In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k$$\Omega$$ are used. The figure of merit of the galvanometer is 60 $$\mu$$A division$$^{-1}$$. In the absence of shunt resistance, the galvanometer produces a deflection of $$\theta$$ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $$\frac{\theta}{2}$$, is closest to:

We have been told that the figure of merit of the galvanometer is $$k = 60\;\mu\text{A division}^{-1}$$. By definition,

$$I_g = k\,\theta$$

where $$I_g$$ is the current through the galvanometer and $$\theta$$ is the scale deflection in divisions. Substituting $$\theta = 9$$ divisions,

$$I_g = 60\;\mu\text{A}\times 9 = 540\;\mu\text{A}= 5.4\times10^{-4}\ \text{A}$$

In the first observation (no shunt connected) the 6 V battery drives this current through the series combination of the high resistance $$R = 11\ \text{k}\Omega = 11000\ \Omega$$ and the unknown resistance $$G$$ of the galvanometer itself. Applying Ohm’s law,

$$V = I_g\,(R+G)$$

Substituting $$V = 6\ \text{V}$$ and $$I_g = 5.4\times10^{-4}\ \text{A}$$,

$$6 = 5.4\times10^{-4}\,(11000 + G)$$

$$\Rightarrow\; 11000 + G = \frac{6}{5.4\times10^{-4}}$$

$$\Rightarrow\; 11000 + G = 11111.11$$

$$\Rightarrow\; G = 11111.11 - 11000 = 111.11\ \Omega$$

So, the resistance of the galvanometer is approximately $$G \approx 111\ \Omega$$.

We now connect a shunt resistance $$S$$ in parallel with the galvanometer. We wish the new deflection to be $$\dfrac{\theta}{2}$$, i.e. the current through the galvanometer must fall to

$$I_g' = \frac{I_g}{2} = \frac{5.4\times10^{-4}}{2}= 2.7\times10^{-4}\ \text{A}$$

Let $$R_p$$ be the equivalent resistance of $$G$$ in parallel with $$S$$. The formula for two resistances in parallel is first stated:

$$R_p = \frac{G\,S}{G+S}$$

The total resistance in the circuit after the shunt is connected is $$R + R_p$$, so the total current delivered by the battery is

$$I = \frac{V}{R + R_p} = \frac{6}{11000 + R_p}$$

Current division tells us that the current through the galvanometer branch is

$$I_g' = I\;\frac{S}{G+S}$$

We require $$I_g' = 2.7\times10^{-4}\ \text{A}$$. Substituting all relations step by step,

$$2.7\times10^{-4} = \frac{6}{11000 + R_p}\;\frac{S}{G+S}$$

Because $$R_p = \dfrac{G\,S}{G+S}$$, we replace it and obtain one single equation containing only the unknown $$S$$:

$$2.7\times10^{-4} = \frac{6}{11000 + \dfrac{G\,S}{G+S}}\;\frac{S}{G+S}$$

To eliminate the fractions we multiply both sides by the two denominators:

$$\Bigl(2.7\times10^{-4}\Bigr)\Bigl(11000 + \dfrac{G\,S}{G+S}\Bigr)(G+S) = 6\,S$$

It is more convenient, however, to work with the ratio $$\dfrac{I_g'}{I_g}$$, because $$I_g'/I_g = 1/2$$ by design. Starting from

$$\frac{I_g'}{I_g} = \frac{V}{R+R_p}\;\frac{S}{G+S}\;\frac{R+G}{V}= \frac{R+G}{R+R_p}\;\frac{S}{G+S}$$

and demanding $$\dfrac{I_g'}{I_g}= \dfrac12$$, we write

$$\frac{1}{2} = \frac{R+G}{R+R_p}\;\frac{S}{G+S}$$

Now substitute the known numerical values $$R = 11000\ \Omega$$, $$G = 111.11\ \Omega$$ and $$R_p = \dfrac{G\,S}{G+S}$$:

$$\frac12 = \frac{11000+111.11}{11000+\dfrac{111.11\,S}{111.11+S}}\;\frac{S}{111.11+S}$$

The numerator $$R+G = 11000+111.11 = 11111.11$$, so a single explicit equation in $$S$$ is obtained:

$$\frac12 = \frac{11111.11}{11000+\dfrac{111.11\,S}{111.11+S}}\;\frac{S}{111.11+S}$$

Multiply both sides by the denominators to clear the fractions. First multiply by 2, then by the two bracketed terms:

$$2\times 11111.11\,S = \Bigl(11000+\dfrac{111.11\,S}{111.11+S}\Bigr)\,(111.11+S)$$

Compute the left-hand side:

$$2\times11111.11\,S = 22222.22\,S$$

For the right-hand side we distribute carefully:

$$\Bigl(11000+\dfrac{111.11\,S}{111.11+S}\Bigr)\,(111.11+S)=11000(111.11+S)+111.11\,S$$

Now insert this in the equality:

$$22222.22\,S = 11000(111.11+S)+111.11\,S$$

Expand the product containing 11000:

$$11000\times111.11 = 1222210$$

and

$$11000\times S = 11000\,S$$

so that the right-hand side becomes

$$1222210 + 11000\,S + 111.11\,S = 1222210 + 11111.11\,S$$

Our equation is now purely linear in $$S$$:

$$22222.22\,S = 1222210 + 11111.11\,S$$

Bring the term $$11111.11\,S$$ to the left to collect like terms:

$$(22222.22 - 11111.11)\,S = 1222210$$

$$11111.11\,S = 1222210$$

Finally divide to obtain $$S$$:

$$S = \frac{1222210}{11111.11} \approx 110\ \Omega$$

The answer that is the closest to this calculated value is 110 $$\Omega$$.

Hence, the correct answer is Option D.