# Practice Problems 0502 ```Practice Problems 0502
The current through a lamp is 0.5 amps (amperes). The lamp is turned on for a total time of 30
seconds. The potential difference across the lamp is 1.2 volts. Calculate the electric potential energy
that is transferred to the lamp in 30 seconds.
I = 0.5 amps = 0.5 coulombs/sec
t = 30 sec
∆V = 1.2 volts
Step 1: Solve for Q (charge)
I=
Q
t
0.5 =
Q
30
30(0.5) = Q
15 Coulombs = Q
Step 2: Solve for E (electric potential energy)
V=
E
Q
1.2 =
E
15
1.2(15) = E
18 Joules = E
Batteries used in cell phones or notebooks sometimes have a number on them, which for example may
be 650 milliamperes-hours. What is a milliampere-hour a measure of?
Milliampere is a measure of current
Hours is a measure of time
(I)
(t)
When these quantities are multiplied to produce the unit milliamperes-hours, we have the following
product
(I)(t)
Charge Q is the product of current and time
Q = (I)(t)
Therefore a milliampere-hour is a measure of charge. For cell phone batteries, the larger the number
of milliampere-hours, the longer the battery lasts before it needs to be re-energized. Check your cell
phone battery for this number.
```