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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 …

But what if, instead of two resistors, you wanted to use only one? … in parallel: $$ Rt = \frac{R1 \times R2}{R1+R2} $$ So you can see that it's really impossible for the total resistance of resistors in parallel to be greater than the value of the smallest. …

Whatever the voltage difference between A and B, because R1 and R2 are equal the voltage at the R1 R5 R2 junction will be half of that difference. Then, since R3 and R4 are equal, the voltage at th …

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 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 …

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 …

LEDs have a negative temperature coefficient of resistance, so in order to prevent current hogging You shouldn't use a single resistor to ballast parallel strings of resistors because of the different …

You'll let the magic smoke out.

According to the data printed on the driver, it's a constant-current supply and it's rated to supply about 600mA into a load that can drop anywhere between 16 and 28 volts with 600mA through it, which …

The ballast resistor must drop the difference between the supply voltage and the LED forward voltage with the LED current through it so, consulting Ohm's law, we have: $$ R = \frac{Vsupply-Vled}{Iled …

Q2 is a P-channel MOSFET, so yes, R4 is a pulldown and part of the load Q2 is working against.

Handy formula, and it works a treat!

If you visualze a digital pot as a conventional pot with a stepping motor coupled to its shaft, then it's easy to see that resistance values which fall between the stepper's resting points cant be acc …

Each resistor would dissipate about: $$ P = IE = 333mA \times 8.9V \approx 3\text{ watts,} $$ so it would be prudent to use 27 ohm, 5 watt resistors for your ballasts. …

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 …