Power (watts) = current (amperes) * voltage (volts)
Current (amperes) = voltage (volts)/resistance (ohms)
120 watts = current * 120 volts
current = 1 ampere
1 ampere = 120 volts/resistance
resistance = 120 ohms
The formula you are looking for is R = E squared/W
The formulas you are looking for is I = E/R.
You can measure the current and power of a 'power supply', using an ammeter and a wattmeter. With the power supply connected to its load, the ammeter must be connected in series with the power supply's input. The wattmeter's current coil must also be connected in series with the power supply's input, and its voltage coil must be connected in parallel with the supply, taking the instrument's polarity markings into account.
Well, first of all, if the resistance of the circuit is 10 ohms and you connect 10 volts to it,then the current is 1 Amp, not 2 . So either there's something else in your circuit thatyou're not telling us about, or else the circuit simply doesn't exist.-- If you connect some voltage to some resistance, then the resistance heats up anddissipates (voltage)2/resistancewatts of power, and the power supply has to supply it.-- If there is some current flowing through some resistance, then the resistance heats up anddissipates (current)2 x (resistance)watts of power, and the power supply has to supply it.-- If there's a circuit with some voltage connected to it and some current flowingthrough it, then the resistance of the circuit is (voltage)/(current) ohms, the partsin the circuit heat up and dissipate (voltage) x (current) watts of power, andthe power supply has to supply it.There's no such thing as "the power of a circuit". The power supply supplies thecircuit with some amount of power, the circuit either dissipates or radiates someamount of power, and the two amounts are equal.
In an uninterrupted power supply there has to be a source of power when the electrical utility is not there. The only supply that is available is a battery that is in or connected to the UPS.
-- a power supply -- a load -- low-resistance material to connect the load to the power supply
We use the 250 ohms with the power supply because the internal resistance of a DC power supply is insufficient to develop a resistance.
ANSWERS; Transparent
it determines how well the current flows through the wires. ANSWER: When there is no outside power connected to it. But some power is necessary to read the resistance so the meter battery will supply the current necessary to measure the IR drop and translate that to resistance
The formulas you are looking for is I = E/R.
It depends on the resistance of everything connected between the terminals of the power supply. If the resistance is infinite or very high, there is little or no current. As the resistance becomes less, the current becomes greater. In general, the current through a circuit with 90 volts applied to it is [ 90 / R ], where 'R' is the resistance of everything across the 90-volt power supply.
You can measure the current and power of a 'power supply', using an ammeter and a wattmeter. With the power supply connected to its load, the ammeter must be connected in series with the power supply's input. The wattmeter's current coil must also be connected in series with the power supply's input, and its voltage coil must be connected in parallel with the supply, taking the instrument's polarity markings into account.
power factor meters are connected across the supply
Well, first of all, if the resistance of the circuit is 10 ohms and you connect 10 volts to it,then the current is 1 Amp, not 2 . So either there's something else in your circuit thatyou're not telling us about, or else the circuit simply doesn't exist.-- If you connect some voltage to some resistance, then the resistance heats up anddissipates (voltage)2/resistancewatts of power, and the power supply has to supply it.-- If there is some current flowing through some resistance, then the resistance heats up anddissipates (current)2 x (resistance)watts of power, and the power supply has to supply it.-- If there's a circuit with some voltage connected to it and some current flowingthrough it, then the resistance of the circuit is (voltage)/(current) ohms, the partsin the circuit heat up and dissipate (voltage) x (current) watts of power, andthe power supply has to supply it.There's no such thing as "the power of a circuit". The power supply supplies thecircuit with some amount of power, the circuit either dissipates or radiates someamount of power, and the two amounts are equal.
They are resistance connected in parallel with high voltage power supply for the purpose of discharging the energy stored in filter capacitance when the equipment is turned off.
The length ,thickness, and alloy of the filament determines its resistance. The lower the resistance, the lower the voltage required to power it.
The circuit voltage or the resistance of the individual bulb is needed to answer this question. Divide the total power (400 W) by the supply voltage.
Power supply units are rated based on their output and efficiency. When more equipment is connected, a higher output power supply is needed.