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Electricity and Magnetism

Secondary Cells and Accumulators

Secondary cells are electrochemical storage devices that can be electrically recharged after discharge by passing current in the reverse direction. They function based on reversible redox reactions, allowing repeated cycles of conversion between chemical and electrical energy.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Working principle of secondary cells, lead-acid accumulator, EMF and internal resistance, terminal potential difference, grouping of cells.

PGT extension

Faraday's laws of electrolysis applied to charging/discharging, capacity of accumulators (Ampere-hour efficiency), internal resistance variation with state of charge, thermodynamic considerations of reversible cells.

Prerequisites

Primary cells and electromotive force (EMF)
Internal resistance of a cell
Basic electrochemistry and redox reactions

Principle of Secondary Cells

Secondary cells act as galvanic (voltaic) cells during discharge, converting chemical energy into electrical energy.
They act as electrolytic cells during charging, where an external DC source drives non-spontaneous reverse chemical reactions.
The reversibility of the chemical reactions is the defining characteristic that permits multiple cycles of use.

Lead-Acid Accumulator Structure

The positive plate consists of lead dioxide (PbO₂) and the negative plate consists of spongy lead (Pb).
The electrolyte is an aqueous solution of sulfuric acid (H₂SO₄) with a specific gravity of approximately 1.250 to 1.280 when fully charged.
Separators made of porous insulating material prevent direct short-circuiting between adjacent positive and negative plates.

Chemical Reactions during Discharge and Charge

At the anode (negative plate during discharge): Pb + SO₄^2- → PbSO₄ + 2e^-.
At the cathode (positive plate during discharge): PbO₂ + 4H⁺ + SO₄^2- + 2e^- → PbSO₄ + 2H₂O.
Overall discharge reaction: Pb + PbO₂ + 2H₂SO₄ ⇒ 2PbSO₄ + 2H₂O, which reverses during charging.

Characteristics and Efficiency

Internal resistance of secondary cells is extremely low (typically a fraction of an ohm), allowing them to deliver very high surge currents.
Ampere-hour (Ah) efficiency is defined as the ratio of discharge current multiplied by discharge time to charging current multiplied by charging time, typically ranging from 80% to 90%.
Energy efficiency is always lower than Ah efficiency due to the higher cell voltage during charging compared to discharging.

Derivations

Terminal Potential Difference During Charging and Discharging

Conditions: Cell of EMF E and internal resistance r connected in an external circuit.

During discharge, current I flows out of the positive terminal. By Ohm's law, the potential drop across internal resistance is Ir.
The terminal potential difference V is given by $V = E - Ir$.
During charging, current I is forced into the positive terminal by an external source. The potential drop adds to the EMF.
The terminal potential difference during charging becomes $V = E + Ir$.

Result: Discharge: $V = E - Ir; Charge: V = E + Ir$

Power Transfer and Maximum Power Theorem for a Secondary Cell

Conditions: Cell of EMF E and internal resistance r connected to external load resistance R.

Circuit current is given by $I = E / (R + r)$.
Power P delivered to the external load is $P = I^{2}R = E^{2}R / (R + r)^{2}$.
To find the condition for maximum power, differentiate P with respect to R and set dP/dR = 0.
Differentiating yields (R + r)² - 2R(R + r) = 0, which simplifies to R = r.

Result: Maximum power is transferred to the load when external resistance equals internal resistance (R = r).

