How do I step down 5V to exactly 3.7V with resistors?
Nothing in the analog world is ever “exact”. The concept for getting the voltage you want is simple, but there are a number of details. Conceptually, put a 3.7 ohm & a 5–3.7 = 1.3 ohm resistor in series, with the 1.3 ohm resistor connected to the +5 V supply, and the 3.7 ohm resistor connected to ground (the - connection on the 5V supply). That will get you 3.7 V across the 3.7 ohm resistor.But:How much current can the source supply? This example requires at least 1 amp. If you make the resistors 1.3 kohm and 3.7 kohm, now the current is only 1 mA.With 1 A of current, the resistors will dissipate a lot of heat—3.7 W for the 3.7 ohm resistor, and 1.3 W for the 1.3 ohm resistor. Standard resistors can tolerate 1/4 watt, and it’s better to not get near the power rating. There are resistors with higher power ratings, but they are significantly more expensive.What tolerance of resistors do you need, and what’s available. Note that 1.3 and 3.7 ohms aren’t standard sizes. Are you willing to make your own resistors, or build a series/parallel network to get the right values? Or search through standard sizes to get something with the right ratio?What current does the load require? Is it constant? Do you want to correct for this current in the resistor calculation to get closer to 3.7 V? If the load isn’t constant, how much are you willing to let the voltage vary? Higher values of resistor increase the source resistance of the 3.7 V, which means the voltage will vary more with different load currents.Instead of 2 resistors, you could use a potentiometer, and adjust the output voltage to be very close to the desired 3.7 V.Why not just use a power supply chip? That gets around all of these problems, and they don’t cost very much. The resistor solution is cheap, and can make sense in some cases—like when there is very low load current, and some tolerance in the exact voltage.
How do I write answers in the CBSE class 12th physics and chemistry board exams so as to get the maximum marks?
Well I have been through this phase and studying physics was a tough job for me back then , but I did improved myself by the time I reached university for M.Sc . Obviously it took a lot of effort on my part as there is no substitute for hard work.So here is what you can do to improve your scoresLook for your complete syllabus and refer to the updated syllabus for your class from the CBSE website just so as to ensure that you have not missed out on any topic.Study with your NCERT textbook. Being completely familiar with all question types from the NCERT text book is good way to study while preparing for your exams, particularly Board Exams.Solve previous years’ Board Papers. Lay your hands on Board Papers from the previous years and practice. Do make sure that the publisher is authenticUnderstand the marking scheme followed by CBSE. Your familiarity with the marking scheme will help you work along by keeping in mind the marks allotted for the various steps in the correct answer.Avoid last minute cramming. It is important to understand your concepts well if you really want to score high in your exams.Sleep well the day before exams. Getting a good night’s sleep will only help you remember all that you have studied and will help you focus betterWhile writing the exam, pay attention to the number of marks allotted to a question and write your answers accordingly. Just cos’ you may know a lot more than is asked, writing it all down will not fetch you extra marks!Pay particular attention to the key terms used in the question and organize your answer accordingly. For example if the question has ‘Differentiate’ your answer should look different from the answer when the question is ‘Explain’. Simple stuff, yet critical to get those marks!I am not saying that this is the proven method or something you need to find what works best for you. Hope it helped.
What is the difference between a rheostat and a variable resistor? I am sure that there’s a difference between the two. The internet said the rheostat uses 3 terminals, while a variable resistor uses two.
In simple words Both are variable resistor. The output of Rheostat is Resistance. The output of Potentiometer is Voltage. Now let see a detailed explanation. First let us consider a resistor with two terminals A & B. Divide the register into 2 parts. Now there are two resistors R1 and R2. The value of R1 and R2 are based on the position of C which is variable. R = R1 + R2. There are two applications for this circuit. First one is using it as resistor as show below. Which is called Rheostat. The value of R1 is based on the position of C. So value of R1 is between 0 and R. In the first application the resister R1 is used in series with the bulb. The value of R1 controls the current I, which interns control the power of the bulb. In the other case the output voltage can be used as reference voltage. The application of Potentiometer is as input to an Electronics Circuit as reference voltage. Input resistance should be much larger than R2. I hope this explanation helps you to understand the difference between Rheostat and Potentiometer. If you still need some explanation write a comment I will explain.
Can you use a potential divider for obtaining variable DC supply from a fixed ac supply?
In my opinion you are asking that if you have a set ac source can you create a variable voltage DC power supply with a voltage divider network? My answer is Yes in a way. You could take say an AC source of 30 volts AC RMS and run it through a bridge rectifier, and filter the pulsating DC to get rid of the ripple to DC quality, and then you have a DC signal. That DC signal can be ran through a voltage divider network and depending on the values of the component you could pick tap points with different voltage outputs. For example two equal value resistors would provide equal voltage drop across both resistors with each one dropping half the voltage across it. If you had 30 volts dc each resistor in series would drop 15 volts. If you change the values you could change the amount that drops across each resistor. Another method would be to use set value Zener diodes in series to create set value tap points like a 12 volt zener in series with a 9 volt zener in series with a 6 volt zener in series with a 3 volt zener. You would be able to tap voltage off from each zener that was at the voltage potential of the zeners set value. Each zener diode would have a set voltage across it. For example just run a tap wire from the leads of the 12 volt zener to get a 12 volt source. You could substitute resistors in series to use up the extra voltage if you did not need all the tap values. Another interesting method might be to start with a plain germanium signal diode at the ground and alternately stack a series of germanium and silicon diodes on up to the highest value you want to tap, and use a resistor to take up the remaining voltage. germanium diodes drops .3 volts in forward bias, and silicon diodes drops .7 volts in forward bias. In theory if you were adding a germanium and silicon diode in series their voltage drop across them would add up to 1 volt. In that way you could step up in 1 volt increments on your voltage divider network. There are many ways to get there, but you just have to be aware of current and power ratings as well as setting your voltage. I do not know how big the load you are working with is, but you may just want to use a variable voltage regulator chip like the LM317. If that has too low of current you can use it to drive a larger power transistor to handle more current. If you are working with small currents of around 1 amp. and use a good heat sink the LM317 can handle it by itself.
Why isn't my NOT logic gate working? How can I get the output through the LED?
First and Foremost check if the transistor is in working condition !! If the base is out of commission by some reason it will not work as a NOR gate.If the transistor is fine, try adding a resistor(minimum 2 k) along with your LED. My analysis would be that the LED is in forward bias initially, the resistance across LED junction is lesser than the resistance across the transistor when it is Conducting(ie in Saturation region). So the current favours that path to the ground rather than through the transistor even when the Base current is provided to put the Transistor in Saturation, so it remains ON the whole time. You should put a resistor with the LED.