DC Circuits Study Notes: How Batteries Store and Deliver Electrical Energy

Electrochemical energy: what a battery really stores

A battery is a device that converts chemical potential energy into electrical energy. The most important idea to get straight at the start is this: a battery does not “store electricity” the way a tank stores water. Instead, it stores energy in the arrangement of chemicals—then uses chemical reactions to separate charge and maintain a potential difference (voltage) between its terminals.

In DC electronic circuits, batteries matter because they are the most common portable DC source. Many circuit rules you learn (Ohm’s law, series/parallel behavior, power) assume there is some source that can provide energy to charges moving around the circuit. A battery is one practical way to do that.

The core pieces inside a battery

Inside a battery cell (a single electrochemical unit) you typically have:

• Two electrodes: an anode and a cathode (the names relate to the reaction type happening there—more on that soon).
• An electrolyte: a medium that allows ions to move (it can be liquid, gel, or solid).
• A separator: prevents the electrodes from touching (which would short-circuit the cell) while still allowing ions to pass.

The key is that electrons are pushed through the external circuit (your wires and components), while ions move inside the battery through the electrolyte to keep charge balanced.

Why chemical reactions can create voltage

In many electrochemical reactions, one material more readily loses electrons (is oxidized), and another more readily gains electrons (is reduced). If you physically separate these half-reactions into two electrodes and connect them through an external wire, electrons will tend to flow from the electrode where electrons are produced toward the electrode where electrons are consumed.

That tendency corresponds to a potential difference. The battery’s chemistry “wants” the reaction to proceed—so it can do work on charges, pushing them “uphill” in electrical potential inside the cell.

Discharge vs charge: what “direction” means

When a battery is discharging, it is delivering energy to the circuit. When a rechargeable battery is charging, the circuit forces current in the opposite direction, driving the chemical reactions backward (as much as the chemistry allows).

A common misconception is to think the battery “creates electrons.” It doesn’t. Electrons already exist in the materials. The chemical reactions simply move electrons from one place to another and maintain an electric field that drives them around the circuit.

Exam Focus

• Typical question patterns:
  • Explain (in words) how chemical energy becomes electrical energy in a battery.
  • Identify what moves inside vs outside the battery (electrons vs ions).
  • Describe what it means for a battery to discharge and what changes when it charges.
• Common mistakes:
  • Saying a battery “stores charge” or “stores electrons” instead of storing chemical energy.
  • Mixing up where electrons flow (external circuit) and where ions move (inside the battery).
  • Treating voltage as something that “flows” like current.

How a battery stores energy: separating charge and building electrical potential

To understand “storage,” focus on what the chemistry accomplishes when the battery is ready to use: it creates and maintains charge separation.

Charge separation is the essential stored state

In a “ready” battery, the chemical system is in a higher-energy arrangement that can lower its energy by allowing a redox reaction to proceed. The battery’s internal structure keeps the two half-reactions apart so the only easy route for electrons is through the external circuit.

This separation produces:

• An excess of electrons at one terminal (making it relatively negative)
• A deficit of electrons at the other terminal (making it relatively positive)

That imbalance sets up an electric field and therefore a potential difference.

Electromotive force (emf): the battery’s “ideal” voltage

The electromotive force (emf), usually written as $\mathcal{E}$, is the energy supplied per unit charge by the source.

Formally:

$\mathcal{E}=\frac{W}{Q}$

where:

• $\mathcal{E}$ is emf in $\text{V}$
• $W$ is energy in $\text{J}$
• $Q$ is charge in $\text{C}$

This equation is powerful because it links the chemical side (energy available) to the electrical side (charge moved). If a battery supplies $1\,\text{J}$ of energy for every $1\,\text{C}$ of charge pushed through it, then $\mathcal{E}=1\,\text{V}$.

A misconception to avoid: emf is not a “force” in the mechanical sense. It’s a potential difference produced by chemical processes.

What changes as a battery is used

As the battery discharges:

• Reactants are consumed and products build up.
• The chemical “push” that maintains charge separation weakens.
• The terminal voltage under load tends to drop.

This is why batteries go “flat”: not because electrons run out, but because the chemistry can no longer maintain the same potential difference.

Example: energy per charge

If a cell has $\mathcal{E}=1.5\,\text{V}$, then each coulomb of charge receives about:

$W=\mathcal{E}Q=1.5\,\text{V}\times 1\,\text{C}=1.5\,\text{J}$

So delivering $10\,\text{C}$ corresponds to about:

$W=1.5\times 10=15\,\text{J}$

This does not mean the battery is 100% efficient—real batteries lose some energy as heat inside due to internal resistance (covered later).