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Discharging Terminal Voltage	$$V = E - Ir$$	V = Terminal voltage, E = EMF, I = Current, r = Internal resistance Volts (V)	When current is drawn from the cell
Charging Terminal Voltage	$$V = E + Ir$$	V = Terminal voltage, E = EMF, I = Charging current, r = Internal resistance Volts (V)	When current is forced into the cell
Ampere-Hour Capacity	$$C = I \times t$$	C = Capacity, I = Discharge current, t = Time Ampere-hours (Ah)	Constant current discharge
Ampere-Hour Efficiency	$$\eta_{Ah} = \frac{I_{d} t_{d}}{I_{c} t_{c}} \times 100$$	I_d, $t_d = Discharge current$ and time; I_c, $t_c = Charging current$ and time Percentage (%)	Complete charge-discharge cycle
Energy Efficiency	$$\eta_{E} = \frac{\int V_{d} I_{d} dt_{d}}{\int V_{c} I_{c} dt_{c}} \times 100$$	V_d, $V_c = Discharge$ and charge voltages Percentage (%)	Accounting for varying voltages
Internal Resistance via External Shunt	$$r = R \left(\frac{E}{V} - 1\right)$$	r = Internal resistance, R = External resistance, E = EMF, V = Terminal voltage Ohms (\Omega)	Measured with a known resistance R
Total EMF of n Identical Cells in Series	$$E_{net} = nE, \quad r_{net} = nr$$	n = Number of cells, E = EMF of one cell, r = Internal resistance of one cell Volts (V), Ohms (\Omega)	Same polarity connection
Total Internal Resistance of m Identical Cells in Parallel	$$E_{net} = E, \quad r_{net} = \frac{r}{m}$$	m = Number of parallel branches, E = EMF, r = Internal resistance Volts (V), Ohms (\Omega)	All positive terminals connected together

Graphs and experiments

Discharge Curve of a Secondary Cell

Draw: Plot of terminal voltage V (y-axis) versus discharge time t (y-axis) showing an initial rapid drop, a long plateau, and a steep final fall.

Interpret: The long flat region indicates stable operating voltage due to constant chemical composition of active materials. The sharp drop marks exhaustion of H₂SO₄.

Exam use: Questions testing the behavior of terminal voltage over time and identification of the plateau region.

Specific Gravity vs. State of Charge

Draw: Linear or near-linear curve of electrolyte specific gravity (y-axis) against state of charge percentage (x-axis) ranging from 1.100 (discharged) to 1.280 (charged).

Interpret: As H₂SO₄ is consumed during discharge, density of electrolyte decreases directly reflecting the state of charge.

Exam use: Hydrometer test principles used in practical physics and conceptual questions.

V-I Characteristic Graph for a Secondary Cell

Draw: Straight line plot of terminal potential V against current I, starting at V = E when I = 0 and sloping downwards for discharge ($V = E - Ir)$ and upwards for charging ($V = E + Ir)$.

Interpret: The intercept on the voltage axis gives the EMF E, and the slope of the line magnitude represents the internal resistance r.

Exam use: Graphical determination of EMF and internal resistance in experimental physics problems.

Solved numericals

TGT core

A storage battery of EMF 12 V and internal resistance 0.5 \Omega is being charged by a DC source of 20 V using a series resistor of 1.5 \Omega. Find the charging current and the terminal potential difference of the battery.

Identify the total EMF in opposition: Source EMF = 20 V, Battery EMF = 12 V.
Total resistance in the circuit: $R_total = R_external + r = 1.5 \Omega + 0.5 \Omega = 2.0 \Omega$.
Calculate charging current: $I = (V_source - E_battery) / R_total = (20 - 12) / 2.0 = 8 / 2 = 4 A$.
Calculate terminal potential difference during charging: $V = E + Ir = 12 + (4 \times 0.5) = 12 + 2 = 14 V$.

Answer: Charging current = 4 A, Terminal potential difference = 14 V

PGT extension

A lead-acid accumulator has an ampere-hour capacity of 100 Ah. If it is discharged at a constant current of 5 A for 16 hours and subsequently charged at 10 A for 9 hours, calculate its ampere-hour efficiency and energy efficiency assuming average discharging and charging voltages of 1.9 V and 2.3 V respectively.

Calculate discharge output (A$h_out) = I_d \times t_d = 5 \text{ A} \times 16 \text{ h} = 80 \text{ Ah}$.
Calculate charge input (A$h_in) = I_c \times t_c = 10 \text{ A} \times 9 \text{ h} = 90 \text{ Ah}$.
Calculate Ampere-hour efficiency: \et$a_{Ah} = (80 / 90) \times 100 = 88.89%$.
Calculate energy output ($E_out) = V_d \times Ah_out = 1.9 \text{ V} \times 80 \text{ Ah} = 152 \text{ Wh}$.
Calculate energy input ($E_in) = V_c \times Ah_in = 2.3 \text{ V} \times 90 \text{ Ah} = 207 \text{ Wh}$.
Calculate Energy efficiency: \et$a_{E} = (152 / 207) \times 100 = 73.43%$.

