how to check using a voltmeter if the output current is positive? how does it flow?
107334
This question needs a little work. By convention, current flows from + to - although in real life, current is electrons which go from - to +. But only physics types do it that way. Electrical engineers say + -> -.\n\nNow, that being said, if you have a non-zero resistance across a voltage source, you can find the current flow by measuing the voltage across the load (resistance). The lower  the resistance, the lower your voltmeter has to read. The sign of the voltage would indicate the current flow. \n\nSo say you have a 10V battery and two 5 ohm resistors in series across it. The total current will be 1A (ohm's law). Now if I measure the voltage across one resistor, I should read 5V (kirchoff's law by the way) and the sign (+ or - 5V) will tell me the current direction. Using conventional + to - flow, if the voltage is + the current is flowing from the red lead to the black lead.
326286
Positive cable  = + = red cable. look for a plus + sine on the battery.
399258
just put a limiting resistor. 60mA and 12V. Limiting resistor =R=(V-Vled)/I. assuming the extreme, 0V for Vled. R=12/6e-2=200ohm. that way the current will never exceed 60mA.
99663
cathodes are positive
185952
a diode blocks current from flowing in a certain direction. current will flow from the non striped side to the striped side
1186419
That would depend on the voltage. Current = Watts / Volts\nA couple of examples:\n110V - 90.9 Amps\n220V - 45.45 Amps\n240V - 41.67 Amps\n440V - 22.7 Amps\n10,000V - 1 Amp
1095335
As said above but to expand a little a voltage is a potential difference between two points so when the resistance is right and a closed circuit is created the current flows allowing the voltage (potential difference) to be measured.\n\nSee http://en.wikipedia.org/wiki/Voltage\n\nEnjoy, \n\nSi.
964185
I sense that you have no clue what you're doing. Tell me I'm wrong.\n\nFor DC circuits, Ohm's law says: V = I * R, voltage equals current times resistance.\n\nYour circuit will consist of a battery (voltage source) and a load (resistance) plus the wires to connect them and the meter(s).\n\nIdeally you will test one battery at a time. That way the load will be identical (including the small resistance of the wires).\n\nBatteries typically have two ratings: voltage and capacity.\nVoltage is measured in volts and this decreases as the battery discharges. Capacity is measured in amp hours (current multiplied by time) which specifies how long the battery can deliver a current.\n\nAssuming the circuit resistance is constant (reasonable assumption for your purposes), when you connect the battery to the load, the voltage (independent variable) will cause a current (dependent variable) to flow. As the battery discharges, the voltage will decrease, and the current will correspondingly decrease. You will need a watch or timer as well. Time zero will be when you first connect the battery. Measure the battery voltage at that instant and at intervals. Depending on the load and how fast the battery discharges should determine how frequently you take measurements.\n\nmsg me when you have the data if you need help analyzing it.
774566
There are a couple of ways that I can think of right off hand.  They depend on what the surrounding circuitry is like.  \n \nRemember, a static discharge is a fairly high voltage arc.  So one possibility is to measure the resistance at the transistor and compare it to a known good board.  The low voltage used to measure the resistance shouldn't hurt anything, but like I pointed out, this depends on the circuitry.  One company I am familiar with used to do this to trouble shoot a circuit.  They had a whole list of resistance readings for various points in a circuit.  It was quite helpful.  \n  \nYou might also power the board and use a scope to measure wave forms.  Or even a DMM to measure voltages.  \n  \nAnother way is to use a curve tracer.  Again, depending on the circuitry.
1169017
A voltmeter and an ammeter will provide the input power. You will need some form of dynamometer to measure output power.\n\nOutput power divided by input power yields efficiency.
266980
4. The current does not change.  I=E/R so multiplying the numerator and denominator by 0.50 results in the same answer, does it not?
1121007
yes. electricity flows one direction,comes to a stop, then flows in the  other direction. the alternator pushes the electrons in one direction for 1/120 of a sec.then the other direction for the same amt. of time.( for 60 cycle current ) one direction is the pos.1/2 cycle the other is the neg. 1/2 cycle.  the zero crossing point is the instant between one direction and the next, when current flow is stopped.
240898
This is stuff I have not worked with in 20 years. LOL. I can not draw this for you here, you can figure it maybe from what I tell you. \n(1) You want a circuit that has say 10V steered to four outputs depending on the control inputs. We will say the control inputs are 0 & 5 volts we are steering 10 volts. You are looking at a circuit with 7 lines. 1 input, 2 control and 4 outputs. The voltage to your controls has 4 possibilites.\n0&0 control volts input 10 volts to output line #1\n0&5 control volts input 10 volts to output line #2\n5&0 control volts input 10 volts to output line #3\n5&5 control volts input 10 volts to output line #4\nThat about all I can say here. May use AND, OR, NOR Gates??\n(2) Is even more complicated and I can not even begin to tell this in words. Hope this helps, I doubt it will, but, I tried.
86126
Suppose you are using an instrument to measure say voltage. The instrument is a voltmeter. Now, you need to be sure that if you apply a 1 volt input, you get a 1 volt output. IF your instrument does not give you an output of 1 Volt, then you need to 'calibrate' it. In other words, you must adjust certain settings of the instrument so that 1 volt input gives you 1 volt output. The same also applies to other kinds of measuring instruments
971534
Well, u can assume that on every section of wire there is an amount of current flowing say i1,i2.... etc. and give any direction to them.\nKirchoff has two laws:\n1st law(current law) : At any junction the incoming current is equal to the outgoing current. So at any junction u can add all the incoming currents and equate it to the addition of all outgoing currents.\n2nd law (voltage law) : it states that the total voltage in a closed loop or mesh is zero. so if u consider first loop on left and the direction of current in both sections of wire clockwise. Then u can write -i1R1-i1R2+Ub1=0. if u r tracing the loop in the direction of current then voltage on resistance is negative(-iR) or u can take the vice versa it doesn't matter.For voltage source if the current is moving from positive to negative plate then voltage is termed as positive and the other way negative.\n\nSo form that many number of equation, that suffice u in finding the unknowns. And then by applying simple mathematics find the current in each segment and then the voltage across each resistor. U need to consider here the concept of voltage drop too.
813641
using KCL (current law), A is corrent.\nusing KVL (voltage law), B is correct\nC doesn't make any sense, and you can support that with the definition of the KCL.
1054541
It is a capacitor tester.  It charges the capacitor and then measures the current that leaks through it.