Physics Class 12 Part I…
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Emf Internal Resistance Cells

🎓 Class 12 Physics CBSE Theory Ch 3 – Current Electricity ⏱ ~14 min
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Emf Internal Resistance Cells

3.11 Electrical Energy & Power

When a current I flows for time Δt across a potential difference V, the energy delivered is W = VIΔt. The rate of doing work — power — is

\[ P = V\,I = I^2 R = \dfrac{V^2}{R} \]

SI unit: watt (W). The energy dissipated as heat in a resistor (Joule heating) is

\[ H = I^2 R\,t \]

This is why incandescent bulbs and toasters work — current through a high-resistance filament heats it until it glows or radiates heat.

3.12 Cells, EMF and Internal Resistance

A cell is a device that maintains a steady potential difference between its terminals using internal chemical reactions. The two terminals are called the positive (+) and negative (−) electrodes, dipped in an electrolyte. Two basic quantities describe a real cell.

EMF (Electromotive Force): The EMF ε of a cell is the potential difference between its terminals when no current flows. It equals the work done by the cell's chemical reaction per unit positive charge.

Internal Resistance: The internal resistance r is the resistance to current flow offered by the electrolyte and electrodes inside the cell.
Real cell ε r B A R I
Fig. 3.8: A real cell modelled as an ideal EMF source ε in series with internal resistance r, connected to an external resistor R.

3.12.1 Terminal Voltage

When a current I is drawn from the cell, there is a voltage drop Ir across the internal resistance. The terminal voltage V (between A and B externally) is

\[ V = \varepsilon - I r \quad\text{(discharging)}\qquad V = \varepsilon + I r \quad\text{(charging)} \]

Note V < ε while the cell delivers current; V → ε only in open-circuit (I = 0). For an external resistor R, the circuit equation \(\varepsilon = I R + I r\) gives

\[ I = \dfrac{\varepsilon}{R + r} \]
Maximum power transfer: The power delivered to R is \(P = I^2 R = \varepsilon^2 R / (R+r)^2\). Setting dP/dR = 0 gives R = r. So a cell delivers maximum power to the load when the external resistance equals the internal resistance.

Interactive Simulation: Terminal Voltage Explorer L4 Analyse

Vary EMF, internal resistance, and external load. See how the terminal voltage and current respond. Try R = r — power delivered is maximum.

Current I = ε/(R+r) = 1.00 A
Terminal voltage V = ε − Ir = 5.00 V
Power to load P = I²R = 5.00 W
Power dissipated in r = I²r = 1.00 W

3.13 Cells in Series and Parallel

Real circuits often need higher voltage or longer life than a single cell can provide. We connect cells in two basic ways:

3.13.1 Cells in Series

Two cells of EMFs ε₁, ε₂ and internal resistances r₁, r₂ joined so that the negative terminal of one connects to the positive of the next:

\[ \varepsilon_\text{eq} = \varepsilon_1 + \varepsilon_2,\qquad r_\text{eq} = r_1 + r_2 \]

If the cells are connected the wrong way (one reversed), \(\varepsilon_\text{eq} = \varepsilon_1 - \varepsilon_2\) (the larger one wins).

ε₁,r₁ A ε₂,r₂ C ≡ ε₁+ε₂, r₁+r₂
Fig. 3.9: Two cells in series — EMFs and internal resistances both add (when oriented same way).

3.13.2 Cells in Parallel

If two cells of the same EMF but different internal resistances are connected in parallel (positive to positive), the equivalent EMF and internal resistance for n identical cells (each ε, r) are:

\[ \varepsilon_\text{eq}=\varepsilon,\qquad r_\text{eq} = \dfrac{r}{n} \]

For two cells of different parameters (ε₁, r₁) and (ε₂, r₂):

\[ \varepsilon_\text{eq}=\dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2},\qquad \dfrac{1}{r_\text{eq}}=\dfrac{1}{r_1}+\dfrac{1}{r_2} \]
B A ε₁,r₁ r₁ ε₂,r₂ r₂ ≡ ε_eq, r_eq
Fig. 3.10: Two cells in parallel — net EMF is a weighted average; internal resistances combine like parallel resistors.

Worked Example 1: Internal Resistance from Two Loads

A storage cell has EMF 6.0 V. When connected to a 6 Ω external resistor, the current is 0.95 A. Find the internal resistance and the terminal voltage.

\[ I = \dfrac{\varepsilon}{R+r}\Rightarrow 0.95 = \dfrac{6.0}{6+r}\] \[ 6+r = 6.0/0.95 = 6.316 \Rightarrow r \approx \boxed{0.316\ \Omega}\] Terminal voltage: V = ε − Ir = 6.0 − 0.95 × 0.316 ≈ 5.70 V.

Worked Example 2: Two Cells in Series

Two cells of EMFs 1.5 V and 2.0 V with internal resistances 0.5 Ω and 1.0 Ω respectively are connected in series with their positive terminals joined alternately to negative ones (i.e. aiding each other). The combination drives current through an external resistor of 2.0 Ω. Find the current.

ε_eq = 1.5 + 2.0 = 3.5 V; r_eq = 0.5 + 1.0 = 1.5 Ω.
I = ε_eq/(R + r_eq) = 3.5/(2.0 + 1.5) = 3.5/3.5 = 1.0 A.