Answer: Ampere-hour efficiency = 88.89%, Energy efficiency = 73.43%

TGT core

A secondary cell gives a balancing length of 70 cm on a potentiometer wire. When the cell is short-circuited through a 2 \Omega resistor, the balancing length falls to 50 cm. Find the internal resistance of the cell.

Potentiometer balancing length is directly proportional to potential difference: l_1 \propto E and l_2 \propto V.
Relation between EMF and terminal voltage with external resistance R: E / $V = l_1 / l_2$.
Substitute given values: E / $V = 70 / 50 = 1.4$.
Use internal resistance formula: $r = R ((E / V) - 1)$.
Calculate r: $r = 2 \times (1.4 - 1) = 2 \times 0.4 = 0.8 \Omega$.

Answer: Internal resistance = 0.8 \Omega

Assumptions and limitations

Internal resistance is assumed to be constant during moderate discharge cycles, although in reality, it increases as the specific gravity of the sulfuric acid decreases and lead sulfate accumulates on the plates.

Common misconceptions

Common error: Terminal voltage of a secondary cell is always equal to its EMF regardless of operating conditions.

Correct: Terminal voltage equals EMF only when no current is drawn (open circuit). It is $V = E - Ir during discharge$ and $V = E + Ir during charge$.

Internal resistance causes an internal potential drop proportional to the current flowing through the cell.

Common error: Charging a secondary cell increases its internal resistance permanently.

Correct: Charging restores the chemical composition of the active plates and electrolyte, thereby reducing the internal resistance back to its normal low state.

Reversible electrochemical reactions regenerate the original low-resistance reactants (Pb, PbO₂, and strong H₂SO₄).

Quick revision

Secondary cells are rechargeable cells utilizing reversible redox reactions like the lead-acid accumulator.
Discharging Terminal Voltage: $V = E - Ir$
Charging Terminal Voltage: $V = E + Ir$

Must remember

Secondary cells act as galvanic cells during discharge and as electrolytic cells during charging due to the reversibility of their chemical reactions.
In a fully charged lead-acid accumulator, the positive plate is $PbO_2$ and the negative plate is spongy $Pb$, immersed in an aqueous $H_2SO_4$ solution.
The terminal voltage during discharge is given by $V = E - Ir$, while during charging it is $V = E + Ir$.
Internal resistance of secondary cells is extremely low, enabling them to deliver very high surge currents.
Ampere-hour (Ah) efficiency of lead-acid cells typically ranges from $80\%$ to $90\%$, and energy efficiency is always lower than Ah efficiency.

Glossary

Secondary CellAn electrochemical cell that can be electrically recharged after discharge by passing current in the reverse direction.

AccumulatorA storage battery, most commonly the lead-acid type, used to store electrical energy in chemical form.

Ampere-Hour EfficiencyThe ratio of the product of discharge current and time to the product of charging current and time, expressed as a percentage.

Internal Resistance ($r$)The opposition offered by the electrolyte and electrodes of a cell to the flow of electric current, measured in ohms ($\Omega$).

Discharge vs. Charging State

Feature	Discharge	Charging
Cell Function	Galvanic (Voltaic) cell	Electrolytic cell
Terminal Voltage ($V$)	$V = E - Ir$	$V = E + Ir$
Chemical Reaction	Spontaneous (forms $PbSO_4$)	Non-spontaneous (regenerates $Pb$, $PbO_2$)
Energy Conversion	Chemical to Electrical	Electrical to Chemical

Practice Quiz

Choose an option to reveal the explanation.

1. What is the function of a secondary cell during the discharge process?

Explanation: During discharge, a secondary cell functions as a galvanic (voltaic) cell, converting stored chemical energy into electrical energy.

2. During the charging process, a secondary cell operates as a:

Explanation: An external DC source drives non-spontaneous reverse chemical reactions during charging, making the cell act as an electrolytic cell.

3. What are the active materials on the positive and negative plates of a fully charged lead-acid accumulator?

Explanation: In a lead-acid accumulator, the positive plate consists of lead dioxide ($PbO_2$) and the negative plate consists of spongy lead ($Pb$).

