Physics · Current Electricity · NEET
Balance length is proportional to the potential the potentiometer taps. When the cell is open (switch to R is open), no current flows inside it, so it shows its full EMF -> balance length l1. When you close R across the cell, current flows, the internal resistance r drops some voltage, so the potentiometer now balances the smaller terminal voltage V -> length l2. Two states, two lengths. The gap between them is caused by r.
l1 = balance length for EMF (E), taken with the external resistance R disconnected (open circuit, so terminal voltage = EMF). l2 = balance length for terminal voltage V, taken with R connected across the cell. Since length is proportional to voltage, E/V = l1/l2.
R forces a known current through the cell so the internal resistance actually drops voltage. Without R, the cell carries no current and terminal voltage always equals EMF - you could never see r. R is the tool that makes r reveal itself as a voltage drop.
Yes. l1 measures EMF, l2 measures terminal voltage, and terminal voltage V = E - I*r is always less than E when current flows. Smaller voltage means shorter balance length, so l1 > l2 always. If you get l2 > l1, you made a wiring or reading error.
E = I(R + r) and V = I*R, so E/V = (R + r)/R. But E/V = l1/l2 (lengths track voltages). So l1/l2 = (R + r)/R. Solve: R*l1 = R*l2 + r*l2, giving r = R(l1 - l2)/l2. Same as r = R(l1/l2 - 1).
EMF comparison uses TWO cells and finds E1/E2 = l1/l2. The internal-resistance experiment uses ONE cell, and gets two lengths from the SAME cell (open vs loaded with R). Do not confuse the two lengths here as two different cells.
The EMF of a cell (internal resistance 1 ohm) balances at 330 cm on a potentiometer. With a 2 ohm external resistance across the cell, the balance length is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
r = R(l1 - l2)/l2, where R is the external resistance across the cell, l1 is the balance length for EMF (open circuit) and l2 is the balance length for terminal voltage (R connected).
At the balance point no current is drawn from the test cell, so it measures the true EMF and true terminal voltage without loading errors. A voltmeter always draws some current and gives slightly wrong values. This makes the potentiometer more accurate.
It is the point on the wire where the galvanometer shows zero deflection. At that spot the potentiometer's potential drop exactly equals the cell's voltage, so no current flows through the galvanometer branch.
The formula assumes the potential gradient (volt per cm) stays constant. If the driver battery current drifts, l1 and l2 are no longer comparable. Keep the driving current steady and use a fresh, stable driver cell.
No. Physically r is positive, so l1 must be greater than l2. A negative result means you swapped the readings or the driver polarity is wrong.
