A galvanometer of resistance R, is converted into an ammeter by using a shunt of resistance S to measure current I in the circuit. The resistance of the ammeter will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → slightly less than the resistance of the shunt S
An ammeter is made by connecting a small shunt S in parallel with the galvanometer of resistance R, so its resistance is that of the parallel combination.
\( \displaystyle R_A = \frac{RS}{R + S} \)
\( \displaystyle R_A = \frac{S}{1 + \frac{S}{R}} \)
Since S is much smaller than R, the term S/R is small but positive, so the denominator is slightly greater than 1 and \( R_A \) is slightly less than S.
Why the other options are incorrect
Option (2) slightly higher than the resistance of the shunt S — Incorrect. A parallel combination is always smaller than the smaller of the two resistances, never larger.
Option (3) equal to the resistance of galvanometer, R — Incorrect. R is much larger than S, while the ammeter resistance is even smaller than S.
Option (4) equal to R+ S — Incorrect. R + S is the resistance of a series combination; the shunt is connected in parallel.
NCERT (2026-27):
The Moving Coil Galvanometer – “… To overcome these difficulties, one attaches a small resistance \( r_s \), called shunt resistance, in parallel with the galvanometer coil; so that most of the current passes through the shunt. The resistance of this arrangement is,
\( \displaystyle R_G\, r_s / (R_G + r_s) \;\simeq\; r_s \qquad \text{if } R_G \gg r_s \)”