4. What is the electrolyte used in a lead-acid accumulator and its typical specific gravity when fully charged?

Explanation: The electrolyte in a lead-acid cell is an aqueous solution of sulfuric acid ($H_2SO_4$) having a specific gravity of $1.250$ to $1.280$ when fully charged.

5. What is the correct expression for the terminal voltage $V$ of a secondary cell when it is being discharged with current $I$ and internal resistance $r$?

Explanation: When current is drawn from the cell (discharge), the terminal voltage is reduced by the internal drop, giving $V = E - Ir$.

6. What is the terminal voltage formula when a secondary cell is being charged by a current $I$?

Explanation: When current is forced into the cell during charging, the terminal voltage must overcome the EMF, resulting in $V = E + Ir$.

7. Why do secondary cells have extremely low internal resistance?

Explanation: Extremely low internal resistance (typically a fraction of an ohm) allows secondary cells to deliver very high surge currents without excessive voltage sagging.

8. What is the primary definition of Ampere-hour (Ah) efficiency ($\eta_{Ah}$) of a secondary cell?

Explanation: Ampere-hour efficiency is defined as $\eta_{Ah} = \frac{I_d t_d}{I_c t_c} \times 100$, comparing output Ah to input Ah.

9. Why is the energy efficiency of a secondary cell always lower than its Ampere-hour efficiency?

Explanation: Energy efficiency accounts for voltage ($V_c > V_d$), and since charging voltage exceeds discharging voltage, energy efficiency is lower than Ah efficiency.

10. What is formed at both the anode and cathode plates of a lead-acid accumulator at the end of a complete discharge?

Explanation: During discharge, both active materials (Pb and $PbO_2$) are converted into lead sulfate ($PbSO_4$).水的化学反应

11. If $n$ identical secondary cells, each of EMF $E$ and internal resistance $r$, are connected in series with the same polarity, what is the net EMF and net internal resistance?

Explanation: For $n$ identical cells in series with same polarity, EMFs add up ($nE$) and internal resistances also add up ($nr$).

12. If $m$ identical secondary cells are connected in parallel (all positive terminals together), what is the total EMF and total internal resistance of the combination?

Explanation: In parallel combination of $m$ identical cells, the net EMF remains equal to that of a single cell ($E$), while internal resistance becomes $\frac{r}{m}$.

13. How can the internal resistance $r$ of a secondary cell be determined using an external shunt resistance $R$, EMF $E$, and terminal voltage $V$?

Explanation: Using voltage divider relations with external load $R$, the internal resistance formula is $r = R \left(\frac{E}{V} - 1\right)$.

14. Which of the following statements about the internal resistance of a secondary cell is correct?

Explanation: Secondary cells are characterized by very low internal resistance, allowing high current delivery.

15. What purpose do separators made of porous insulating material serve in a lead-acid accumulator?

Explanation: Porous insulating separators keep the positive and negative plates physically apart to prevent short-circuits while permitting ion flow through the electrolyte.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. A lead-acid accumulator of EMF $2.0\text{ V}$ and internal resistance $0.05\text{ \Omega}$ is being charged with a DC source of $10\text{ V}$ through a series resistor of $1.95\text{ \Omega}$. What is the terminal voltage of the cell during charging?

Explanation: Charging current $I = \frac{V_{source} - E}{R + r} = \frac{10 - 2.0}{1.95 + 0.05} = \frac{8}{2} = 4\text{ A}$. Terminal voltage during charge $V = E + Ir = 2.0 + (4)(0.05) = 2.0 + 0.2 = 2.2\text{ V}$.

2. If a storage battery delivers a discharge current of $5\text{ A}$ for $10\text{ hours}$ and requires a charging current of $4\text{ A}$ for $15\text{ hours}$ to restore its original state, what is its Ampere-hour efficiency?

Explanation: Ah output = $5\text{ A} \times 10\text{ h} = 50\text{ Ah}$. Ah input = $4\text{ A} \times 15\text{ h} = 60\text{ Ah}$. Efficiency $\eta_{Ah} = \frac{50}{60} \times 100 = 83.3\%$.

