QJEEM24Phd3f4: JEE Main 2024 Physics - Class 12 JEEM24Phd3f4 Solution: In an ammeter, of the main current passes through the…

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: $\frac{G}{20}$ .