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You'll need to keep the reverse voltage to the LED at less than about 5 volts, and an easy way to do that is to connect a diode in parallel opposition to the LED across the LED. Using resistive drop …

For the lamp current: $$I = \frac{P}{E} = \frac{60W}{12V} = 5 \text{ amperes} $$ For the ballast resistor value: $$ R =\frac {Vt-Vl}{I} =\frac {220V - 12V}{5A} \approx 42 \text{ ohms} $$ For th …

\$V1 = 60V \$ \$ V2 = 10V\$ \$ It = \frac {V1 -V2}{R1} =12.5mA \$ \$ Il = \frac{V2}{R2} = 5mA \$ \$ Iz = It - Il = 7.5mA \$

You misunderstand how an LED works in that Vf isn't the voltage you put across an LED to make it work, it's the voltage that appears (is dropped) across an LED when current is forced through it. If …

The 20k and 6k resistors are in parallel, yielding an equivalent resistance of 4.6k ohms. with the switch open,20mA through 4.6k will drop 92 volts, not 300 volts. …

Looking just at B, C, and E ... simulate this circuit – Schematic created using CircuitLab

Q1 is a P-channel MOSFET used to connect/disconnect the PV from the circuitry on the downstream side of its [Q1's] drain. R3 functions to pull Q1's gate up to its source, turning Q1 OFF when T1 is tu …

It appears to be a wirewound power resistor, and I believe it's 10 ohms +/- 5% with the 5th band representing the temperature coefficient or reliability.

You'll let the magic smoke out.

With the silver and gold bands being where they are, it appears to be a four-band color coded resistor. With the first band being green, the second band brown or blue, the third band silver and the f …

It's called a "switched pot." Google "switched pot" for a zillion hits

Power is the product of voltage and current and is the work done in moving a quantity of charge through a resistance. For example, If a 1 volt difference exists across a 1 ohm resistor, then the char …

As for the resistors themselves, the 91 ohm resistor will dissipate: P = (Vs - Vf) * If ~ 40 milliwatts, while the 100 ohm one will dissipate about 36 milliwatts, so a +/-5%, 1/10th watt (or larger … As an aside, you mentioned that you'll be using four tricolor lamps, so you'll need to use three separate resistors for each lamp, or a total of 12 for the array. …

Two 10ks in parallel are equal to 5k, so if you did it that way your voltage divider would look like this:

It's accepted engineering practice - in switching circuits - to force transistors' betas to 10, so with that in mind, if Q2's collector current is 300 mA, then its base current should be set at 30 mA. …