JEE Main 2024 — Current Electricity Question with Solution

When an ammeter is designed, a shunt resistor ($R_s$) is placed in parallel with the galvanometer to ensure that only a small fraction of the total current passes through the galvanometer itself. This is done because the galvanometer is usually a sensitive instrument designed to measure small currents, and passing a large current through it could damage it.

In this scenario, we are told that $5\%$ of the main current passes through the galvanometer. This means that the remaining $95\%$ of the current must pass through the shunt. Let's denote the total current as $I$, the current through the galvanometer as $I_g$, and the current through the shunt as $I_s$. Therefore, we have:

$$I_g=\frac{5}{100}I$$

$$I_s=I-I_g=I-\frac{5}{100}I=\frac{95}{100}I$$

Since the galvanometer and shunt are in parallel, the voltage across each must be the same:

$$V_g=V_s$$

According to Ohm's law, $V=IR$, where $V$ is the voltage, $I$ is the current, and $R$ is the resistance. Hence, for the galvanometer and the shunt:

$$I_{g}G=I_s{R}_s$$

By substituting $I_g$ and $I_s$ from the above proportionality, we get:

$$\left(\frac{5}{100}I\right)G=\left(\frac{95}{100}I\right)R_s$$

Let's solve for $R_s$:

$$R_s=\frac{5}{95}G$$

Further simplifying this:

$$R_s=\frac{G}{19}$$

The total resistance of the ammeter $R_a$ can be found using the parallel resistance formula:

$$\frac{1}{R_a}=\frac{1}{G}+\frac{1}{R_s}$$

Substitute $R_s$ with $\frac{G}{19}$:

$$\frac{1}{R_a}=\frac{1}{G}+\frac{19}{G}$$

$$\frac{1}{R_a}=\frac{20}{G}$$

Thus the resistance of the ammeter $R_a$ is:

$$R_a=\frac{G}{20}$$

Hence, the correct answer is Option C: $\frac{G}{20}$.