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# Same question again?

Hey Guys,

Here I am with the same question. Sorry for asking the same thing over and over again but im not very bright... Ok so I'm going to attach three pictures can anyone of you tell me which one will be brighter for longer?

## Discussions

Best Answer 4 years ago

First let's make some assumptions.

1. The LEDs have a 1.8V drop across them. (Typical for red LEDs)

2. The more current through the LED the brighter it will be.

3. The right node of each circuit is ground or 0V.

So the question then becomes, which LED will has the most current passing through it?

One key thing to understand is that LEDs have a constant voltage drop across them no mater the current flowing through them. Resistors do not behave this way. They follow Ohm's law, E=IR. As the current increases so does the voltage drop across them.

Top Circuit:Since the 3 legs are identical they will all have the same current flowing through them. So let's just worry about one of them. To calculate the current through the LED we must calculate the current through the resistor. The current through the resistor will be the same for the LED because of Kirchoff's law. To calculate the current through the resistor we use Ohm's law. (Thanks Mr. Ohm! Where would we be without you?)E=IR

9-1.8 = I * 330

7.2/300 = I

I = 22mA

Total current would be 3 times this, 22mA for each leg.

It=66mA

To find the current in the resistor we must have the voltage drop across it. This will be 9V minus the drop across the LED which is 1.8. We know the drop across the LED AND the resistor must be 9V. We know the voltage drop across the LED is 1.8. That means 9-1.8 must be the drop across the resistor. This gives us a current of 22mA.

Left Circuit:We do the same for this one except we have to calculate both legs.Top Leg: 9-1.8-1.8 = I * 150

I = 36mA

Bottom Leg: 9 - 1.8 = I * 330

I = 22mA

It=58mA

Right Circuit:This is the same as the first with just the values changed.4.5 - 1.8 = I * 68

I = 40mA

It = 120mA

This tells me that the Right Circuit would have the brightest LEDs at 40mA but this circuit uses the most current total and would last the least amount of time on a battery. The left circuit would last the longest because it uses the least total current.

Answer 4 years ago

Forget current, think about POWER which is what discharges the battery capacity.

For circuit top, 9x0.066= 0.594 W, on a 9V battery

For circuit left, P=9 x0.058 = 0.496W, on a 9V battery

For circuit right, though P=4.5 x 0.040 = 0.540W, on three AA batteries.

Capacity of AA batteries is >> 9V, so the right circuit wins4 years ago

IF we assume the battery capacity was the same, and we assume the LEDs are all the same colour, lets say red, then the

one with two leds in series, and one in parallel with that is marginally

ahead.

BUT the capacity of a 9V battery is much less than three AA cells, so your 4.5 volt circuit lasts a lot longer

Answer 4 years ago

The leds are blue and use 3.3v. As the AA cells get used wouldnt there voltage drop below 3.3v before they are fully used resulting in dim leds

Answer 4 years ago

Yes, but the time the battery is alive for is STILL a lot longer than a 9V. The batteries are dead by the time you have 3.6V from the cell.

A 9V battery is dead when you reach 7.2V. It holds around 1/10th of the energy that a 1.5V cell holds

Answer 4 years ago

+1

Answer 4 years ago

+1