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For a d.c. circuit, you divide its supply voltage by the resistance of its load. For an a.c. circuit, you divide its supply voltage by the impedance of its load.
That would depend on the power supply.
The purpose of the battery in a circuit is to wive energy to the circuit
A device or circuit that provides power to the rest of the circuit or system is called a power supply,
it is a radio frequency choke which connects the dc supply to the circuit but isolate the dc supply from the high frequency oscillations generated in the feedback circuit
Ring main
-- a power supply -- a load -- low-resistance material to connect the load to the power supply
You would not want to connect Vcc and Gnd together. That would short circuit the power supply.
First you connect a trnsformer to the main supply and then a rectfier .youmay connect a zener diode or a resistor or a capacitor as a filter circuit
pressure coil or voltage coil across the supply and the current coil in series.
To answer this question the wattage of the flood lights must be stated.
A D cell battery supplies nominally 1.5V. Connecting a 6V supply in it's place would supply an extra 4.5V and could potentially damage the circuit components.
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 the short circuit test we applied supply voltage on L.V side and short circuit the H.V side and connect the ammeter in H.V circuit to measure the short circuit current. with the help of s.c test we measure the copper losses in the transformers.
The first thing is to see if there is a spare N.O. set of contacts on the alarm relay. If there is a spare set of contacts connect your horn supply to one side of the N.O. contact terminals. Connect the horn to the other side of the N.O. contact terminals. Energize the horn circuit's supply. Now when the alarm circuit relay activates the horn will sound.
ac supply is given and then in the circuit rectifier converts ac to dc
For a d.c. circuit, you divide its supply voltage by the resistance of its load. For an a.c. circuit, you divide its supply voltage by the impedance of its load.