using a step-up transformer. ****warning***** will stress a power supply
The question doesn't make sense. Watts are the product of volts and amps so you could have 1 V with a current of 1 amp = 1 watt or 10V and a current of 0.1 amps = 1 watt or 100V and a current of 0.01 amps = 1 watt etc.
The GCF is 2.
The Amazing Grace - 2006 is rated/received certificates of: South Africa:10V
10v In series, just add them together.
Cape of Good Hope - 2004 is rated/received certificates of: USA:PG-13
If we apply Ohm's law, which is E = I x R and we have a voltage (E) of 110 volts and a current (I) of 10 amps, we can use the variation of the formula to solve. That variation is R = E / I and the resistance (R) is discovered by dividing the voltage by the current. R = E / I = 110 / 10 = 11 ohms
use kvl or kcl
It is: -10v+6v = -4v
The question doesn't make sense. Watts are the product of volts and amps so you could have 1 V with a current of 1 amp = 1 watt or 10V and a current of 0.1 amps = 1 watt or 100V and a current of 0.01 amps = 1 watt etc.
100v2 - 220v + 121 = (10v - 11)2
Voltage = Current * Resistance Current = Voltage / Resistance 2.5ma = 10v / 4K
you should specify: - circuit topology, I assume a series connection. - diode allows current flow? It depends how it's connected - diode forward voltage drop value if diode is in forward conduction, you have VR=10V - VDIODE and, thus, I = VR/R=(10-Vdiode)/1200.
A: Since you know the current flow you need the voltage drop for this particular led at 20ma. All LED require a current and voltage to operate properly assuming a voltage drop of 2v then for a 12v source it becomes 12-2=10v 10v/.02=500 ohms in series to limit the current is required
A zener diode is designed to be operated in reverse bias, and have a specified breakdown voltage. For example, if you wanted a part that is supposed to get 10V applied to it, and you only have a 12V rail, you can put a resistor and a 10V zener diode in series, and the zener diode will make sure that only 10V get dropped across it. Any more and it will begin conducting. It will draw enough current to drop the excess voltage across the series resistor. If the load on the output terminals increases, tending to reduce the output voltage, the Zener just takes less current. The idea is that the total current through the series resistor is enough to stabilise the output at 10V. ANSWER: A zener is a diode that has the property that when reversed bias is applied to exibit a breakdown of a voltage with a sharp knee. This property can be used as a regulator since the voltage will not change it the current trough the zener changes.
The rms of 10V is 6.02V. Take the peak voltage of the sine wave and multiply it by 0.707.
Ohms are the unit of resistance you find in Ohms LAw which says Volts = Amps x Ohms. You can get a voltage drop across a resistance, but would have to know what current is being used and you would have a potentiometer in effect. You are not "converting 12V" to 10V, your are essentially loosing two volts through a resistor.
Direct Conversion: X.VIII.MMVIIM=1000X=10V=5I=1Improved Answer:-10-8-2008 = X-VIII-MMVIII in Roman numerals