If I0 = V/R, then Inew = (2*V)/(.5*R) = (2 / .5) * (V/R) = 4 *V/R = 4 * I0
Ohms law, V = IR
R = resistance, this will stay the same unless the resistor is changed.
If the voltage is doubled, the current is doubled.
Ohm's law states that the voltage across a resistor is the product of the current times the Resistance or V=I x R (I times R). V is Voltage, R is Resistance, and I is Current or Amperage. So if the Voltage is doubled and Resistance stays the same, the Current will be doubled.
leakage resistance of a resister is type of fault which occurs at a specific voltage across the resister which are undertesting.
The current should be high enough to maintain the voltage at each division. Generally, the current should be ten times the load current or the voltage will be across the voltage divider. If possible use regulators or zeners or regular diodes. I'm not saying dividers are bad but there are less current sensitive solutions.
The reason an AC voltage applied across a load resistance produces alternating current is because when you have AC voltage you have to have AC current. If DC voltage is applied, DC current is produced.
voltage across inductor create a flux. because of variation current developes an opposite emf.
Ohm's law states that the voltage across a resistor is the product of the current times the Resistance or V=I x R (I times R). V is Voltage, R is Resistance, and I is Current or Amperage. So if the Voltage is doubled and Resistance stays the same, the Current will be doubled.
it will cause a Short Circuit
Nothing. But the current is halved.
By Ohm's Law, current is voltage divided by resistance, so if you double both the voltage and the resistance, the current would remain the same.
leakage resistance of a resister is type of fault which occurs at a specific voltage across the resister which are undertesting.
The current should be high enough to maintain the voltage at each division. Generally, the current should be ten times the load current or the voltage will be across the voltage divider. If possible use regulators or zeners or regular diodes. I'm not saying dividers are bad but there are less current sensitive solutions.
The capital Greek letter Ω (omega) is the symbol for Ohms - the unit of resistance. A 1.5Ω resister is a resistor with quite a low resistance. If a voltage of 1.5 Volts is applied across it, a current of 1 Amp[ere]s will flow through it.
It may be better to say that a resistor allows current flow through itself rather than to say that a resistor is a device that will "use" current. It does "resist" current flow, and thus limits it to some degree depending on its resistance. (More resistance means more limiting of current flow.) The resistor "drops voltage" as well limits current. A resistor "feels voltage" from some source, and the voltage it "feels" is said to be the "voltage drop" of the resistor. The voltage drop is the voltage that could be measured across that resistor with a meter.
Load resistors are connected across the circuit to limit the current flowing through the load.
Resistance = voltage / current = 12V / 2.5mA = 12V / (2.5 x 10-3 A) = 4.8 x 103 ohm
To measure the value of a resistor, apply a voltage and measure the voltage across the resistor and the current through the resistor. Use Ohm's law: Resistance equals Voltage divided by Current. Start with a small voltage and increase gradually until a reading is obtained, but be careful that the power dissipation (watts = volts times amperes) of the resistor is not exceeded. Simpler solution: Use an ohmeter.
The reason an AC voltage applied across a load resistance produces alternating current is because when you have AC voltage you have to have AC current. If DC voltage is applied, DC current is produced.