Q.What should be the value of the shunt to be connected in parallel to a galvanometer of resistance G, so that 1/n part of the main current will pass through the shunt? OR Equal number of identical cells are joined in series and again in parallel. Under what condition, will the currents in both the cases be the same?

The instrument's full-scale current is 0·5 A. To read up to 8 A it needs a low shunt in parallel: S = I_g R_g/(I − I_g) = 20 Ω. Option (a).

Step 1 — full-scale deflection current of the meter:

I_g = V_max/R = 150 V/300 Ω = 0·5 A.

Step 2 — to convert a galvanometer/meter into an ammeter of range I = 8 A, connect a shunt S in parallel so the excess current bypasses the meter:

I_g·R_g = (I − I_g)·S.

Step 3 — solve for S: …

A shunt in parallel converts a galvanometer to an ammeter.

To read a larger current than the galvanometer's own full-scale current $I_g$ can handle, a small resistance (shunt) is connected in parallel with the coil, so that most of the current bypasses the sensitive coil through the shunt, while only $I_g$ passes through the galvanometer it …

Connecting a very low resistance (shunt) in parallel with the galvanometer coil diverts most of the current through the shunt, letting the combination measure large currents.

A galvanometer is a sensitive device that gives full-scale deflection for a very small current $I_g$, and it has a coil resistance $G$. To use it as an ammeter capable of reading currents much larger than $I_g$, a low-value resistance called a shunt ($S$) is connected in parallel with the galvanometer.

The shunt provides a low-resistance bypass path so that only the small fraction $I_g$ flows through the galvanometer coil while the rest of the current $(I - I_g)$ flows through the shunt. Since the galvanometer and shunt are in parallel, the potential drop across each is equal: …