the melting point, so as to achieve the purpose of automatically cutting off the circuit
So the answer is: strings; A fuse wire with the same length and cross-sectional area has a greater resistance than a wire and a lower melting point than a wire
q=w=pt=u²/r×t∵t1:t2=1:2,r1:
r2=2:1, so 1min in parallel, q1=1 10q total, q2=1 20q total, t=q total (q1+q2)=1 (3 20)=20 3min >>>More