emf, the volt, equals joules per coulomb. You can think of a nine-volt battery as doing nine joules of work on each coulomb of charge that flows
through it.
Batteries also are defined by how much energy they can supply over their lifetimes. AAA and D batteries are both 1.5-volt batteries, but the
larger D battery supplies more energy over its lifetime. This total available energy is measured in watt-hours. A small battery for watches will
have about 0.1 watt-hours of total energy, while a car battery has a total available energy of about 500 watt-hours.
When current flows through a battery, it encounters resistance. This is called a battery’s internal resistance. This means that when placed in a
circuit, a battery’s emf and the potential difference across its terminals are not the same. The emf is greater than the potential difference
(unless the battery is being charged, in which case the emf can be greater). The internal battery resistance is usually minor, however. Many
times in this chapter we will treat it as zero, and consider the emf and the potential difference to be the same.
27.3 - Energy and electric potential in a circuit
We will use the simple circuit shown to the right as our starting point for discussing how
to analyze circuits. Specifically, it is important to correctly determine the changes in
electric potential across various components in the circuit.
The circuit in Concept 1 contains a resistance-free battery and a resistor: a flashlight
bulb. As the diagram reflects, the resistor has a resistance of 5.0 ohms, and the battery
has an emf of 1.5 volts. We want to determine the potential difference across the
resistor.
To analyze the changes in electric potential occurring around the circuit, consider some
charge conducting a hypothetical journey around the circuit loop, as shown in Concept
- Imagine 0.5 coulombs of positive charge traveling one complete loop in the direction
of conventional current (clockwise in this case). What happens to the potential energy of
this charge as it passes through each component í the battery and the resistor í in
completing a round trip around the loop?
For starters, the potential energy must be the same at the beginning and end of the
closed path, because the electrostatic force is a conservative force. Just as your
gravitational PE is the same after you take a round trip walk, even if you go uphill and
down on the way, so too the charge’s electrostatic PE is the same when it returns to its
starting point.
To determine the changes in potential energy as the charge makes its journey through
each component, we will use an equation that relates the change in potential energy to
charge and potential difference: ǻPE= qǻV.
We will start with the charge at the negative terminal of the battery, and have it flow
through the battery to the positive terminal. The change in potential energy for a
positive charge traveling across the battery in this direction is positive. The battery must
do work on this positive charge to move it toward the positive terminal. The charge’s
change in PE is the product of the charge and the potential difference, (0.5 C)(1.5 V),
which equals +0.75 J. The electric potential energy of the charge increases and the
chemical potential energy of the battery decreases correspondingly.
Next, the charge flows through the wire around the circuit. We assume that the wire has
negligible resistance. To complete a circuit, the charge must pass through theresistor.
This is the one location in this simple circuit where the charge loses energy before
returning to its starting point. The resistor heats up, giving off this energy as heat. The
charge’s change in potential energy in the resistor must equal í0.75 joules, because the
sum of the changes in potential energy around the complete loop must equal zero.
Now that we have assessed the potential energy changes around a circuit loop, we can
also assess the changes in electric potential. Concept 3 shows the changes in electric
potential around a complete circuit loop. The changes in electric potential can be
determined using the equation ǻV=ǻPE/q.
Concept 3 again shows a clockwise path around the circuit. Moving in this direction, the
electric potential increases by +1.5 V as the battery is traversed. The potential
difference across the battery traveling in this direction is equal to the change in PE,
+0.75 J, divided by the amount of charge, 0.5 C. Across the resistor, since the change
inPE is í0.75 J in this direction, the change in potential is í1.5 V. The electric potential
is greater on the “upstream” end of the resistor than on the “downstream” end. The
charge moves across the resistor from a region of a higher electric potential to one of
lower electric potential.
What we have shown is fundamental and important. In this two-component circuit, the
change in electric potentialacross one component í the battery í is equal to but
opposite the change in electric potentialacross the other component: the resistor. The changes sum to zero. We have shown this in a specific
case, but the rule holds in general for any closed loop around a circuit.
The direction we chose to travel the circuit was arbitrary. Had we chosen to travel the circuit in a counterclockwise direction, the change in
electric potential would be negative as we crossed the battery and positive as we crossed the resistor. We also chose to consider a positive
Battery-resistor circuit
Circuit energy analysis
PE of charge is unchanged after any
round-trip
·Energy increases in battery
·Energy decreases in resistorElectric potential changes in
circuit
Potential changes around circuit sum to
zero