1. Calculate the total resistance of the series circuit by adding the individual resistances of the resistors together. Let's call this value Rt. 2. Apply Ohm's law to calculate the current through the circuit. The voltage across the circuit is equal to the voltage of the power source, and the resistance of the circuit is equal to the total resistance of the series resistors. Let's call the current through the circuit I. I = V / Rt 3. Since the resistors are in series, the same current flows through each resistor. Therefore, the current through each resistor can be calculated using Ohm's law and the voltage drop across each resistor. I1 = V1 / R1 I2 = V2 / R2 I3 = V3 / R3 . . . where V1, V2, V3, etc. are the voltage drops across each resistor. It is important to note that the voltage drops across each resistor can be calculated using the voltage divider rule, which states that the voltage across a resistor in a series circuit is proportional to its resistance. The voltage divider rule can be useful in cases where the voltage supplied to the circuit is divided among the series resistors.
True you made a voltage divider. However, unless anything you plan to use the voltages for is really very little current drain, those resistors will burn open... Take the input voltage and divide the resistor(s) value(s) into that voltage, you will see the current capability. And than the wattage of each resistor needs to be high enough to handle that current...
Dear garymucher thanks for comment Series resistors have an educational aspect and two resistors with suitable power can be used to reach the desired voltage
so, can I hook the red and black to a 5w solar panel rated at 17 volts and just tap off the resistor i want to charge my 4.3 outdoor light so i never have to remove the batteries to charge it again. is the panel going to be fine circulating that current through the resistors?
Good tip! Thank you. I'm quite new to electronics, starting to learn now, and this must be a stupid question. Sorry. I have a 230VAC to 12VAC little transformer and I need reduce it to 9VAC. Would your project work in this case (AC to AC), or it works only for DC/DC? Thanks again.
You can add a resistor in series with the load to drop the voltage. However, this method is inefficient as it dissipates power as heat. Use Ohm's Law (V = IR) to calculate the required resistance. For example, if your load draws 100mA, the resistance needed to drop 3V (12V - 9V) would be R = V/I = 3V / 0.1A = 30 ohms. You can use a voltage divider circuit with two resistors to divide the voltage. Calculating Resistor Values: The voltage divider formula is Vout = Vin * (R2 / (R1 + R2)). You can choose suitable resistor values based on this formula.
I am an absolute Zero when it comes to electronic ! So I have 7.4 V 6000mah Lipo Battery and I want to run a 6 V Motor which I don't know how many Amperes he pules ( it's a 380 brushed motor in a Ship Model ). Would the simple 100 ohm Resistor survive that or do I need something different / bigger / with cooling ??? As said I have no Idea
Resistors are not designed to handle the kind of current a motor draws, especially a motor that's pulsing. Using a resistor to power a motor is highly inefficient and can lead to overheating and damage to both the resistor and the motor. A voltage regulator is a much better solution for powering your 6V motor from a 7.4V battery. It will efficiently reduce the voltage to 6V while supplying the necessary current to the motor.
Dear seyedrauf🌹 thanks for comment❤ The use of a voltage divider with four series resistors of 100 ohms can be a good or bad choice depending on the specific application and requirements. A voltage divider is a simple and commonly used circuit that can be used to obtain a lower voltage from a higher voltage source. The voltage divider circuit using four series resistors of 100 ohms can be a good choice if the input voltage is relatively constant, the output voltage is not required to be very precise, and the load impedance is relatively high. However, there are some potential issues with using a voltage divider for certain applications. One of the main issues is that the output voltage of a voltage divider circuit may vary with changes in the load impedance. The output voltage may also be affected by changes in the input voltage, temperature, and other factors. Additionally, the voltage drop across each resistor in the voltage divider circuit can generate heat, which can be a concern in high-power applications. In summary, the use of a voltage divider with four series resistors of 100 ohms can be a good or bad choice depending on the specific application and requirements.🌹
Yes, it is possible to reduce the voltage with a series diode. However, it is not a very efficient way to do it, and it is not recommended for most applications. When a diode is forward biased, it has a voltage drop of about 0.6 volts for silicon diodes and 0.3 volts for germanium diodes. This means that if you connect a diode in series with a power supply, the voltage at the output of the diode will be reduced by the diode's forward voltage drop. For example, if you connect a silicon diode in series with a 12 volt power supply, the voltage at the output of the diode will be about 11.4 volts. If you connect two silicon diodes in series, the voltage at the output will be about 10.8 volts, and so on. The main problem with using a series diode to reduce voltage is that the diode's forward voltage drop is not constant. It varies depending on the current flowing through the diode. This means that the output voltage of the circuit will also vary depending on the current draw. Another problem with using a series diode is that it wastes power. The diode's forward voltage drop is dissipated as heat. This means that the power supply has to provide more power than the circuit actually needs. For these reasons, it is not recommended to use a series diode to reduce voltage for most applications. There are better ways to do it, such as using a voltage regulator or a resistor divider. However, there are some cases where using a series diode to reduce voltage may be acceptable. For example, it may be used in a low power circuit where the power loss is not significant.
