What it does
A real cell is not a perfect source of voltage. Inside it there is some resistance, the internal resistance r. When a current I flows, a voltage I r is “lost” inside the cell, so the voltage at its terminals is less than its emf:
V = E − I r, with I = E / (R + r) for a load R.
This page shows a cell (the dashed box: an ideal emf E with r beside it) driving a rheostat, with an ammeter and a voltmeter across the cell. Change E, r and R and the meters, the formula with your numbers and the facts update at once. The graph shows either V against I, a straight line that falls from E with slope −r, or the power in the load against R, which peaks when R = r (maximum power transfer). In the last part you can record readings like a real practical and let the page find E and r from the best straight line.
It follows the Class 12 Physics chapter on current electricity (emf, internal resistance, terminal voltage, cells in circuits) and the practical on finding a cell’s internal resistance. Start with the ohm's law lab; for circuits with several cells use the kirchhoff's rules solver, and to combine loads, series and parallel.
How to use it
- Set the cell’s emf E and internal resistance r.
- Change the load resistance R. Open the switch to see the open-circuit voltage; Short circuit sets R to 0.
- Pick the graph: V against I or Power against R. The red dot is where you are now.
- Press Record this reading at three or more loads, then read the fitted E and r below the table. Changing E or r clears the table, because it is then a different cell.
Worked example
A 1.5 V cell with r = 0.5 Ω drives a 2.5 Ω load. Then I = 1.5 / (2.5 + 0.5) = 0.5 A, the voltage lost inside is 0.5 × 0.5 = 0.25 V and the terminal voltage is V = 1.5 − 0.25 = 1.25 V. The load takes 0.625 W and the cell wastes 0.125 W, so the efficiency is 1.25 / 1.5 = 83.3%. The most power any load could take is E² / 4r = 1.125 W, with R = 0.5 Ω.
Record readings at R = 2.5 Ω (0.5 A, 1.25 V) and R = 1 Ω (1 A, 1 V): the line through them meets the V axis at 1.5 V and has slope −0.5 Ω, which gives back E and r.
Tips
- A voltmeter across a cell on open circuit reads (very nearly) the emf, because no current flows.
- At maximum power transfer only half the energy reaches the load; power stations and batteries aim for high efficiency instead, with R much bigger than r.
- Old or cold cells have a bigger internal resistance, so their voltage sags more under load.
FAQ
Why does a car’s headlights dim when it starts? The starter motor draws a huge current, so I r is large and the battery’s terminal voltage drops.
Can the terminal voltage be more than the emf? Yes, while the cell is being charged: the current flows in backwards and V = E + I r. The kirchhoff's rules solver shows a cell being charged.
Is this an official NCERT resource? No. fibub is independent; the page follows the topics in the syllabus in its own words.
Where the numbers come from
- Terminal voltage V = E − I r with I = E / (R + r), the power E² R / (R + r)² in the load and its maximum E² / (4 r) at R = r: standard circuit results.
- Least-squares straight line through the recorded (I, V) readings: intercept E, slope −r.
- Checked by hand (1.5 V, 0.5 Ω, 2.5 Ω: 0.5 A, 1.25 V, 0.625 W; maximum 1.125 W at 0.5 Ω), by numerically scanning R for the power peak, and by fitting exact readings back to E and r.
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.