9.2 The Power of Electricity
Name: _________________
Power
• P o w e r is the rate of change in e n e r g y , also known as the rate at which w o r k is done.
•
The u n i t of measurement for p o w e r is the w a t t (W).
•
P o w e r , c u r r e n t and v o l t a g e are related using a mathematical e q u a t i o n :
•
•
P o w e r (P) = C u r r e n t (I) x V o l t a g e (V)
Example Problems:
1. A 40.0 W light bulb is connected to a 15 V battery. What
is the current running through the bulb?
𝐼=
𝑃 40.0 W
=
= 2.7 A
𝑉
15 V
2. A current of 5.0 A flows through a toaster that is connected to 110 V. Determine the
power being used.
3. A 220 W radio is connected to 2.0 A of current. Find the voltage of the circuit.
𝑉=
𝑃 220 W
=
= 110
𝐼
2.0 A
Science 9 ● Physics● Notes
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Energy
• E n e r g y is the a b i l i t y to do w o r k .
•
The u n i t of measurement for e n e r g y is the j o u l e (J).
•
An e q u a t i o n can be used to relate e n e r g y (joules) with
p o w e r (watts) and t i m e (seconds):
•
•
E n e r g y (E) = P o w e r (P) x T i m e (t)
Example Problem:
1. How much energy, in joules, does a 250 W stereo consume if it is left on for 10
minutes?
𝐸 = 𝑃𝑡 = (250 W)(600 s) = 150 000 J
Relationship Between Energy and Power
• P o w e r (watts) is m e a s u r e d as units of e n e r g y (joules) p e r unit of
t i m e (seconds).
•
E l e c t r i c a l power is the r a t e of change in electrical e n e r g y .
•
Eg. A 25 W fluorescent bulb converts 25 j o u l e s per second of
e l e c t r i c a l energy into l i g h t energy.
•
If the v o l t a g e and c u r r e n t flowing through a device are known, the
p o w e r of the device can be c a l c u l a t e d .
•
The amount of t i m e the d e v i c e is in u s e , along with the p o w e r will
determine how much e n e r g y it consumes.
Science 9 ● Physics● Notes
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Energy on a Large Scale
• A j o u l e is a very s m a l l amount of e n e r g y .
•
Energy c o n s u m p t i o n in homes is often measured using k i l o w a t t •h o u r s (kw•h).
•
When using kilowatt•hours for e n e r g y , p o w e r and t i m e must be in terms of
k i l o w a t t s and h o u r s , respectively.
Paying for Power
• The p o w e r c o m p a n y can determine how many kilowatt•hours of
e n e r g y have been c o n s u m e d .
•
Consumption is m e a s u r e d on an e l e c t r i c m e t e r .
•
The e n e r g y is then m u l t i p l i e d by the c o s t per kilowatt•hour.
•
Example Problem:
1. A house consumes 1500 kWh of energy. The cost of energy is
$0.07 per kilowatt—hour. Determine the cost of the energy bill.
𝐶𝑜𝑠𝑡 = (1500 kW • h)($0.07 ) = $105.00
Science 9 ● Physics● Notes
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