In an ammeter, $5 \%$ of the main current passes through the galvanometer. If resistance of the galvanometer is $\mathrm{G}$, the resistance of ammeter will be :

<p>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.</p> <p>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:</p> <p>$I_g = \frac{5}{100} I$</p> <p>$I_s = I - I_g = I - \frac{5}{100} I = \frac{95}{100} I$</p> <p>Since the galvanometer and shunt are in parallel, the voltage across each must be the same:</p> <p>$V_g = V_s$</p> <p>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:</p> <p>$I_g G = I_s R_s$</p> <p>By substituting $I_g$ and $I_s$ from the above proportionality, we get:</p> <p>$\left(\frac{5}{100} I\right) G = \left(\frac{95}{100} I\right) R_s$</p> <p>Let's solve for $R_s$:</p> <p>$R_s = \frac{5}{95} G$ </p> <p>Further simplifying this:</p> <p>$R_s = \frac{G}{19}$ </p> <p>The total resistance of the ammeter $R_a$ can be found using the parallel resistance formula:</p> <p>$\frac{1}{R_a} = \frac{1}{G} + \frac{1}{R_s}$ </p> <p>Substitute $R_s$ with $\frac{G}{19}$:</p> <p>$\frac{1}{R_a} = \frac{1}{G} + \frac{19}{G}$</p> <p>$\frac{1}{R_a} = \frac{20}{G}$</p> <p>Thus the resistance of the ammeter $R_a$ is:</p> <p>$R_a = \frac{G}{20}$</p> <p>Hence, the correct answer is Option C: $\frac{G}{20}$.</p>