Answer to Question 3
|The maximum current must
be 15 mA. The current flowing through the series resistor is the
same as the current in the diode.
The voltage drop across the resistor is 12 V - 3.4 V = 8.6 V
Resistance = 8.6 V ÷ 15 × 10-3 A = 573.3 W = 570 W to 2 s.f.
The nearest E24 series resistor is 560 W, but that would allow too big a current to flow ( 15.4 mA). The next one up is 620 W. The LED might not glow with quite full brightness, but the difference will be negligible. More importantly, it will last.