Wheatstone Bridge Balance and Imbalance
For arms P, Q in one branch and R, S in the other, with corresponding midpoints joined by the detector, balance requires P/Q = R/S. Off balance, solve the full network or use a Thevenin equivalent across the detector terminals.
Why this shows up in the exam
Measuring an unknown resistance · Post-office box circuits · Temperature-sensitive bridge output
Learn the idea
Bridge balance gives equal midpoint potentials; off balance, their divider-voltage difference drives the detector. At balance, the galvanometer connects equal-potential points and does not disturb the two divider branches. If one arm changes, first find the two midpoint potentials; their difference is the bridge output and can drive a finite detector current.
🧠 Memory hook: At balance cross products match; off balance compare midpoint voltages.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- P/Q = R/S — balanced bridge for the stated corresponding arms
- I_g = 0 and V_mid1 = V_mid2 — null condition independent of galvanometer resistance
- I_g = V_th/(R_th+G) — off-balance detector current using the bridge Thevenin equivalent
How to approach it
- 1Label four arms and detector terminals
- 2Test the balance ratio first
- 3If unbalanced, find midpoint voltage or the detector Thevenin equivalent
Common slip-ups that cost marks
- •Memorizing a ratio without matching diagram arms
- •Assuming galvanometer current is zero off balance
- •Ignoring finite detector resistance in an unbalanced bridge
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 2 ohm resistor. Find the circuit current.
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