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EMF, Internal Resistance, Power and Combination of Cells

A cell does work on charges; the work per coulomb is its emf ε. Inside the cell there is a small internal resistance r. With current I, the terminal voltage is V = ε − Ir and I = ε/(R + r). Electrical power P = VI = I²R = V²/R, and energy W = Pt. Power delivered to R is largest when R = r. Cells in series: εeq = ε1 + ε2, req = r1 + r2. Cells in parallel: εeq = (ε1r2 + ε2r1)/(r1 + r2), 1/req = 1/r1 + 1/r2.

🎬 Step-by-step story

  1. The switch is open, so no current flows. The voltmeter reads the full emf ε = 6 V. emf is the energy the cell gives each coulomb of charge.
  2. Close the switch. Current I = ε/(R + r) flows. Some voltage, Ir, is used up inside the cell, so the terminal voltage V = ε − Ir is less than ε.
  3. Make R smaller. The current grows, the inside loss Ir grows, and V drops further.
  4. Now watch power. The green bar is I²R, the useful power in R. The red bar is I²r, wasted as heat inside the cell. The green bar is tallest when R = r.
  5. Join two cells in series: the emfs add and the internal resistances add. In parallel, identical cells keep the same emf but share r.
  6. Free play: change R, the number of cells and series or parallel. Predict V and I first.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Is emf a force?

No. It is the work the cell does per coulomb, in volts. The 'force' in its old name is only history.

Why does the voltmeter read less once current flows?

Part of the emf is used to push current through the cell's own resistance r. What reaches the terminals is ε − Ir.

What happens if I short-circuit a cell?

R ≈ 0, so I = ε/r, which is large, V ≈ 0 and the cell heats up and wears out fast.

Why is maximum power at R = r and not at R = 0?

At R = 0 the current is big but R gets no voltage; at very large R the current is tiny. The best balance is R = r.

When should cells be put in series and when in parallel?

Series when the external R is large (you need more push). Parallel when R is small compared with r (you need less internal resistance).

Does a cell's emf depend on its size?

No. emf depends on the chemicals. Size changes the internal resistance and how long the cell lasts.

What is emf?

A cell has two electrodes in an electrolyte. Chemical reactions push positive charge to one electrode and negative to the other. The cell does work to move charge against the field inside it, like a pump lifting water.

emf (ε) = work done by the cell per unit charge in taking it once round the complete circuit. Unit: volt. emf is not a force, despite the old name 'electromotive force'.

emf equals the potential difference across the terminals when no current flows (open circuit).

Internal resistance and terminal voltage

Current must also flow through the electrolyte inside the cell. The electrolyte resists it: this is the internal resistance r.

  1. Total resistance in the loop = R + r
  2. Current: I = ε /(R + r)
  3. Voltage used inside the cell = Ir
  4. Terminal voltage: V = ε − Ir = IR

So V < ε while the cell gives current. When the cell is being charged (current forced in the other way), V = ε + Ir, more than ε.

Finding r: r = (ε/V − 1) R = (ε − V)/I.

r depends on the electrolyte, the distance between plates, their area and the age of the cell.

emf versus potential difference

Electrical energy and power

When charge q moves through a potential drop V, the energy given is W = qV = VIt. Power P = W/t = VI.

For a resistor, using V = IR: P = I²R = V²/R. This becomes heat (Joule heating).

Energy bills use the kilowatt hour: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J (one 'unit').

Power in a whole circuit: the cell supplies εI. Of this, I²R goes to R and I²r is lost inside: εI = I²R + I²r.

Maximum power: P = ε²R/(R + r)². This is largest when R = r, and then Pmax = ε²/4r (efficiency only 50%).

Why power lines use high voltage: for the same power P = VI, higher V means smaller I, so the loss I²Rline in the cables becomes much smaller.

Cells in series

Join + of one cell to − of the next. The same current flows through both.

  1. Potential rises: ε1 − Ir1 and ε2 − Ir2
  2. Total: V = (ε1 + ε2) − I(r1 + r2)
  3. Compare with V = εeq − I req

εeq = ε1 + ε2, req = r1 + r2. If one cell is reversed, εeq = ε1 − ε2 but req still adds. For n identical cells: nε and nr, I = nε/(R + nr). Series is best when R is much larger than r.

Cells in parallel

Join all + terminals together and all − terminals together. Same V across both, currents add: I = I1 + I2.

