As shown in the figure, the resistance of a galvanometer G can be found by the half-deflection method. Here the resistance R₂ is adjusted such that when the key K is closed the deflection in the galvanometer becomes half of the value as compared to when K is open.

Half-deflection is obtained at R₂ = 4 Ω and thus the galvanometer resistance is found to be 6 Ω. In this half-deflection condition the current (in mA) through the resistor R₁ is:

The correct answer is 694.44.

Step-by-step Explanation:

From the given circuit diagram, we can observe a voltage source of potential difference $V=10\text{ V}$ connected in series with a resistor $R_1$ and a galvanometer $G$. A resistance box $R_2$ is connected in parallel with the galvanometer through a key $K$.

Step 1: Deflection when key K is open
When the key $K$ is open, no current flows through resistor $R_2$. The initial current passing through the galvanometer is given by:

$I=\frac{V}{R_1+G}$

Step 2: Deflection when key K is closed
When key $K$ is closed, the parallel combination of the galvanometer resistance $G$ and shunt resistance $R_2$ gives an equivalent resistance $R_{\text{eq}}$:

$R_{\text{eq}}=\frac{G·R_2}{G+R_2}$

The total current flowing through the circuit (which is also the current through resistor $R_1$, denoted as $I_1$) is:

$I_1=\frac{V}{R_1+R_{\text{eq}}}=\frac{V}{R_1+\frac{G{R}_2}{G+R_2}}$

By using the current division rule, the new current flowing through the galvanometer $I^{\&\#prime;}$ becomes:

$I^{\&\#prime;}=I_1·\left(\frac{R_2}{G+R_2}\right)$

According to the half-deflection condition, $I^{\&\#prime;}=\frac{I}{2}$. Therefore:

$\frac{V}{R_1+\frac{G{R}_2}{G+R_2}}·\frac{R_2}{G+R_2}=\frac{1}{2}·\frac{V}{R_1+G}$

Step 3: Finding the value of $R_1$
We are given:

$G=6\text{ Ω},R_2=4\text{ Ω},V=10\text{ V}$

Substitute these values into the half-deflection condition equation:

$\frac{4}{R_1(6+4)+6·4}=\frac{1}{2(R_1+6)}$

Simplifying the denominator on the left-hand side:

$\frac{4}{10R_1+24}=\frac{1}{2R_1+12}$

Cross-multiplying to solve for $R_1$:

$4(2R_1+12)=10R_1+24$

$8R_1+48=10R_1+24$

$2R_1=24\Rightarrow R_1=12\text{ Ω}$

Step 4: Calculating current through resistor $R_1$
Now, calculate the equivalent resistance $R_{\text{eq}}$ of $G$ and $R_2$ in parallel:

$R_{\text{eq}}=\frac{6·4}{6+4}=\frac{24}{10}=2.4\text{ Ω}$

The total resistance of the circuit in the half-deflection condition is:

$R_{\text{total}}=R_1+R_{\text{eq}}=12+2.4=14.4\text{ Ω}$

The current through resistor $R_1$ (in Amperes) is:

$I_1=\frac{V}{R_{\text{total}}}=\frac{10}{14.4}\text{ A}=\frac{100}{144}\text{ A}$

Converting this current to milliamperes (mA):

$I_1\text{ (in mA)}=\frac{100}{144}·1000=\frac{100000}{144}\approx 694.44\text{ mA}$

Thus, the current through resistor $R_1$ in half-deflection condition is 694.44 mA.