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Thank you Bob,What you need is a resistor in series with the bulb.
If the bulb draws 300mA at 1.5V we can use Ohms law to find the resistance of the bulb when it is lit. This is different than the resistance when it is not lit because, as the filament heats up, its resistance increases. So, using Ohms law:
V = I * R
R = V / I
R = 1.5 / 0.3 = 5 Ohms.
Now, we want to make a voltage divider using the bulb itself as the lower resistor so that the voltage across the bulb is 1.5V.
Using the equation of a voltage divider:
Vout = Vin * R2 / (R1 + R2)
replacing R2 with the bulb resistance, and the two voltages we get:
1.5 = 9 * 5 (R1 + 5)
1,5 * (R1 + 5) = 9 * 5
1.5 * R1 + 1.5 * 5 = 9 * 5
1.5 * R1 = 9 * 5 - 1.5 * 5
1.5 * R1 = 35
R1 = 35 / 1.5 = 25
Use a 27 Ohm resistor in series with your bulb and your bulb will light. With a fresh battery it might light for 20 minutes.
Edit: If you have successfully used LED, you should have already known how to calculate this.
The way it is usually expressed for LEDs is
R = (Vin - Vf) / If
in this case:
R = (9 - 1.5) / 0.3 = 25
Edited to fix original calculation and show alternate calculation.
Bob
Only if you include the current through the ground or adjust pin. For some parts that is uA.A linear voltage regulator is less efficient than a simple resistor.
Thank you Bob I will get a 3 watt resistor. I will study your calculations. Much appreciatedWell, I didn’t do the power calculation.
P = I * V
The current is 0.3 and the voltage across the resistor is 9- 1.5 = 7.5
So the power is 2.25 Watts. You need a resistor rated at 3W or more.
Bob
Hello Bob,What you need is a resistor in series with the bulb.
If the bulb draws 300mA at 1.5V we can use Ohms law to find the resistance of the bulb when it is lit. This is different than the resistance when it is not lit because, as the filament heats up, its resistance increases. So, using Ohms law:
V = I * R
R = V / I
R = 1.5 / 0.3 = 5 Ohms.
Now, we want to make a voltage divider using the bulb itself as the lower resistor so that the voltage across the bulb is 1.5V.
Using the equation of a voltage divider:
Vout = Vin * R2 / (R1 + R2)
replacing R2 with the bulb resistance, and the two voltages we get:
1.5 = 9 * 5 (R1 + 5)
1,5 * (R1 + 5) = 9 * 5
1.5 * R1 + 1.5 * 5 = 9 * 5
1.5 * R1 = 9 * 5 - 1.5 * 5
1.5 * R1 = 35
R1 = 35 / 1.5 = 25
Use a 27 Ohm resistor in series with your bulb and your bulb will light. With a fresh battery it might light for 20 minutes.
Edit: If you have successfully used LED, you should have already known how to calculate this.
The way it is usually expressed for LEDs is
R = (Vin - Vf) / If
in this case:
R = (9 - 1.5) / 0.3 = 25
Edited to fix original calculation and show alternate calculation.
Bob
@stspringer: I think it is very admirable that you are taking a "hands on" experimental approach to learning electricity and electronics. You are in the process of learning some valuable lessons about voltage dividers and practical values of the resistance associated with them. Take careful notes, and try to keep a journal or a notebook of what you measure, lest you forget what you did before and keep repeating the same thing while expecting different results. Burning up resistors is all part of the learning experience. We have all done it at one time or another. Just determine what went wrong, correct it, and move on.
Voltage dividers as used today are very seldom used to provide significant levels of power to a load, such as your 1.5 V incandescent lamp for example. The main reason is they are terribly inefficient for that purpose. Instead, most applications of voltage dividers use them as a low-level signal attenuator, the volume control in a radio for example, or to create a low-level, relatively high impedance, control signal for some other circuit, such as a user-adjustable power supply output voltage for a programmable power supply.
As you can imagine, while the basic voltage divider equation hasn't changed, the actual values selected for R1 and R2 can be all over the map while still providing the same attenuation ratio. The only thing that changes is how the voltage divider appears electrically to circuit with which it is connected.
In days of old, when incandescent lighting was first used to illuminate theater stage lighting, physically very large variable resistances (called rheostats) were used in series with the lights to control their brightness. These variable resistances were typically capable of dissipating hundreds, or even thousands, of watts of power. The entire back-stage, behind-the-scenes area, was kept toasty warm in the winter from the heat given up by immense banks (often hundreds in a Broadway production) of these rheostats. Of course the heat during a summer production was pretty much unbearable.
That all changed, practically overnight, during the middle of the twentieth century with the advent of solid-state lighting controls. Some prior efforts had used mercury-vapor thyratrons to control lighting intensity, which had the advantage of allowing separation between the control electronics and the thyratron, but it was all pretty much a kludge.
Good hunting on your electronic adventure! Make sure to nail the basics before moving further into the jungle.
If the 9V battery kept its voltage at 9V (it won't) when overloaded with the high current, then the total resistance is 27 ohms + 5 ohms= 32 ohms and the current is 9V/32 ohms= 281mA.
The heating is called Power Dissipation and the formula is (P= I squared x R) so the 27 ohms resistor heated with 2.13W.
You need a huge 5W resistor. What is the power rating of your smoking resistor?
You are killing the 9V battery with such a high current so it also got hot.
I don't use Algebra, instead I use simple Ohm's Law and simple Math Fractions.Thank you,
Can you help me by explaining Bobs post,on how he found R1 value using algebra? Post #22
I don't use Algebra, instead I use simple Ohm's Law and simple Math Fractions.
1) What is the resistance of the white-hot bulb? It is spec'd at 1.5V/0.3A and OHM'S LAW is R= V/I. Then 1.5V/0.3A= 5 ohms. .
2) What value resistor will allow the bulb to work from a very strong 9V battery? This resistor will have 9V - 1.5V= 7.5V across it and will have 0.3A in it and OHM'S LAW is R= V/I. Then 7.5V/0.3A= 25 ohms. But 25 ohms is not a standard value so use 27 ohms.
3) Heating power? P= V x I or I squared x R. Then the heating is a little less than 2.25W. A 3W resistor will get hot enough to melt nearby plastic so I would use a 5W resistor.
No, you are using algebra, you just let someone else do the work.I don't use Algebra, instead I use simple Ohm's Law and simple Math Fractions.
Oh, BORSCHT !I don't use Algebra.
Yes, a smaller resistance and smaller physical size. But the high current of 0.3A will quickly kill a little 9V battery, you will see the lights dim more and more as they kill the battery..In my circuit , if I put 3 of my 1.5 volt bulbs in series, I could then use a smaller resistor right?
Please advise
Thanks