A 50 $$\Omega$$ resistance is connected to a battery of 5 V. A galvanometer of resistance 100 $$\Omega$$ is to be used as an ammeter to measure current through the resistance, for this a resistance $$r_S$$ is connected to the galvanometer. Which of the following connections should be employed if the measured current is with in 1% of the current without the ammeter in the circuit?

We have a battery of e.m.f. $$E = 5\text{ V}$$ connected to a load resistance $$R = 50\;\Omega$$.

If no ammeter is present, the only resistance in the circuit is this 50 $$\Omega$$ resistor, so the current is obtained from Ohm’s law $$I_0 = \dfrac{E}{R}$$.

Substituting the given numbers,

$$I_0 = \dfrac{5\text{ V}}{50\;\Omega} = 0.10\ \text{A}$$

So the true current is $$0.10\ \text{A} = 100\ \text{mA}$$.

We now examine each proposed ammeter construction. In every case the ammeter is inserted in series with the 50 $$\Omega$$ resistor, so the total circuit resistance becomes

$$R_{\text{total}} = R + R_{\text{ammeter}}.$$

The requirement is that the current with the ammeter, $$I,$$ must differ from $$I_0$$ by less than 1 %, i.e.

$$\left|\dfrac{I - I_0}{I_0}\right| < 0.01.$$

The galvanometer resistance is $$R_g = 100\;\Omega.$$ We consider the four possibilities one by one.

Option A: $$r_S = 0.5\;\Omega$$ in series with galvanometer

Series combination: $$R_{\text{ammeter}} = R_g + r_S = 100 + 0.5 = 100.5\;\Omega.$$

Total resistance: $$R_{\text{total}} = 50 + 100.5 = 150.5\;\Omega.$$

Current: $$I = \dfrac{E}{R_{\text{total}}} = \dfrac{5}{150.5}\ \text{A} \approx 0.0332\ \text{A}.$$

Percentage error: $$\dfrac{0.10 - 0.0332}{0.10}\times100\% \approx 66.8\%.$$ This is far larger than 1 %, so Option A is rejected.

Option B: $$r_S = 1\;\Omega$$ in series with galvanometer

Series combination: $$R_{\text{ammeter}} = 100 + 1 = 101\;\Omega.$$

Total resistance: $$50 + 101 = 151\;\Omega.$$

Current: $$I = \dfrac{5}{151}\ \text{A} \approx 0.0331\ \text{A}.$$

Error again exceeds 66 %, so Option B fails the requirement.

Option C: $$r_S = 1\;\Omega$$ in parallel with galvanometer

For two resistances in parallel the equivalent resistance is given by

$$R_{\text{eq}} = \dfrac{R_g\,r_S}{R_g + r_S}.$$

Substituting $$R_g = 100\;\Omega$$ and $$r_S = 1\;\Omega,$$

$$R_{\text{ammeter}} = R_{\text{eq}} = \dfrac{100 \times 1}{100 + 1}\;\Omega = \dfrac{100}{101}\;\Omega \approx 0.990\;\Omega.$$

Total resistance: $$R_{\text{total}} = 50 + 0.990 = 50.990\;\Omega.$$

Current: $$I = \dfrac{5}{50.990}\ \text{A} \approx 0.0981\ \text{A}.$$

Percentage error: $$\dfrac{0.10 - 0.0981}{0.10}\times100\% \approx 1.9\%,$$ which is >1 %, so Option C is not acceptable.

Option D: $$r_S = 0.5\;\Omega$$ in parallel with galvanometer

Again using the parallel-resistance formula,

$$R_{\text{ammeter}} = \dfrac{R_g\,r_S}{R_g + r_S} = \dfrac{100 \times 0.5}{100 + 0.5}\;\Omega = \dfrac{50}{100.5}\;\Omega \approx 0.4975\;\Omega.$$

Total resistance: $$R_{\text{total}} = 50 + 0.4975 = 50.4975\;\Omega.$$

Current: $$I = \dfrac{5}{50.4975}\ \text{A} \approx 0.0990\ \text{A}.$$

Percentage error: $$\dfrac{0.10 - 0.0990}{0.10}\times100\% \approx 0.99\%.$$

This error is less than 1 %, fulfilling the stated condition.

Hence, the correct answer is Option D.