With low-voltage DC power supply, the resistance of electrical appliances is constantly changing between 2000 ~ 6000 Ω. How to make a circuit with stable output current of 1mA? It is planned to use a single 9V or 12V battery as the power supply. The resistance value of the electric appliance changes greatly, so it is necessary to make 1mA current continuously pass through the electric appliance. It is better to add a regulating device so that the current can be adjusted between 1mA and 5mA, but when it is not adjusted, the current should be stable at a certain value, and the smaller the fluctuation is, the better. Under the premise of ensuring the current stability, the simpler the circuit is, the better Can lm334z with resistor and adjustable potentiometer achieve the goal? Why? Please explain the principle of constant current output of the circuit. If it is not convenient to explain, please give tips, such as telling me what concepts and principles I should understand to make this circuit

With low-voltage DC power supply, the resistance of electrical appliances is constantly changing between 2000 ~ 6000 Ω. How to make a circuit with stable output current of 1mA? It is planned to use a single 9V or 12V battery as the power supply. The resistance value of the electric appliance changes greatly, so it is necessary to make 1mA current continuously pass through the electric appliance. It is better to add a regulating device so that the current can be adjusted between 1mA and 5mA, but when it is not adjusted, the current should be stable at a certain value, and the smaller the fluctuation is, the better. Under the premise of ensuring the current stability, the simpler the circuit is, the better Can lm334z with resistor and adjustable potentiometer achieve the goal? Why? Please explain the principle of constant current output of the circuit. If it is not convenient to explain, please give tips, such as telling me what concepts and principles I should understand to make this circuit

This load is quite light. You can search "constant current source circuit" on the Internet

A DC power supply with electromotive force of 2V is connected with a 9 Ω resistor to form a closed circuit, and the voltage between the two poles of the power supply is 1.8V

According to Ohm's law of closed circuit, u = E-IR = E-Er / (R + r = Er / (R + R), u = E-IR = E-Er / (R + R),
Substituting the known data, we get: 1.8 = 2 * 9 / (9 + R)
R = 1 ohm
The internal resistance of the power supply is 1 ohm

If the solar panel is connected with a resistor with a resistance of 60 Ω to form a closed circuit, find: (1) the current through the resistor; (2) the terminal voltage of the solar panel; (3) the power consumption ratio of the internal and external circuits

When the power supply is not connected to the external circuit, the terminal voltage is equal to the electromotive force, then the electromotive force of the battery board is e = 800mv. According to Ohm's law of closed circuit, the short circuit current is I short = Er, then the internal resistance of the power supply is R = EI short = 800mv40ma = 20 Ω

There is a solar cell with a voltage of 800mv and a short circuit current of 40mA without external circuit
1. What is the electromotive force and internal resistance of the battery? If it is connected with a 20 ohm resistor to form a closed circuit, what is its terminal voltage?

Internal resistance R = E / I '= 0.8/0.04 = 20 Ω,
If the load r = 20 Ω, which is exactly equal to the internal resistance, a closed loop is formed, and the output voltage is exactly half of the electromotive force, i.e. 400mV
Calculation formula:
I=E/(r+R)＝0.8/(20+20)=0.02mA；
V=E-Ir=0.8-0.02*20=0.4V.