According to Q = I RT, when the resistance and current-on time of the conductor are constant, the greater the secondary current, the more internal energy will be generated. If you want to protect the circuit, the current must not be too large. At this time, the fuse can play the role of protecting the circuit-as long as the current is too large and the internal energy is too large, it will reach the melting point of the fuse, and the fuse will blow, thus disconnecting the circuit and protecting it. I wrote a lot in one breath. I am doing a task, hoping to get the best answer.