Exam Focus

• Typical question patterns:
  • Define emf and relate it to energy per unit charge.
  • Explain why a battery’s voltage falls as it discharges.
  • Use $W=\mathcal{E}Q$ to calculate energy transferred.
• Common mistakes:
  • Confusing emf with current (emf is a voltage-like quantity).
  • Assuming emf equals terminal voltage in all situations (it’s only equal when no current flows).
  • Saying “charge is used up” rather than energy being transferred.

How a battery disperses energy in a circuit: from chemical reactions to current

A battery “disperses” (delivers) energy when it is connected into a complete circuit and current flows. The process has two linked parts:

1. Inside the battery: chemical reactions move charges in a direction they would not move purely due to electric forces.
2. Outside the battery: the electric field in the circuit causes electrons to drift through components, transferring energy.

Step-by-step picture of discharge

When you connect a load (like a resistor, lamp, or motor):

1. The circuit becomes a closed loop.
2. The battery’s electric field drives electrons from the negative terminal through the external circuit.
3. In the load, electrons collide with the material’s lattice and transfer energy—often as heat, light, or mechanical work.
4. Electrons return to the positive terminal.
5. Inside the battery, chemical reactions at the electrodes move charge internally so the terminals do not immediately neutralize.

A useful analogy: think of the battery as a “charge pump.” The load is where the energy is spent, but the battery provides the energy that keeps the charges moving around the loop.

Electron flow vs conventional current

In metal wires, electrons are the moving charges, drifting from negative to positive terminal. However, by convention, current is defined in the direction positive charge would move—so conventional current goes from positive to negative terminal in the external circuit.

Both descriptions predict the same circuit behavior as long as you stay consistent.

Power delivery: how fast energy is transferred

The electrical power delivered to a load is:

$P=IV$

where:

• $P$ is power in $\text{W}$
• $I$ is current in $\text{A}$
• $V$ is the potential difference across the load in $\text{V}$

This tells you the rate at which the battery’s chemical energy is being converted to other forms in the circuit.

You can combine this with time to get energy:

$E=Pt$

where $t$ is time in $\text{s}$.

Worked example: energy delivered over time

A battery provides $V=9.0\,\text{V}$ across a device drawing $I=0.20\,\text{A}$ for $t=300\,\text{s}$.

1) Power:

$P=IV=0.20\times 9.0=1.8\,\text{W}$

2) Energy transferred:

$E=Pt=1.8\times 300=540\,\text{J}$

Interpretation: during those 5 minutes, about $540\,\text{J}$ of chemical energy is converted into other forms (plus some internal heating losses).

Exam Focus

• Typical question patterns:
  • Describe the energy pathway during discharge (chemical to electrical to thermal/light/mechanical).
  • Distinguish electron flow direction from conventional current direction.
  • Calculate power and energy using $P=IV$ and $E=Pt$.
• Common mistakes:
  • Saying energy “flows” with electrons; it’s the electric field and potential difference that represent energy per charge.
  • Mixing up volts across the load with the battery’s labeled emf when current is flowing.
  • Using $P=IV$ with the wrong voltage (not the voltage across the component).

The real battery model: internal resistance and terminal voltage

In ideal circuit diagrams, a battery is often drawn as a perfect voltage source. Real batteries are not perfect. When current flows, some energy is dissipated inside the battery due to its internal resistance.

Internal resistance: what it is and why it matters

Internal resistance (often written $r$) represents all the resistive and kinetic limitations inside the cell: resistance of materials, ion mobility limits, electrode processes, and contact resistances. You can model a real battery as:

• an ideal source of emf $\mathcal{E}$
• in series with a resistor $r$

This matters because it explains:

• why the terminal voltage drops when you draw large currents
• why batteries heat up under heavy load
• why a “short circuit” is dangerous (very large current limited mainly by $r$)

Terminal voltage under load

If a current $I$ is being delivered, the **terminal voltage** $V_{\text{terminal}}$ is:

$V_{\text{terminal}}=\mathcal{E}-Ir$

Interpretation: $Ir$ is the “lost volts” inside the battery—the energy per charge converted to heat internally.

When no current flows (open circuit), $I=0$ and:

$V_{\text{terminal}}=\mathcal{E}$

So the voltage you measure with a high-resistance voltmeter across an unloaded battery is close to the emf.

Short-circuit current (conceptually)

If you connect the terminals with a near-zero external resistance, the current can become very large. In the simple model, the current is limited mostly by $r$:

$I\approx \frac{\mathcal{E}}{r}$

This is why batteries can overheat, leak, or fail violently when shorted—the internal power loss is:

$P_{\text{internal}}=I^2r$

Worked example: internal resistance and terminal voltage

A battery has $\mathcal{E}=12.0\,\text{V}$ and internal resistance $r=0.50\,\Omega$. It supplies $I=4.0\,\text{A}$.

