Physics · Current Electricity · NEET
The driver cell sends a steady current through the long potentiometer wire, so the potential falls evenly along it. This fall per unit length is the potential gradient, k (in V/cm), and it is the same at every point. At the balance point the cell's EMF exactly equals the potential drop across the balanced length: E1 = k L1 and E2 = k L2. Divide the two equations and k cancels, giving E1/E2 = L1/L2. So you never even need to know k to compare two cells.
It measures true EMF, not terminal voltage. At the balance point the galvanometer shows zero, which means no current is drawn from the test cell. With zero current, there is no drop across the cell's internal resistance (I r = 0), so the terminal voltage equals the EMF. This is the whole reason a potentiometer is more accurate than a voltmeter, which always draws a small current and reads terminal voltage instead.
The test cell's EMF and the potential drop across the balanced wire length oppose each other. When they are exactly equal, there is no net voltage to push current through the galvanometer, so it reads zero. This 'null' condition is key for NEET: no current means no I r loss, so you are reading the real EMF. Adjust the jockey until the galvanometer needle sits at zero — that contact point gives the balance length.
Connect the two cells so they help each other (aiding), and the net EMF is E1 + E2, balanced at length L_sum. Then reverse one cell so they oppose, giving net EMF E1 - E2, balanced at length L_diff. Because balance length is proportional to net EMF: (E1 + E2)/(E1 - E2) = L_sum/L_diff. Solve this to get E1/E2. This is exactly what NEET 2016 asked.
The potentiometer can only balance a cell if the wire can supply a potential drop equal to that cell's EMF. The maximum drop available is across the full wire, set by the driver cell. If a test cell's EMF is greater than the total drop across the wire, no balance point exists anywhere and the galvanometer never reaches zero. So the driver (primary) cell must always have a higher EMF than any cell you compare.
A potentiometer wire is 100 cm long with a constant PD across it. Two cells in series give balance at 50 cm (aiding) and 10 cm (opposing). The ratio of their EMFs (E1 : E2) is:
In a potentiometer, a cell of EMF 1.5 V balances at 36 cm. If a 2.5 V cell replaces it, the balance length is:
A potentiometer is an accurate and versatile device to measure EMF because the method involves:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The potentiometer works on the principle that the potential drop across a uniform wire carrying a steady current is proportional to its length. Because balance length is proportional to EMF, comparing two cells reduces to comparing their balance lengths: E1/E2 = L1/L2.
E1/E2 = L1/L2, where L1 and L2 are the balance lengths for the two cells measured with the same potential gradient. The potential gradient cancels out, so you never need its value to compare cells.
A voltmeter always draws a small current, so it reads terminal voltage (EMF minus the I r drop). A potentiometer draws zero current at balance, so there is no I r drop and it reads the true EMF.
One-sided deflection with no null point usually means a wrong connection or that the driver cell EMF is smaller than the test cell EMF. Check the polarity of the cells and make sure the driver (primary) cell has the larger EMF so a balance point exists on the wire.
Aiding gives net EMF E1 + E2 balanced at L_sum; opposing gives E1 - E2 balanced at L_diff. Then (E1 + E2)/(E1 - E2) = L_sum/L_diff, which you solve to get E1/E2.