These electrical components oppose the flow of current. By adding a resistor in series with your load (the motor in your case), you can drop the voltage across the load. Ohm's Law: This fundamental law of electricity states that V = IR, where V is voltage, I is current, and R is resistance. To calculate the required resistance, you'll need to know the current your motor draws. Let's assume your motor draws 1 Ampere at 40V.
Suppose you want to reduce the voltage (V_total) by half. In this case, you'd choose R_out to be equal to the total resistance of the two resistors connected after the tap point (assuming resistors are arranged sequentially). Therefore, R_out = 2 resistors * 100 kΩ/resistor = 200 kΩ Using the voltage divider formula: V_out = (200 kΩ / 400 kΩ) * V_total = (1/2) * V_total This will give you an output voltage (V_out) that's half of the original voltage (V_total).
Ok, I will comment since I noticed a few things and have been doing some math to fill in the gaps. However, I noticed that the resistors were not all facing the same direction for this video.
the resistance of a material itself does not have a direction. It opposes the flow of current equally regardless of which direction the current flows through the material. This is because resistance is a property of the material itself, like its length, width, or conductivity. However, when analyzing circuits, we often refer to a direction of current flow. This is a convention to simplify calculations and circuit analysis. It doesn't affect the resistance value itself.
Reducción de voltaje con resistencia Mediante 4 resistencias de 100 ohmios y voltaje de entrada de 12 voltios, crea voltajes de aproximadamente 3, 6 y 9 voltios En este video se usan 4 voltímetros simultáneamente Compare instantáneamente el voltaje de entrada con otros 3 voltajes resistencia 👉 100 ohmios *4 cable y voltaje variable de entrada
An uninteligable tangle of wires with illegible meter readout digits to anyone who doesn't yet understand what is going on I'm afraid. A simple sketch with explanation would be far better!
Dear kuku 🌹To create a voltage divider using four resistors in series, each with a resistance of 100 ohms, you can follow these steps: 1. Connect the resistors in series: Connect the four resistors end to end, so that the current flows through them one after another. 2. Identify the input and output: Determine which end of the resistor chain will be the input and which will be the output. The input will be the end where you apply the input voltage, and the output will be the end from which you measure the divided voltage. 3. Calculate the total resistance: To find the total resistance, add up the resistance values of all the resistors in series. In this case, since you have four resistors of 100 ohms each, the total resistance (R_total) will be 4 * 100 ohms = 400 ohms. 4. Apply the input voltage: Connect the input voltage source (V_in) to the input end of the resistor chain. 5. Calculate the output voltage: The output voltage (V_out) can be calculated using the voltage divider formula: V_out = (R2 / (R1 + R2)) * V_in In this case, since you have four resistors in series, you can consider the first three resistors as R1, and the last resistor as R2. So, R1 = 300 ohms and R2 = 100 ohms. The input voltage (V_in) will be the voltage you apply at the input end. V_out = (100 ohms / (300 ohms + 100 ohms)) * V_in = (100 ohms / 400 ohms) * V_in = (1/4) * V_in Therefore, the output voltage will be one-fourth (1/4) of the input voltage. 6. Measure the output voltage: Connect a voltmeter between the output end of the resistor chain and the common ground to measure the output voltage (V_out). Please note that in this voltage divider configuration, the output voltage will be one-fourth of the input voltage regardless of the magnitude of the input voltage, as long as the resistors remain the same.