Worked Example 3: Two Cells in Parallel (Different EMFs)

Cells of ε₁ = 2.0 V, r₁ = 1.0 Ω and ε₂ = 1.5 V, r₂ = 0.5 Ω are connected in parallel (same polarity to the same node). Find equivalent EMF and internal resistance, and the current through an external resistor of 1.0 Ω.

\[ \varepsilon_\text{eq} = \dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1+r_2} = \dfrac{2.0(0.5)+1.5(1.0)}{1.5} = \dfrac{2.5}{1.5} \approx 1.67\text{ V} \] \[ \dfrac{1}{r_\text{eq}} = \dfrac{1}{1.0}+\dfrac{1}{0.5}=3 \Rightarrow r_\text{eq} = 1/3 \approx 0.33\text{ Ω} \] \[ I = \dfrac{1.67}{1.0+0.33} \approx \boxed{1.25\ \text{A}} \]
Activity 3.3 — Detecting Internal ResistanceL4 Analyse

Materials: 1.5 V dry cell, voltmeter, ammeter, variable resistor (rheostat), key, wires.

Procedure:

  1. Read the cell's open-circuit voltage with a voltmeter (no current). Call it ε.
  2. Connect a small external resistor R via the rheostat. Read I (ammeter) and V (voltmeter across cell).
  3. Reduce R; read another (I, V) pair.
  4. Plot V (y-axis) vs I (x-axis).
Predict: What slope and intercept will the V–I graph have?

Observation: V = ε − Ir is a straight line; intercept on the V-axis = ε; slope = −r.

Conclusion: A simple V–I plot directly reveals both the EMF (intercept) and the internal resistance (slope) of the cell.

Competency-Based Questions

A torch uses two 1.5 V dry cells of internal resistance 0.4 Ω each, connected in series, to drive a bulb of resistance 4.0 Ω. As the cells age, their internal resistance increases. The student also tries connecting the cells in parallel.

Q1. The current through the bulb when the cells are fresh and in series is: L3 Apply

  • (a) 0.6 A
  • (b) 0.625 A
  • (c) 0.75 A
  • (d) 1.0 A
(b) ε_eq = 3.0 V, r_eq = 0.8 Ω. I = 3.0/(4.0+0.8) = 3/4.8 = 0.625 A.

Q2. Short Answer: Why does the brightness of the torch decrease as the dry cells age? L4 Analyse

As the cells age, their internal resistance r increases due to changes in the electrolyte and electrodes. For the same ε, the current I = ε/(R+r) decreases, the terminal voltage V = ε − Ir drops, and so the power IV delivered to the bulb falls — the bulb glows dimmer.

Q3. To get maximum power transferred to the bulb, the student should choose: L3 Apply

  • (a) external R much larger than r
  • (b) external R much smaller than r
  • (c) external R equal to r
  • (d) external R = 0
(c). dP/dR = 0 gives R = r — the maximum-power-transfer theorem.

Q4. Fill in the blank: The terminal voltage of a cell becomes equal to its EMF only when ____. L2 Understand

no current is drawn from it (i.e., open circuit, I = 0). Then the Ir drop vanishes and V = ε.

Q5. HOT: A cell of EMF 12 V and internal resistance 2 Ω is being charged by a 16 V supply. Find (i) the charging current and (ii) the terminal voltage of the cell during charging. L6 Create

(i) Driving voltage = 16 − 12 = 4 V acts across the internal resistance only (assume no series resistor). I = 4/2 = 2 A.
(ii) During charging, current flows INTO the + terminal, so V = ε + Ir = 12 + 2 × 2 = 16 V. Notice V > ε while charging — opposite to the discharging case.

Assertion–Reason Questions

Choose: (A) Both A and R true; R explains A. (B) Both true; R does not explain A. (C) A true, R false. (D) A false, R true.

Assertion (A): The terminal voltage of a cell is always less than its EMF when the cell delivers current to an external circuit.

Reason (R): A potential drop equal to Ir occurs across the internal resistance.

(A) Both true and R explains A. V = ε − Ir.

Assertion (A): When two identical cells of EMF ε are connected in parallel, the equivalent EMF is 2ε.

Reason (R): Cells in parallel share the load.

(D). A is FALSE — equivalent EMF for identical parallel cells equals ε (not 2ε); only the internal resistance halves. R is essentially true.

Assertion (A): A car battery delivers thousands of amperes when starting the engine, even though its EMF is only 12 V.

Reason (R): Car batteries are designed with very low internal resistance (~0.01 Ω).

(A) Both true and R explains A. With small r, I = ε/(R+r) can be huge for a small starter resistance.

Frequently Asked Questions - Emf Internal Resistance Cells

What is the main concept covered in Emf Internal Resistance Cells?
In NCERT Class 12 Physics Chapter 3 (Current Electricity), "Emf Internal Resistance Cells" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Emf Internal Resistance Cells useful in real-life applications?
Real-life applications of "Emf Internal Resistance Cells" from NCERT Class 12 Physics Chapter 3 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Emf Internal Resistance Cells?
Key formulas in "Emf Internal Resistance Cells" (NCERT Class 12 Physics Chapter 3 Current Electricity) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 3?
NCERT Class 12 Physics Chapter 3 (Current Electricity) is structured so each part builds on the previous one. "Emf Internal Resistance Cells" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Emf Internal Resistance Cells?
CBSE board questions from "Emf Internal Resistance Cells" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Emf Internal Resistance Cells" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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