Answer:
(a) P₁ = 1320 W
(b) P₂ = 480 W
(c) [tex]\frac{E_1}{E_2} = 0.46[/tex]
Explanation:
(a)
The power consumed by the blow dryer can be calculated as follows:
[tex]P_1 = VI[/tex]
where,
P₁ = Power Consumed by the blow dryer = ?
V = Voltage = 120 V
I = Current Rating of the blow dryer = 11 A
Therefore,
[tex]P_1 = (120\ V)(11\ A)\\[/tex]
P₁ = 1320 W
(b)
The power consumed by the blow dryer can be calculated as follows:
[tex]P_2 = VI[/tex]
where,
P₂ = Power Consumed by the vacuum cleaner = ?
V = Voltage = 120 V
I = Current Rating of the vacuum cleaner = 4 A
Therefore,
[tex]P_2 = (120\ V)(4\ A)\\[/tex]
P₂ = 480 W
(c)
[tex]\frac{E_1}{E_2} = \frac{P_1t_1}{P_2t_2}[/tex]
where,
E₁ = Energy used by the blow-dryer
E₂ = Energy used by the vacuum cleaner
t₁ = time of use of the blow-dryer = (15 min)(60 s/1 min) = 900 s
t₂ = time of use of the vacuum cleaner = (1.5 h)(3600 s/1 h) = 5400 s
Therefore,
[tex]\frac{E_1}{E_2} = \frac{(1320\ W)(900\ s)}{(480\ W)(5400\ s)}\\\\\frac{E_1}{E_2} = 0.46[/tex]
