Date: _______________________________
Grade 9 AP Electricity: Additional Formulas
So far you have seen the following formulas:
(1) Ohm’s Law
V = IR
(2) Energy
E = Pt
(3) Price
Price = Cost x E
(4) Efficiency % efficiency = Euseful/Ein × 100%
Often when we speak about current electricity, we talk of “flowing electrons”. However, when we use
quantitative measurements to describe and relate the values in a circuit, we are actually measuring the flow of
charges – specifically, the amount of negative charge that is carried by the flowing electrons.
Current and Charge
Recall the definition of current – the amount of charge that flows past a point, in one second.
 Charge (Q) is expressed in coulombs (C).
 Current (I) is expressed in amperes (A).
 Therefore, 1 A = 1 C/s. These units can be used interchangeably, since they are
equivalent. However, you may find it easier to express current in C/s if you are
performing calculations involving charge.
Equation 1
Q
=
charge,
in coulombs (C)
Practice Problems:
1. Re-arrange Equation 1 to isolate:
a) current, I
I=
I
current,
C/s
×
Figure 1 One coulomb is the amount of
charge carried by 6.24 x 1018 electrons!
t
time,
seconds (s)
b) time, t
t=
2. When a circuit is switched on, 0.25 A of current runs through a resistor. If the circuit is on for 45 seconds,
how much charge passes through this resistor?
3. Henry uses an electric blanket that draws 2 A of current. While it is on, 14 400 C of charge passes through it.
a) Determine the amount of time, in hours, that the resistor is on.
b) How many electrons have passed through this resistor, during this time?
Energy and Charge
In an electrical circuit, there is a transfer of energy from flowing charges to other
components (for example, resistors) in the circuit. The amount of energy that is
transferred by each coulomb of charge is expressed as the potential difference, or voltage,
of that resistor.
 Energy (E) is typically expressed in Joules (J). When dealing with consumer
electricity costs, energy is also often expressed in kilowatt-hours, or watt-seconds.
 Potential difference, aka Voltage (V) is expressed in volts (V), where 1 V = 1 J/C.
Equation 2
E
=
Energy,
in Joules (J)
Q
×
V
charge,
coulombs
(C)
Potential
difference,
volts (V)
Figure 2 When I passes through the
resistor, each coulomb of charge
transfers a certain amount of energy.
This is measured as the voltage, which
is the energy difference, per coulomb.
In this case, each coulomb transfers 9 J
of energy to the resistor (9 V = 9 J/C).
By combining Equations (1) and (2):
Equation 3
E
=
Energy,
in Joules (J)
Practice Problems:
4. Re-arrange Equation 2 to isolate:
a) charge, Q
I
×
t
current,
A
I=
V
Potential
difference,
volts (V)
b) voltage, V
Q=
5. Re-arrange Equation 3 to isolate:
a) current, I
×
time,
seconds
(s)
V=
b) time, t
t=
c) voltage, V
V=
6. Virginia turns on her laptop computer, which operates at a potential difference of 19 V. If 95 J of energy is
consumed by the computer, how many coulombs of charge has passed through it?
7. Justin toasts his bread for 45 seconds, using a toaster that operates at 120 V. During this time, the toaster
consumes 54 000 J of energy.
a) How much current runs through the toaster?
b) If the toaster has an efficiency of 76%, how much energy actually goes into toasting the bread (find Euseful)?
Power
The power rating of a device describes the rate at which it consumes energy.
 Power (P) is expressed in Watts (W), where 1 W = 1 J/s
 Since current is the amount of coulombs per second, and voltage is joules per coulomb:
Equation 4
P
=
Power, in watts
(W) or
joules/second (J/s)
8. Re-arrange Equation 4 to isolate:
a) current, I
I=
I
current,
A
×
V
Potential
difference,
volts (V)
b) voltage, V
V=
9. Joanna charges her cell phone, which draws 0.7 A of current, and operates at a potential difference of 5 V.
a) Calculate the power rating of the cell phone charger, in watts, and kilowatts.
b) If it takes 2 hours to fully charge the phone, and the cost of electricity is 14.4 ¢ /kW•h. Determine the
price of the electricity used to charge her phone.
c) Use Ohm’s Law to find the internal resistance of the cell phone charger.