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(1) When the LED works normally, UL=2.0V,

From the figure, we can know the current of the series circuit:

I=IL=20mA=0.02A，

Voltage across the constant-value resistor:

UR=U-UL= 12V-2V= 10V，

r = UI = 10v 0.02 a = 500ω，

(2) According to the meaning of the question, the power supply voltage used is 3V, so for the sake of safety and accuracy, the voltmeter should choose the range of D and 3V;

As can be seen from the figure, the maximum current is about 40mA, so the ammeter should be F; So choose d, f;

So the answer is: (1) series, 500ω? (2)D、F