Converting a Galvanometer into an Ammeter
For galvanometer resistance G and full-scale current I_g converted to ammeter range I, the shunt S satisfies I_g G = (I - I_g)S, so S = I_g G/(I - I_g).
Why this shows up in the exam
Designing ammeter ranges · Finding shunt current · Calculating effective instrument resistance
Learn the idea
A low shunt resistance bypasses most current so a galvanometer can measure a larger current. Only the safe full-scale current passes through the delicate coil; the rest takes a parallel low-resistance path. Equal voltage across the two parallel branches determines the shunt.
🧠 Memory hook: Ammeter means a small shunt in parallel.
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) — required parallel shunt for ammeter range I
- I_g G = (I - I_g) S — equal branch-voltage condition
- R_A = G S/(G + S) — effective ammeter resistance
How to approach it
- 1Mark coil and shunt currents
- 2Equate their voltage drops
- 3Solve S and, if needed, the parallel equivalent
Common slip-ups that cost marks
- •Putting the shunt in series
- •Using I rather than I - I_g in the shunt branch
- •Assuming the finished ammeter has zero resistance
🌟 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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