To answer this question a voltage must be stated.
Current (amps)=Watts/Volts =2000/120 =16.75 =16.75 amps
The electric heater is basically a resistor, designed to have the right resistance to draw the required current. So a 2 kW heater designed for a 230 v supply is really a resistor of 28.8 ohms, so when it's connected the current is 8 amps and the power is 2 kW.
The star configuration has less current draw across the line when starting.
4.12A. V=IR
Divide 25,000 by the supply voltage to get the current.
The formula you are looking for is I = W/E.
Assuming it is also rated for 120V., yes. The wattage doesn't change with an increase or decrease in voltage. However, the current draw does. When you double the voltage a load is hooked up to, the Amperage draw (current) drops in half. Example: if a 240 volt heater draws 6 amps, it will draw 12 amps if connected to 120V. If a 120V heater draws 15 amps, it will draw only 7.5 amps when connected to 240V. But power, or wattage stays the same, regardless. And this is what is used to calculate energy usage and therefore, cost. Please note the above answer says "if it is also rated for"
In a high voltage installation, with the same power, current drawn is small compared to those in low voltage. However it requires a thicker insulation, thus cost of insulation is significant compared to the conductor cost. By using a star connected winding, the insulation voltage required is equal to line to neutral connection, otherwise if it is connected in delta the insulation rating has to be provided for a line to line connection. Despite having a higher current, the total cost is still lower compared to using a higher insulation rating.
A ammeter will tell you how much current draw the load is drawing
In star each phase is 120 degree apart in delta angle between phases(R,Y,B) are 60 degree so star connection draw low current.
The line current would be the same if the motor were connected in delta. The current can be based on the rule of thumb which says 7 amps must be allowed for a 1-HP single-phase motor on 240 v. A 2.2 kW motor is three times as powerful, and on a three-phase supply of the same voltage (240/415) it would draw 7 amps.
24 amps. This could be three 11.54 resistors connected in star, or three 34.6 ohm resistors connected in delta.