Physics · Current Electricity · NEET
Balanced means the galvanometer reads zero. This happens because the two junctions on the galvanometer diagonal (call them B and D) reach the SAME potential. When two points have equal potential, there is no potential difference across the galvanometer, so no current flows through it. The bridge is then said to be at its null point.
Current flows only when there is a potential difference. At balance, the four arms divide the voltage so that V_B = V_D. Since both ends of the galvanometer are at the same potential, the drop across it is zero, so the current through it is zero. This is why a galvanometer (a null detector) is used instead of an ammeter.
At balance no current goes through the galvanometer, so the current I1 flows straight through P then Q, and I2 flows through R then S. Equal potentials at B and D give two loop equations: I1·P = I2·R (top loop) and I1·Q = I2·S (bottom loop). Divide the first by the second: P/Q = R/S. That is the balance condition.
Write the condition as P·S = Q·R. Here P and Q are the two arms in the first branch (in series), and R and S are the two arms in the second branch. The condition links the ratio of the top-to-bottom resistances in one branch to the same ratio in the other branch. Keep the diamond order fixed so you do not swap arms by mistake.
No. At balance the galvanometer current is zero, so its resistance never enters the equations. The condition P/Q = R/S depends only on the four arm resistances. The galvanometer resistance only affects sensitivity (how sharply the null appears), not the balance point itself. The same is true if you interchange the cell and galvanometer positions.
A Wheatstone bridge is used to determine an unknown resistance X by adjusting the variable resistance Y. For the most precise measurement of X, the resistances P and Q should be:
If the galvanometer G shows no deflection in the circuit shown (a 2 V cell with galvanometer in one branch, a 400 Ω resistor and unknown R in the main loop across 10 V), the value of R is:
A uniform metallic wire of resistance 4 Ω is bent to form a square loop ABCD. A resistance of 2 Ω is connected between B and D, and a 2 V battery is connected across A and C. The current i drawn from the battery is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It works on the null (balance) principle: when the four arms satisfy P/Q = R/S, the two galvanometer junctions are at equal potential, so the galvanometer shows zero deflection. This lets you find an unknown resistance accurately without reading currents, which removes errors from the cell voltage.
P/Q = R/S, which is the same as P·S = Q·R. Only the four arm resistances appear; the galvanometer and cell resistances do not.
It is a null method. At balance you only need to detect zero current, not measure a current or voltage value. Since it needs no reading of meter scales and does not depend on the exact cell EMF, it avoids calibration and voltage-drop errors.
A metre bridge is a practical Wheatstone bridge where two of the arms are replaced by a 1 m uniform wire. The balance point (jockey position) sets the ratio of these two arms as l/(100 − l), and the balance condition becomes R/S = l/(100 − l).
No. Because the galvanometer current is zero at balance, swapping the cell and galvanometer positions gives exactly the same condition P/Q = R/S, so the balance point does not shift.
