A galvanometer has a full-scale deflection current of 1.0 mA and a coil resistance of 50 Ω. A student wishes to convert it into an ammeter that reads up to ...

Marking Scheme

• (a) V = IR = 0.0010 × 50 [1]; = 0.050 V [1]
• (b) 0.050 V (same as across the galvanometer since they are in parallel) [1]
• (c) I_S = 5.0 - 0.0010 = 4.999 A (accept 5.0 A) [1]
• (d) S = V / I_S = 0.050 / 4.999 [1]; = 0.010 Ω [1]
• (e) S must carry almost all the current (4.999 A out of 5.0 A) [1]; a low resistance diverts most current through S rather than through the delicate galvanometer [1]

Explanation

(a) At full-scale deflection, the galvanometer carries 1.0 mA = 0.0010 A through its 50 Ω coil:

\( V = I_G \times R_G = 0.0010 \times 50 = 0.050 \text{ V} \)

(b) Since S is connected in parallel with the galvanometer, both have the same potential difference across them. Therefore V_S = 0.050 V.

(c) The total ammeter current is 5.0 A, of which the galvanometer carries 0.0010 A. By Kirchhoff's first law, the shunt carries the remainder:

\( I_S = 5.0 - 0.0010 = 4.999 \text{ A} \)

(d) Using Ohm's law for the shunt:

\( S = \frac{V_S}{I_S} = \frac{0.050}{4.999} = 0.01000 \; \Omega \approx 0.010 \; \Omega \)

This is 5000 times smaller than the galvanometer resistance.

(e) The shunt provides a low-resistance bypass for almost all the current. Of the 5.0 A total, only 0.001 A (0.02%) goes through the galvanometer, while 99.98% goes through S. If S were not much smaller than the galvanometer resistance, too much current would flow through the galvanometer coil, overloading it. The ratio of resistances determines how the current splits: since the components are in parallel, the lower-resistance path carries proportionally more current. This is the fundamental principle behind extending an ammeter's range.