Und was bringt mir dieses Video? Ich weiß wie groß die Widerstände sind, das hier ein mehrfach Spannungsteiler entstanden ist, ich weiß auch welche Eingangsspannung angelegt wurde aber ich weiß nicht welche Sröme fließen und ich weiß auch nicht an welchem Abgriff ich welchen Strom entnehmen kann. Einfach nur was zusammenlöten bringts doch nicht, da helfen auch keine braunen Hände mit abgefressenen Fingernägeln!
How is the current passing through several resistors in series calculated?
1. Calculate the total resistance of the series circuit by adding the individual resistances of the resistors together. Let's call this value Rt.
2. Apply Ohm's law to calculate the current through the circuit. The voltage across the circuit is equal to the voltage of the power source, and the resistance of the circuit is equal to the total resistance of the series resistors. Let's call the current through the circuit I.
I = V / Rt
3. Since the resistors are in series, the same current flows through each resistor. Therefore, the current through each resistor can be calculated using Ohm's law and the voltage drop across each resistor.
I1 = V1 / R1
I2 = V2 / R2
I3 = V3 / R3
.
.
.
where V1, V2, V3, etc. are the voltage drops across each resistor.
It is important to note that the voltage drops across each resistor can be calculated using the voltage divider rule, which states that the voltage across a resistor in a series circuit is proportional to its resistance. The voltage divider rule can be useful in cases where the voltage supplied to the circuit is divided among the series resistors.
Hello, you forgot to wipe the solderflux off your finger the last time....🤣🤣
😀
True you made a voltage divider. However, unless anything you plan to use the voltages for is really very little current drain, those resistors will burn open... Take the input voltage and divide the resistor(s) value(s) into that voltage, you will see the current capability. And than the wattage of each resistor needs to be high enough to handle that current...
Dear garymucher
thanks for comment
Series resistors have an educational aspect and two resistors with suitable power can be used to reach the desired voltage
so, can I hook the red and black to a 5w solar panel rated at 17 volts and just tap off the resistor i want to charge my 4.3 outdoor light so i never have to remove the batteries to charge it again. is the panel going to be fine circulating that current through the resistors?
Hi boss! Ilove ur video but as a newvie what is the wattage of 100ohms resistor u use? Thanks and more power! 👍
@@electrofixtips thank you very much! ❤
Good tip! Thank you. I'm quite new to electronics, starting to learn now, and this must be a stupid question. Sorry.
I have a 230VAC to 12VAC little transformer and I need reduce it to 9VAC. Would your project work in this case (AC to AC), or it works only for DC/DC? Thanks again.
Yeah it will work just do the math to figure out what size resistors you need.
You can add a resistor in series with the load to drop the voltage. However, this method is inefficient as it dissipates power as heat.
Use Ohm's Law (V = IR) to calculate the required resistance. For example, if your load draws 100mA, the resistance needed to drop 3V (12V - 9V) would be R = V/I = 3V / 0.1A = 30 ohms.
You can use a voltage divider circuit with two resistors to divide the voltage.
Calculating Resistor Values: The voltage divider formula is Vout = Vin * (R2 / (R1 + R2)). You can choose suitable resistor values based on this formula.
@@electrofixtips Thanks a lot.
We also can use diode. every diode can reduce 0.7 voltage
yes very good❤
please watching this video:
ua-cam.com/video/dl6QpCajRgw/v-deo.html
I am an absolute Zero when it comes to electronic ! So I have 7.4 V 6000mah Lipo Battery and I want to run a 6 V Motor which I don't know how many Amperes he pules ( it's a 380 brushed motor in a Ship Model ). Would the simple 100 ohm Resistor survive that or do I need something different / bigger / with cooling ??? As said I have no Idea
Resistors are not designed to handle the kind of current a motor draws, especially a motor that's pulsing. Using a resistor to power a motor is highly inefficient and can lead to overheating and damage to both the resistor and the motor.
A voltage regulator is a much better solution for powering your 6V motor from a 7.4V battery. It will efficiently reduce the voltage to 6V while supplying the necessary current to the motor.
