The voltage of the lights doesn't matter, the current draw does. Find the current draw of your circuit. Look up the resistance of the wire that you are using, either in the manufacturer's catalog or NEC chapter 9. Remember to figure twice the wiring run length, since you have two wires running out to the lights. The formula: E=I*R E(voltage) equals I(current) times R(wiring resistance) Example: 100 feet from breaker to end of lights, 12A total current draw, 12 AWG copper wire Resistance (from 2005 NEC chapter 9 table 8) = 2.01 ohms per 1000 feet. Total wiring resistance = (200feet / 1000feet) * 2.01ohms per 1Kfeet = 0.402 ohms E=I*R E=12A*0.402ohms E=4.824V The voltage drop to our lightswould be about 4.8V. They would see about 272V.
to get the V.D for street lighting the following EQ. should be used
Vd=(I*K/10*VL){N*X+X1[(N^2-N)/2]}
I=CURRENT PER ONE LUMINARIE
VL=LINE VOLTAGE
N=NO. OF STREET LIGHTING POLE POINTS
X=SPACE FROM THE POWER SUPPLY TO THE NEAREST POLE POINT TO THAT SUPPLY
X1=SPACING BETWEEN POLE POINTS
K=HEIGHT OF POLE LIGHT
it depends on what type of load. Motor amperage will drop off as voltage rises. loads such as lights will increase amperage with voltage rise.
As the resistance in the wire increases due to the longer length the voltage drop across the wire resistance increases. This leaves less voltage across the load. To overcome this voltage drop usually a larger size wire which has less resistance is used. A safe nominal figure for voltage drop is to keep it at 3% of the line voltage.
The voltage drop in a line can be decreased by
The effect of diode voltage drop as the output voltage is that the input voltage will not be totally transferred to the output because power loss in the diode . The output voltage will then be given by: vout=(vin)-(the diode voltage drop).
voltage drop deviding accure
it depends on what type of load. Motor amperage will drop off as voltage rises. loads such as lights will increase amperage with voltage rise.
As the resistance in the wire increases due to the longer length the voltage drop across the wire resistance increases. This leaves less voltage across the load. To overcome this voltage drop usually a larger size wire which has less resistance is used. A safe nominal figure for voltage drop is to keep it at 3% of the line voltage.
the voltage drop means whenever the conductor passing through the supply voltage, according to the resistivity property to reduces the some amount of voltage that drop is known as voltage drop for example the resistance is used to drop the voltage to the circuit.............................................
In all probability you have a bad battery or a loose connection. If the voltage drops with the motor running, you probably have a bad alternator (or generator) or regulator.
Voltage drop is caused by circuit resistance
The voltage drop in a line can be decreased by
The VOLTAGE in a series circuits will be dropped in to the connected lights.So the light after no1 will get much lowyer than the first light.Because voltage drop in first light will cause that problem of light to lighter. But in prallel connection the voltages are divided in to equal levels.So no voltage drops in prallel connection, all lights get full voltage as applied in input mode.SO THE LIGHTS BE LIGHTER IN SERIES CONNECTION.
The effect of diode voltage drop as the output voltage is that the input voltage will not be totally transferred to the output because power loss in the diode . The output voltage will then be given by: vout=(vin)-(the diode voltage drop).
because the voltage likes to drop
The voltage drop in a wire has nothing to do with the insulation. Voltage drop has to do with the cross sectional area of the wire.
Voltage is the potential difference between the source & any point in the circuit. The forward voltage is the voltage drop across the diode if the voltage at the anode is more positive than the voltage at the cathode (if you connect + to the anode). Voltage drop means, amount of voltage by which voltage across load resistor is less then the source voltage.
It will decrease the voltage drop.