Q.The resistance of a voltmeter is 300 Ω. It can measure the highest potential difference of 150 volt. The resistance that should be connected to make it a suitable ammeter to measure current up to 8 Amp is

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: …