Amperes and volts, respectively.
A) Kilogram and D) Gram are both units used to measure mass in the metric system. The kilogram is the base unit, and the gram is a smaller unit of mass. The other options, Anagram and Pound, are not units used in the metric system for measuring mass.
Compound AB would have a total of 9 mass units, calculated by adding the units of element A (4 units) and element B (5 units).
Weight can be measured in metric units also. The most commonly used metric weight unit would probably be kilograms. (1 kilogram = 2.20462262 pounds)CommentA kilogram measures MASS, not weight!!
Measuring cylinder graduation may be marked in milliliters and fluid ounces. Pints are not a common unit used for measuring cylinder graduations.
the power consumption of device is constant (provided no any condition is changed) at low voltage the power draw by motor is same at rated voltage bu according to formula the current is rise .
A diode has a fixed voltage drop (c), and if the voltage is less than that, no (or very little) current flows. Once the voltage increases above the voltage drop, the current increases exponentially as the voltage increases. A function that models this is: I = exp(b*(V-c)) where V is the voltage, I is the current, c is the voltage drop and b is a parameter that determines how fast the current increases; the value of b varies from one diode to another
magnetic core balance test is used to find out the flux distribution. It can conduct either in HV side or LV side. In HV side, applying voltage between 2 phases and measuring the voltage across other phase.(eg.applying b/w U&V, and voltage b/w V&W , W&U measuring.) In LV side, applying voltage b/w any phase....and measuring voltage across other phases. Result: sum of measured voltage will be equal to applied voltage.
while measuring for voltage.
Compute the open load voltage of the current source across its shunt resistance.This voltage becomes the voltage source's voltage.Move the current source's shunt resistance to the voltage source's series resistance.Insert the new voltage source into the original circuit in place of the current source.
I assume a series of amplifiers with one F/B The F/B is the portion of voltage/current F/B from input to output no matter how many amplifiers are in series within the loop or what the gain of each one is.
A triangle with side a: 13, side b: 14, and side c: 15 units has an area of 84 square units.
A normal diode only allows current to flow in one direction, e.g: A to B. However, current will only flow across this diode A to B when there is a voltage ( called the breakdown voltage), and different normal diodes require different voltages to allow current to flow, e.g: silicon diodes allow current to flow A to B when there is a minimum of 0.6V flowing, at which point the resistance drops, and current flows. A zener diode does exactly this, however, it also allows current to flow in an opposite direction once the breakdown voltage for this direction has been reached. e.g: if a Zener diode has a breakdown voltage of 3.6V, at 3.6V its resistance will drop significantly enough so that current can flow. hopes this helps
ANSWER: one way tie a zener x volts from B+ to ground with a resistor. that gives you a reference voltage using an PNP tie the base to the reference voltage collector goes to the load and the emitter is tied to +B trough a resistor Since the emitter follows the base then the current should be +B volts -zener voltage - Vbe or .6 v. that is it.
Both Technician A and Technician B are correct in their statements. Technician A is right that a voltmeter can monitor the action of a switch by measuring the voltage across it when the switch is in operation. Technician B is also correct, as measuring the resistance across the switch can verify its continuity when it is in the closed position.
For any value turn the multimeter to the highest scale in the value that you are measuring. By doing this you will not overload the meter by using a lower scale to measure a higher value.
Both technicians are right, and both technicians are wrong, because not enough information is present in the question, nor in their statements. Given constant impedance, current should decrease as voltage decreases, while given constant power, current should increase as voltage decreases.
In ferromagnetic materials,such as electrical steel,the magnetisation curve is not a straight line,or else the relation B=μH is not linear ( μ is not constant),H is linear function of magnetization current,while B is linear function of input voltage ,so we have : Im=f(μ)U,where f(μ) is not constant,Im=magnetization (magnetizing) current, U=input voltage. I hope this will help.