  1. I1 = (ε1 − V)/r1, I2 = (ε2 − V)/r2
  2. Add and solve for V: V = (ε1r2 + ε2r1)/(r1 + r2) − I r1r2/(r1 + r2)

εeq = (ε1r2 + ε2r1)/(r1 + r2), req = r1r2/(r1 + r2). Handy form: εeq/req = ε1/r1 + ε2/r2. For n identical cells: ε and r/n. Parallel is best when R is much smaller than r.

Try it

In the 3D: open the switch and read V. Close it with R = 5 Ω and predict V before looking (ε = 6 V, r = 1 Ω). Then find the R that makes the green bar tallest.

At home (with an adult): in a car or bike with lights on, watch the headlight brightness when the engine is started. The dimming is the Ir drop of the battery.

Board exam focus

Derive equivalent emf and internal resistance for series and parallel (3–5 marks); V–I graph of a cell (straight line, slope −r, intercept ε); numericals on r, terminal voltage and power; difference between emf and terminal voltage.

Key formulas and definitions

Worked examples

1. A cell of emf 6 V and internal resistance 1 Ω is joined to a 5 Ω resistor. Find I and V.

Step 1: I = ε/(R + r) = 6/(5 + 1) = 1 A. Step 2: V = ε − Ir = 6 − 1 × 1 = 5 V. Check: V = IR = 5 V.

2. A cell reads 1.5 V on open circuit and 1.2 V when a 4 Ω resistor is joined. Find r.

Step 1: I = V/R = 1.2/4 = 0.3 A. Step 2: r = (ε − V)/I = 0.3/0.3. Answer: r = 1 Ω.

3. A 12 V battery with r = 0.5 Ω is charged by a current of 2 A. Terminal voltage during charging?

Step 1: while charging, V = ε + Ir. Step 2: V = 12 + 2 × 0.5 = 13 V.

4. A 1000 W heater runs 3 hours a day for 30 days. Units used and cost at ₹8 per unit?

Step 1: energy = 1 kW × 3 h × 30 = 90 kWh. Step 2: cost = 90 × 8 = ₹720.

5. A 220 V, 100 W bulb: find its resistance and the current it takes.

Step 1: R = V²/P = 220²/100 = 484 Ω. Step 2: I = P/V = 100/220 ≈ 0.45 A.

6. Three identical cells, each 2 V and 0.3 Ω, are in series with a 5.1 Ω resistor. Find the current.

Step 1: εeq = 3 × 2 = 6 V. Step 2: req = 3 × 0.3 = 0.9 Ω. Step 3: I = 6/(5.1 + 0.9) = 1 A.

7. Two cells, 4 V with 2 Ω and 2 V with 1 Ω, are in parallel (like terminals together). Find εeq and req.

Step 1: εeq = (ε1r2 + ε2r1)/(r1 + r2) = (4 × 1 + 2 × 2)/3 = 8/3 ≈ 2.67 V. Step 2: req = (2 × 1)/3 ≈ 0.67 Ω.

8. A cell of emf 6 V and r = 2 Ω. What R draws maximum power, and how much?

Step 1: maximum power when R = r = 2 Ω. Step 2: I = 6/(2 + 2) = 1.5 A. Step 3: P = I²R = 2.25 × 2 = 4.5 W (= ε²/4r = 36/8).

Common mistakes

Practice quiz

1. Terminal voltage of a cell giving current I is:
2. emf equals terminal voltage when:
3. Power delivered to external R is maximum when:
4. n identical cells (ε, r) in parallel give:
5. 1 kWh equals:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is the difference between emf and terminal voltage?

emf is the voltage of the cell when no current flows. Terminal voltage is the voltage across its terminals while current flows: V = ε − Ir, so it is lower.

What is internal resistance of a cell?

It is the resistance offered by the electrolyte and electrodes inside the cell to the current flowing through it.

What is the formula for cells in parallel?

εeq = (ε1r2 + ε2r1)/(r1 + r2) and req = r1r2/(r1 + r2). For identical cells: εeq = ε, req = r/n.

Where this is taught

CBSE (India)Class 12Current Electricity
USA (Common Core, NGSS, AP)Grade 12Electric Circuits
USA (Common Core, NGSS, AP)Grade 12Electric Circuits
Russia10 классDirect current
Russia10 классElectrodynamics: direct current
China高一Compulsory 3 Ch.12 Electric energy

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