Galvanometer Voltage Sensitivity
With galvanometer resistance G and current sensitivity theta/I = NAB/k, the voltage sensitivity is theta/V = NAB/(kG), using V = IG for the bare galvanometer.
Why this shows up in the exam
Comparing voltage sensitivities · Predicting effects of coil resistance · Designing voltmeter ranges
Learn the idea
Voltage sensitivity equals current sensitivity divided by the galvanometer resistance. A sensitive coil may deflect strongly for current, but a large coil resistance reduces the current produced by a given applied voltage. Voltage response therefore depends on both mechanics and resistance.
🧠 Memory hook: Current sensitivity is mechanical; divide by resistance for voltage sensitivity.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- theta/V = N A B/(k G) — voltage sensitivity of the bare galvanometer
- theta/V = (theta/I)/G — relation between voltage and current sensitivity
- V_g = I_g G — full-scale voltage of the galvanometer
How to approach it
- 1Find current sensitivity
- 2Identify the relevant resistance
- 3Divide by resistance and compare ratios
Common slip-ups that cost marks
- •Assuming increased current sensitivity always increases voltage sensitivity
- •Multiplying by G instead of dividing
- •Using total voltmeter resistance as bare galvanometer G
🌟 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 charge of 2 microC moves perpendicular to a 3 T magnetic field at 4 x 10^5 m/s. Find the magnetic force.
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