3. In the lead-acid cell discharge reaction, $Pb + PbO_2 + 2H_2SO_4 arrow 2PbSO_4 + 2H_2O$, what happens to the specific gravity of the sulfuric acid electrolyte as discharge proceeds?

Explanation: During discharge, sulfuric acid ($H_2SO_4$) is consumed to form lead sulfate ($PbSO_4$) and water, causing the specific gravity of the electrolyte to decrease.

4. Four identical secondary cells, each having an EMF of $2\text{ V}$ and internal resistance of $0.1\text{ \Omega}$, are connected in series. If the combination is connected across an external load of $4.6\text{ \Omega}$, what is the circuit current?

Explanation: Total EMF $E_{net} = 4 \times 2 = 8\text{ V}$. Total internal resistance $r_{net} = 4 \times 0.1 = 0.4\text{ \Omega}$. Current $I = \frac{E_{net}}{R + r_{net}} = \frac{8}{4.6 + 0.4} = \frac{8}{5.0} = 1.6\text{ A}$.

5. A secondary cell shows a terminal voltage of $1.9\text{ V}$ when drawing a current of $2\text{ A}$ and $2.1\text{ V}$ when drawing a current of $1\text{ A}$. What is the internal resistance of the cell?

Explanation: Using $V = E - Ir$: $1.9 = E - 2r$ and $2.1 = E - 1r$. Subtracting the two equations gives $0.2 = r$, so $r = 0.2\text{ \Omega}$.

6. Which reaction occurs at the anode (negative plate) of a lead-acid cell during discharge?

Explanation: At the anode (negative plate) during discharge, spongy lead reacts with sulfate ions to form lead sulfate, releasing two electrons: $Pb + SO_4^{2-} arrow PbSO_4 + 2e^-$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What is the primary defining characteristic of secondary cells compared to primary cells?

Explanation: The reversibility of chemical reactions is the defining characteristic that permits secondary cells to be reused through charging.

2. In a lead-acid cell, what material forms the positive plate?

Explanation: The positive plate of a lead-acid accumulator consists of lead dioxide ($PbO_2$).

3. What is the typical Ampere-hour efficiency range for lead-acid accumulators?

Explanation: Ampere-hour efficiency of secondary lead-acid cells typically ranges from $80\%$ to $90\%$.

4. When a secondary cell is in an open circuit (no current flowing), its terminal voltage is equal to:

Explanation: When no current flows ($I = 0$), the internal drop $Ir$ is zero, so terminal voltage equals the EMF ($E$).

5. If 10 identical cells each of EMF $2\text{ V}$ and internal resistance $0.05\text{ \Omega}$ are connected in parallel, what is the equivalent EMF of the combination?

Explanation: For identical cells connected in parallel, the net EMF of the combination is equal to the EMF of a single cell, which is $2\text{ V}$.

6. What happens to the internal resistance of a lead-acid battery as it becomes discharged?

Explanation: As discharge proceeds, insulating lead sulfate accumulates on the plates, increasing the internal resistance of the cell.

7. What is the overall chemical reaction during the discharge of a lead-acid cell?

Explanation: The overall discharge reaction is $Pb + PbO_2 + 2H_2SO_4 arrow 2PbSO_4 + 2H_2O$.

8. Which of the following explains why energy efficiency is lower than Ampere-hour efficiency in storage batteries?

Explanation: Since charging voltage ($V = E + Ir$) exceeds discharging voltage ($V = E - Ir$), the energy input exceeds the energy output proportionally, making energy efficiency lower than Ah efficiency.

9. A cell of EMF $2.1\text{ V}$ and internal resistance $0.1\text{ \Omega}$ is connected to an external resistance of $4.0\text{ \Omega}$. What is the terminal voltage of the cell?

Explanation: Current $I = \frac{E}{R + r} = \frac{2.1}{4.0 + 0.1} = \frac{2.1}{4.1} \approx 0.512\text{ A}$. Terminal voltage $V = E - Ir = 2.1 - (0.512)(0.1) = 2.1 - 0.051 = 2.049\text{ V} \approx 2.05\text{ V}$.

10. What is the primary role of the external DC source when charging a secondary cell?

Explanation: An external DC source forces current backwards through the cell, reversing the chemical reactions and restoring the original active materials.