Terminal voltage:

$V_{\text{terminal}}=\mathcal{E}-Ir=12.0-(4.0)(0.50)=10.0\,\text{V}$

Internal power loss:

$P_{\text{internal}}=I^2r=(4.0)^2(0.50)=8.0\,\text{W}$

Meaning: the battery is delivering power to the external circuit, but it is also heating internally at $8.0\,\text{W}$ while doing so.

Polarization and voltage sag (a practical note)

Besides pure resistance, real cells show voltage sag due to limitations in reaction rates and transport of ions. Under high current, reactants near electrodes can be temporarily depleted, making the voltage dip. When the load is removed, the voltage may recover somewhat. In circuits, this shows up as “it works for a moment then dims.” The simple $\mathcal{E}-Ir$ model captures the main idea but not every detail.

Exam Focus

• Typical question patterns:
  • Use $V_{\text{terminal}}=\mathcal{E}-Ir$ to calculate terminal voltage or internal resistance.
  • Explain why voltage drops under heavy load.
  • Analyze why short circuits cause heating and danger using $P=I^2r$.
• Common mistakes:
  • Using $\mathcal{E}$ as if it is always the same as the voltage across the load.
  • Forgetting that internal resistance is in series, not parallel.
  • Ignoring internal heating when discussing efficiency or battery life.

Capacity and energy delivery: linking chemistry to circuit use

When you buy a battery, you see labels like voltage (e.g., 1.5 V, 9 V) and capacity (e.g., 2000 mAh). These numbers connect the battery’s chemistry to how long it can deliver current.

Charge, current, and capacity

Current is the rate of flow of charge:

$I=\frac{Q}{t}$

so:

$Q=It$

A battery’s capacity is often given in ampere-hours (Ah) or milliampere-hours (mAh). This is essentially a measure of total charge the battery can deliver under specified conditions.

Conversion:

$1\,\text{Ah}=3600\,\text{C}$

So a $2000\,\text{mAh}$ battery is $2.0\,\text{Ah}$, corresponding (ideally) to:

$Q=2.0\times 3600=7200\,\text{C}$

Important limitation: actual delivered capacity depends strongly on discharge rate, temperature, and cutoff voltage. But for basic circuit calculations, the Ah rating is a useful approximation.

Energy from capacity and voltage

If the battery maintains roughly constant voltage $V$ while delivering total charge $Q$, the electrical energy delivered is approximately:

$E\approx VQ$

Using $Q=It$, you also get:

$E\approx VIt$

This ties the chemistry to the circuit in a clean way: voltage tells you energy per charge, and capacity tells you how much charge you can move.

Worked example: estimate runtime and energy

A battery is rated $1500\,\text{mAh}$ at $1.5\,\text{V}$. A device draws $I=0.10\,\text{A}$.

1) Convert capacity:

$1500\,\text{mAh}=1.5\,\text{Ah}$

2) Estimate runtime:

$t\approx \frac{1.5\,\text{Ah}}{0.10\,\text{A}}=15\,\text{h}$

3) Estimate total energy delivered (idealized):

$Q=1.5\times 3600=5400\,\text{C}$

$E\approx VQ=1.5\times 5400=8100\,\text{J}$

Interpretation: the battery’s chemistry can supply on the order of $8.1\times 10^3\,\text{J}$ of electrical energy, ignoring voltage sag and inefficiencies.

Why high current drains batteries faster than you expect

Students often assume: “Double the current, half the time.” Real batteries frequently do worse than that because internal losses and chemical limitations increase at higher currents (effective internal resistance and polarization effects matter more). So battery life is not purely a linear Ah calculation in real-world conditions.

Exam Focus

• Typical question patterns:
  • Convert between $\text{mAh}$, $\text{Ah}$, and coulombs using $1\,\text{Ah}=3600\,\text{C}$.
  • Estimate runtime using $t=\frac{Q}{I}$.
  • Estimate energy using $E\approx VQ$ or $E\approx VIt$.
• Common mistakes:
  • Treating $\text{mAh}$ as energy (it is primarily charge capacity, not joules).
  • Forgetting the factor of $3600$ when converting hours to seconds.
  • Assuming constant voltage and constant capacity regardless of load conditions.

Primary vs secondary cells: how “rechargeable” changes the energy story

The basic storage and delivery mechanism is the same for all batteries—chemical energy becomes electrical energy via redox reactions—but the reversibility of those reactions differs.

Primary cells (non-rechargeable)

A primary cell is designed so that the discharge reaction is not practically reversible. As it discharges, the chemistry changes in ways that you cannot reliably “undo” by forcing current backward.

What this means conceptually:

• The chemical potential energy decreases as reactants are converted to products.
• Once the reactants are mostly consumed, the cell can no longer maintain its emf.

