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
When connected to a 110-volt supply, the 60-watt 220-volt lamp will consume power that is calculated using the formula P = V^2 / R, where P is power, V is voltage, and R is resistance. Since the resistance of the lamp remains constant, the power consumption would be (110^2 / 220) = 55 watts. Thus, the lamp would consume 55 watts of power when connected across a 110-volt 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.
The formulas you are looking for is I = E/R.
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.
If three equal resistors are connected in parallel, the equivalent resistance will be one-third of the resistance in series. This lower resistance will result in a higher current flowing through the resistors when connected in parallel compared to when they are in series. Therefore, the power dissipated by the resistors in parallel will be greater than 10W.
When connected to a 110-volt supply, the 60-watt 220-volt lamp will consume power that is calculated using the formula P = V^2 / R, where P is power, V is voltage, and R is resistance. Since the resistance of the lamp remains constant, the power consumption would be (110^2 / 220) = 55 watts. Thus, the lamp would consume 55 watts of power when connected across a 110-volt supply.
ANSWERS; Transparent
We use the 250 ohms with the power supply because the internal resistance of a DC power supply is insufficient to develop a resistance.
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
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.
The formulas you are looking for is I = E/R.
power factor meters are connected across the supply
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.
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.
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.
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.