Galvanometer Conversion and Half-Deflection
For galvanometer resistance G, ammeter range I needs shunt S such that IgG = (I-Ig)S. Voltmeter range V needs series resistance R = V/Ig - G. Half-deflection methods require the actual circuit, not an automatic S = G shortcut.
Why this shows up in the exam
Converting galvanometers · Choosing instrument ranges · Half-deflection resistance measurement
Learn the idea
A low parallel shunt makes an ammeter; a high series resistance makes a voltmeter. A galvanometer can tolerate only full-scale current Ig. A shunt bypasses excess current for ammeter use, while a series multiplier limits current for voltmeter use.
🧠 Memory hook: Ammeter gets a shunt; voltmeter gets a series multiplier.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- S = I_g G/(I-I_g) — ammeter shunt for range I
- R_series = V/I_g - G — voltmeter multiplier for range V
- V_g = I_g G — full-scale galvanometer voltage
How to approach it
- 1Write full-scale branch currents
- 2Equate parallel voltage or series current
- 3Check the converted instrument's total resistance
Common slip-ups that cost marks
- •Putting an ammeter shunt in series
- •Neglecting G when it is not negligible
- •Assuming half-deflection directly equals G in every topology
🌟 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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