A battery is not a perfect voltage source — it has an internal resistance that causes its terminal voltage to drop under load. The EMF (electromotive force) is the open-circuit voltage; the terminal voltage is always less during discharge because some voltage is "lost" inside the battery. When multiple cells are combined — in series, parallel, or mixed — the resulting EMF and internal resistance depend on the arrangement. Choosing the right combination for a given external load is a classic JEE/NEET problem type. Maximum power transfer (when R = r) is the most frequently tested result from this topic at advanced level.
1. EMF, Internal Resistance and Terminal Voltage
EMF (ε or E) is the work done per unit charge by the battery's chemical energy source to move charge through the entire circuit (including through itself). It is the open-circuit terminal voltage.
Internal resistance (r) is the resistance offered by the electrolyte and electrodes inside the cell.
Terminal Voltage
| Condition | Formula | Comparison to EMF |
|---|
| Discharging (cell driving current) | V = E − Ir | V < E |
| Charging (current forced in) | V = E + Ir | V > E |
| Open circuit (I = 0) | V = E | V = E |
| Short circuit (R = 0) | I_sc = E/r; V = 0 | Maximum current; V = 0 |
Current in Circuit
Example: E = 12 V, r = 0.5 Ω, R = 5.5 Ω:
I = 12/(5.5+0.5) = 12/6 = 2 A; V_terminal = 12 − 2×0.5 = 11 V
2. Cells in Series
n identical cells (each EMF E, internal resistance r) connected in series:
Best for: High external resistance R ≫ nr. In series, total EMF is multiplied — useful when you need high voltage.
Note: If one cell is reversed: E_total = (n−2)E, r_total = nr (EMF of reversed cell subtracts).
3. Cells in Parallel
n identical cells (each EMF E, internal resistance r) connected in parallel:
Best for: Low external resistance R ≪ r/n. In parallel, internal resistance is reduced — useful when you need high current.
Each cell supplies only I/n — parallel combination extends battery life and handles high current loads better.
4. Mixed (Series-Parallel) Combination
m rows of n cells in series each, all m rows in parallel:
Maximum current when R = nr/m (external resistance = effective internal resistance).
5. Maximum Power Transfer
Power delivered to external resistance R:
Taking dP/dR = 0: Maximum power is delivered when R = r:
Verification: E = 10 V, r = 2 Ω: At R = 2 Ω → I = 10/4 = 2.5 A → P = (2.5)² × 2 = 12.5 W = E²/4r = 100/8 = 12.5 W ✓
Practice Questions
Q1 (NEET/Board): A battery of EMF 6 V and internal resistance 1 Ω is connected to an external resistance of 5 Ω. Find: (i) current, (ii) terminal voltage, (iii) power dissipated inside the battery.
(i) I = E/(R+r) = 6/(5+1) = 1 A
(ii) V_terminal = E − Ir = 6 − 1×1 = 5 V
(iii) P_internal = I²r = 1² × 1 = 1 W
Total power = I²(R+r) = 1 × 6 = 6 W = EI ✓
Q2 (JEE Main): Four identical cells each of EMF 1.5 V and internal resistance 0.5 Ω are connected (a) in series and (b) in parallel to an external resistance of 5 Ω. Compare the currents.
(a) Series: E_total = 4×1.5 = 6 V; r_total = 4×0.5 = 2 Ω
I_series = 6/(5+2) = 6/7 ≈ 0.857 A
(b) Parallel: E_total = 1.5 V; r_total = 0.5/4 = 0.125 Ω
I_parallel = 1.5/(5+0.125) = 1.5/5.125 = ≈ 0.293 A
Series gives higher current for R = 5 Ω ≫ r = 0.5 Ω → series combination is better here.
Q3 (MCQ): The internal resistance of a battery can be determined by:
A) Measuring EMF in open circuit and terminal voltage under load, then r = (E−V)/I
B) It cannot be measured
C) r = E × I
D) r = V/I
Answer: A). From V = E − Ir → r = (E − V)/I. Measure EMF (open circuit) with voltmeter, then connect known load R, measure terminal voltage V and current I. Compute r = (E−V)/I.