@@electrofixtips thank you
Excellent
Ooo I am very glad to witness the discovery of a thing people will call a voltage divider in future 😂😂😂😂
Dear seyedrauf🌹 thanks for comment❤
The use of a voltage divider with four series resistors of 100 ohms can be a good or bad choice depending on the specific application and requirements.
A voltage divider is a simple and commonly used circuit that can be used to obtain a lower voltage from a higher voltage source. The voltage divider circuit using four series resistors of 100 ohms can be a good choice if the input voltage is relatively constant, the output voltage is not required to be very precise, and the load impedance is relatively high.
However, there are some potential issues with using a voltage divider for certain applications. One of the main issues is that the output voltage of a voltage divider circuit may vary with changes in the load impedance. The output voltage may also be affected by changes in the input voltage, temperature, and other factors. Additionally, the voltage drop across each resistor in the voltage divider circuit can generate heat, which can be a concern in high-power applications.
In summary, the use of a voltage divider with four series resistors of 100 ohms can be a good or bad choice depending on the specific application and requirements.🌹
Is it possible to reduce the voltage with a series diode?
Yes, it is possible to reduce the voltage with a series diode. However, it is not a very efficient way to do it, and it is not recommended for most applications.
When a diode is forward biased, it has a voltage drop of about 0.6 volts for silicon diodes and 0.3 volts for germanium diodes. This means that if you connect a diode in series with a power supply, the voltage at the output of the diode will be reduced by the diode's forward voltage drop.
For example, if you connect a silicon diode in series with a 12 volt power supply, the voltage at the output of the diode will be about 11.4 volts. If you connect two silicon diodes in series, the voltage at the output will be about 10.8 volts, and so on.
The main problem with using a series diode to reduce voltage is that the diode's forward voltage drop is not constant. It varies depending on the current flowing through the diode. This means that the output voltage of the circuit will also vary depending on the current draw.
Another problem with using a series diode is that it wastes power. The diode's forward voltage drop is dissipated as heat. This means that the power supply has to provide more power than the circuit actually needs.
For these reasons, it is not recommended to use a series diode to reduce voltage for most applications. There are better ways to do it, such as using a voltage regulator or a resistor divider.
However, there are some cases where using a series diode to reduce voltage may be acceptable. For example, it may be used in a low power circuit where the power loss is not significant.
So if I want to reduce the voltage from 120v to 40v how many resistors do I need?
These electrical components oppose the flow of current. By adding a resistor in series with your load (the motor in your case), you can drop the voltage across the load.
Ohm's Law: This fundamental law of electricity states that V = IR, where V is voltage, I is current, and R is resistance.
To calculate the required resistance, you'll need to know the current your motor draws. Let's assume your motor draws 1 Ampere at 40V.
I what to buy a big resistor pack do you know what I'm looking for cuz I don't
You can use 1 watt resistors🌹
So just put a 100k resistor for every .25% voltage you want rediced?
Suppose you want to reduce the voltage (V_total) by half. In this case, you'd choose R_out to be equal to the total resistance of the two resistors connected after the tap point (assuming resistors are arranged sequentially).
Therefore, R_out = 2 resistors * 100 kΩ/resistor = 200 kΩ
Using the voltage divider formula:
V_out = (200 kΩ / 400 kΩ) * V_total = (1/2) * V_total
This will give you an output voltage (V_out) that's half of the original voltage (V_total).
Good Information Thank U
Thank you so much for taking the time to share your thoughts. I really appreciate it!
Nice sharing
Para pelar los cables el alicate se usa del otro lado
Good
Thanks
Would it be possible to use this method to reduce the voltage of a 24V AC transformer down to 14V to charge a car battery?
Hi dear friend😊
Unfortunately no
Use an AC to DC converter
Como se puede utilizar?
Only able for lets or very low consumption if not that resistor will burn 😂
Ok, I will comment since I noticed a few things and have been doing some math to fill in the gaps. However, I noticed that the resistors were not all facing the same direction for this video.
the resistance of a material itself does not have a direction. It opposes the flow of current equally regardless of which direction the current flows through the material. This is because resistance is a property of the material itself, like its length, width, or conductivity.
