The label on a ceiling lighting fixture warns you to use a lightbulb of 60 watts or less. The voltage to the lightbulb is 120 volts. An intern calculated how much amperage a bulb of the maximum allowed wattage will draw. You are checking her work, shown below. If there is an error, what is the first step that has an error, and why is it an error?
Answers
Answer:
The bulb will draw 0.5 amperes of current when the power is maximum.
Step-by-step explanation:
Actual Calculation :
Given: Voltage, V = 120 volts
Maximum power, P = 60 watt
To find: Current, I = ?
We know, Power = Voltage * Current
Substituting the given values in the above formula,
60 = 120 * I
I = 60/120 = 0.5 Amp
Hence, the bulb will draw 0.5 amperes of current when the power is maximum.
Secondary School Math 5+3 pts
The label on a ceiling lighting fixture warns you to use a lightbulb of 60 watts or less. The voltage to the lightbulb is 120 volts. An intern calculated how much amperage a bulb of the maximum allowed wattage will draw. You are checking her work, shown below. If there is an error, what is the first step
Answer:
The bulb will draw 0.5 amperes of current when the power is maximum.
Step-by-step explanation:
Actual Calculation :
Given: Voltage, V = 120 volts
Maximum power, P = 60 watt
To find: Current, I = ?
We know, Power = Voltage * Current
Substituting the given values in the above formula,
60 = 120 * I
I = 60/120 = 0.5 Amp
Hence, the bulb will draw 0.5 amperes of current when the power is maximum.