A vacuum cleaner has a current of 1.5A, and uses 5000J of energy in 2 minutes. What is the resistance of the vacuum cleaner?
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Energy consumed for a resistive device = I²Rt
I = current = 1.5A
R = resistance = ?
t = 2 minutes = 120 seconds
5000 = (1.5)² * R * 120
R = 5000 / [120 * (1.5)²] = 18.52 Ohms
I = current = 1.5A
R = resistance = ?
t = 2 minutes = 120 seconds
5000 = (1.5)² * R * 120
R = 5000 / [120 * (1.5)²] = 18.52 Ohms
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2 minutes = 120 seconds.
5000 J in 120 s is a rate of energy use of 5000 / 120 = 41.67 J/s.
A J/s is a watt. {41.67 W seems a bit modest for a vacuum cleaner! Are you sure of th figures?}
Anyway Power = i² * r
So
r = P/i² = 41.67 / 1.5² = 18.5 ohms
5000 J in 120 s is a rate of energy use of 5000 / 120 = 41.67 J/s.
A J/s is a watt. {41.67 W seems a bit modest for a vacuum cleaner! Are you sure of th figures?}
Anyway Power = i² * r
So
r = P/i² = 41.67 / 1.5² = 18.5 ohms