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16. A DC electric motor transforms 1.50 kW of electrical power into mechanical form. If the
motor's operating voltage is 300 volts,
how much current does it "draw" when operating at full
load (full power output)?



Answer :

GinnyAnswer
To find the current that a DC motor "draws" when operating at full load (full power output), we can use the power equation for DC circuits. The formula that relates power, voltage, and current is given by:

[tex]\[ P = V \times I \][/tex]

Where:
- [tex]\( P \)[/tex] is the power (in watts, W)
- [tex]\( V \)[/tex] is the voltage (in volts, V)
- [tex]\( I \)[/tex] is the current (in amperes, A)

Here are the steps to solve for the current:

1. Identify the given values:
- Power ([tex]\( P \)[/tex]) is given as 1.50 kilowatts (kW)
- Operating voltage ([tex]\( V \)[/tex]) is given as 300 volts (V)

2. Convert the power from kilowatts to watts:
[tex]\[ 1.50 \, \text{kW} = 1.50 \times 1000 \, \text{W} = 1500 \, \text{W} \][/tex]

3. Rearrange the power equation to solve for current ([tex]\( I \)[/tex]):
[tex]\[ I = \frac{P}{V} \][/tex]

4. Substitute the known values into the equation:
[tex]\[ I = \frac{1500 \, \text{W}}{300 \, \text{V}} \][/tex]

5. Perform the division to find the current:
[tex]\[ I = \frac{1500}{300} \, \text{A} = 5 \, \text{A} \][/tex]

So, the current drawn by the motor when operating at full load (full power output) is [tex]\( 5 \)[/tex] amperes (A).

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