However, when analyzing circuits, we often refer to a direction of current flow. This is a convention to simplify calculations and circuit analysis. It doesn't affect the resistance value itself.
Good job
Por favor explicalo en castellano , gracias de Chile 6:22
Reducción de voltaje con resistencia
Mediante 4 resistencias de 100 ohmios y voltaje de entrada de 12 voltios, crea voltajes de aproximadamente 3, 6 y 9 voltios
En este video se usan 4 voltímetros simultáneamente
Compare instantáneamente el voltaje de entrada con otros 3 voltajes
resistencia 👉 100 ohmios *4
cable y voltaje variable de entrada
👌 thanks
RESISTOR VOLTAGE DIVIDER WOW
Your feedback is highly appreciated. Thank you for your comment
Professional
Thanks a lot
Great
If power supply is 14 volt how to get 12.6 v
@@electrofixtips wo sahi ha values kn si ho ghi resistance ki
An uninteligable tangle of wires with illegible meter readout digits to anyone who doesn't yet understand what is going on I'm afraid.
A simple sketch with explanation would be far better!
Thanks for the feedback! I appreciate you taking the time to point that out🌹
Saw your skill in soldering but, the wiring up to the meters wasn’t well explained and needed a diagram.
Mas.... apa tahan sama Amper besar resistor nya...😅😅
Dear ahmadkos 🌹Suitable for low amps
What are the colour code
Kalo bisa lebih detai mase
Dear kuku 🌹To create a voltage divider using four resistors in series, each with a resistance of 100 ohms, you can follow these steps:
1. Connect the resistors in series: Connect the four resistors end to end, so that the current flows through them one after another.
2. Identify the input and output: Determine which end of the resistor chain will be the input and which will be the output. The input will be the end where you apply the input voltage, and the output will be the end from which you measure the divided voltage.
3. Calculate the total resistance: To find the total resistance, add up the resistance values of all the resistors in series. In this case, since you have four resistors of 100 ohms each, the total resistance (R_total) will be 4 * 100 ohms = 400 ohms.
4. Apply the input voltage: Connect the input voltage source (V_in) to the input end of the resistor chain.
5. Calculate the output voltage: The output voltage (V_out) can be calculated using the voltage divider formula:
V_out = (R2 / (R1 + R2)) * V_in
In this case, since you have four resistors in series, you can consider the first three resistors as R1, and the last resistor as R2. So, R1 = 300 ohms and R2 = 100 ohms. The input voltage (V_in) will be the voltage you apply at the input end.
V_out = (100 ohms / (300 ohms + 100 ohms)) * V_in
= (100 ohms / 400 ohms) * V_in
= (1/4) * V_in
Therefore, the output voltage will be one-fourth (1/4) of the input voltage.
6. Measure the output voltage: Connect a voltmeter between the output end of the resistor chain and the common ground to measure the output voltage (V_out).
Please note that in this voltage divider configuration, the output voltage will be one-fourth of the input voltage regardless of the magnitude of the input voltage, as long as the resistors remain the same.
Sir aap vdo me bda asani se dikha dete ho lekin fack
Jab 12v milega n ise to sab k sab jal jayega
using dioda you can so
but, the current (A) decreased
@@electrofixtips Thanks so much for your reply... I meant TOTAL current
Translate in Tamil
Und was bringt mir dieses Video? Ich weiß wie groß die Widerstände sind, das hier ein mehrfach Spannungsteiler entstanden ist, ich weiß auch welche Eingangsspannung angelegt wurde aber ich weiß nicht welche Sröme fließen und ich weiß auch nicht an welchem Abgriff ich welchen Strom entnehmen kann. Einfach nur was zusammenlöten bringts doch nicht, da helfen auch keine braunen Hände mit abgefressenen Fingernägeln!
Demasiado desorden en el vídeo,lo más interesante es lo que menos se ve, debes ser más pulcro al grabar un vídeo.
Dear diogen, thank you for your criticism🌹❤
Waste
Wasting time....
"Thanks for the feedback! I appreciate you taking the time to point that out."🌹
Bakvas😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅
Dear kasam 🌹 thanks for comment
What benefit of this
Có cái D gì hay 😮