Secondary cells (rechargeable)

A secondary cell (rechargeable battery) uses reactions that can be reversed (within design limits). During charging:

• an external power source does electrical work on charges
• the battery stores that energy by driving the chemical system back toward its higher-energy state

In other words, charging is the reverse energy pathway:

• During discharge: chemical $\rightarrow$ electrical $\rightarrow$ other forms in the load
• During charge: electrical (from charger) $\rightarrow$ chemical (stored)

Efficiency and heating during charging

Charging is not 100% efficient. Some energy becomes heat due to internal resistance and side reactions. This is why fast charging can warm a battery and why charge control is important for safety and lifespan.

A common misconception is: “If it’s rechargeable, it must return to exactly the original state each cycle.” In reality, many batteries slowly degrade because not all reactions are perfectly reversible.

Exam Focus

• Typical question patterns:
  • Explain why some cells are rechargeable and others are not (reaction reversibility).
  • Describe energy transformations during charging vs discharging.
  • Link inefficiency to heating (internal resistance, side reactions).
• Common mistakes:
  • Assuming rechargeable means “infinite cycles” or “no energy losses.”
  • Thinking charging creates energy rather than storing it chemically.
  • Confusing the direction of current during charging vs discharging.

Batteries as DC sources in circuits: practical implications for design

In DC electronic circuits, you rarely use a battery in isolation—you combine it with resistors, sensors, semiconductors, and motors. Understanding how batteries disperse energy helps you design circuits that actually work.

Voltage is not guaranteed under load

If your circuit draws more current than expected, terminal voltage can drop because of $Ir$ losses. This can cause:

• microcontrollers to reset (brownout)
• LEDs to dim
• motors to stall

Design implication: consider the battery’s internal resistance and the expected current draw. A “higher capacity” battery is not always “better” if it cannot supply the required current without sag.

Matching battery choice to load type

Different loads stress a battery differently:

• Low, steady current (e.g., remote controls): batteries last close to their rated capacity.
• High pulses (e.g., camera flash, motors): internal resistance and reaction limits dominate performance.

Even without memorizing specific chemistries, the principle is the same: the battery must supply the required power without excessive internal loss.

Simple circuit illustration: battery + resistor

If a battery with emf $\mathcal{E}$ and internal resistance $r$ is connected to a load resistor $R$, then the current is:

$I=\frac{\mathcal{E}}{R+r}$

The voltage across the load is:

$V_R=IR$

and the terminal voltage is the same as $V_R$ in this simple series circuit.

This model makes it clear: increasing load current increases the internal voltage drop $Ir$ and reduces the voltage delivered to the load.

Exam Focus

• Typical question patterns:
  • Analyze a source with internal resistance in series with a load.
  • Predict effects of increased load current on terminal voltage.
  • Explain real-world symptoms of voltage sag in DC devices.
• Common mistakes:
  • Treating the battery as an ideal voltage source regardless of load.
  • Forgetting to include $r$ when calculating current.
  • Assuming capacity alone determines whether a battery can power a high-current device.

What can go wrong: misconceptions and failure modes tied to energy delivery

Understanding how batteries store and disperse energy also means understanding what happens when the intended chemical-to-electrical pathway is disrupted.

Common misconceptions (and the correct thinking)

1. “The battery provides current.”

   • More accurate: the battery provides a potential difference (energy per unit charge). The circuit determines the current according to its total resistance/impedance.

2. “Electrons come out of the battery and get used up in the bulb.”

   • Electrons circulate; energy is transferred to the bulb’s filament (or other load) through electric forces and collisions, not by consuming electrons.

3. “A dead battery has no charge.”

   • It has plenty of charge. It lacks sufficient chemical potential to maintain the necessary potential difference under load.

Failure modes linked to internal energy and heating

• Short circuits: very large current, large internal heating $P_{\text{internal}}=I^2r$.
• Over-discharge (especially in some rechargeables): can cause irreversible chemical changes and reduced capacity.
• Overcharge: forces unwanted side reactions, heat, gas production, and degradation.

Even if your course doesn’t require detailed chemistry, the circuit-level takeaway is consistent: extreme currents and improper charging conditions cause energy to be dissipated internally in unsafe ways.

Exam Focus

• Typical question patterns:
  • Explain why current is determined by the circuit, not “pushed out” as a fixed amount by the battery.
  • Use internal resistance ideas to explain heating, voltage sag, and short-circuit danger.
  • Identify correct statements about what changes when a battery goes flat.
• Common mistakes:
  • Treating current as a battery-set value rather than a load-determined value.
  • Ignoring energy losses inside the battery.
  • Mixing “charge,” “current,” “voltage,” and “energy” as if they are interchangeable.