Current Through Resistor in Galvanometer Half Deflection Method

To find the current through the resistor $R_1$ in the half-deflection condition, we analyze the circuit in both states (when key $K$ is open and when key $K$ is closed).

1. Circuit Parameters from Diagram and Given Text

• Voltage of the source, $V = 10\text{ V}$
• Resistance of the galvanometer, $G = 6\ \Omega$
• Shunt resistance for half-deflection, $R_2 = 4\ \Omega$

---

2. Derivation of Resistance $R_1$

Case 1: Key $K$ is Open

When the key $K$ is open, no current flows through $R_2$. The total resistance of the circuit is $R_1 + G$. The current flowing through the galvanometer is given by: $I_g = \frac{V}{R_1 + G}$

Case 2: Key $K$ is Closed (Half-deflection Condition)

When key $K$ is closed, resistor $R_2$ is connected in parallel with the galvanometer $G$. The equivalent resistance of this parallel combination $R_p$ is: $R_p = \frac{G R_2}{G + R_2}$

The total resistance of the circuit is $R_{\text{total}} = R_1 + R_p$. The total current supplied by the battery (which is the current flowing through $R_1$) is: $I = \frac{V}{R_1 + R_p} = \frac{V}{R_1 + \frac{G R_2}{G + R_2}}$

Using the current divider rule, the current through the galvanometer in this case, $I_g'$, is: $I_g' = I \times \frac{R_2}{G + R_2} = \frac{V}{R_1 + \frac{G R_2}{G + R_2}} \times \frac{R_2}{G + R_2} = \frac{V R_2}{R_1 (G + R_2) + G R_2}$

According to the half-deflection condition, $I_g' = \frac{1}{2} I_g$: $\frac{V R_2}{R_1 (G + R_2) + G R_2} = \frac{1}{2} \left( \frac{V}{R_1 + G} \right)$

Simplifying the equation: $\frac{R_2}{R_1 G + R_1 R_2 + G R_2} = \frac{1}{2(R_1 + G)}$ $2 R_2 (R_1 + G) = R_1 G + R_1 R_2 + G R_2$ $2 R_1 R_2 + 2 G R_2 = R_1 G + R_1 R_2 + G R_2$ $R_1 R_2 + G R_2 = R_1 G$ $R_1 (G - R_2) = G R_2 \implies R_1 = \frac{G R_2}{G - R_2}$

---

3. Calculation of $R_1$ and Total Current $I$

Substitute $G = 6\ \Omega$ and $R_2 = 4\ \Omega$ into the formula for $R_1$: $R_1 = \frac{6 \times 4}{6 - 4} = \frac{24}{2} = 12\ \Omega$

Now, calculate the equivalent parallel resistance $R_p$: $R_p = \frac{6 \times 4}{6 + 4} = \frac{24}{10} = 2.4\ \Omega$

The total resistance of the circuit in the half-deflection condition is: $R_{\text{total}} = R_1 + R_p = 12 + 2.4 = 14.4\ \Omega$

The current through $R_1$ in amperes is: $I = \frac{V}{R_{\text{total}}} = \frac{10}{14.4} = \frac{100}{144} = \frac{25}{36}\text{ A}$

Converting the current to milliamperes ($\text{mA}$): $I = \frac{25}{36} \times 1000\text{ mA} = \frac{25000}{36}\text{ mA} = \frac{6250}{9}\text{ mA} \approx 694.44\text{ mA}$

---

Final Answer

The current through the resistor $R_1$ in the half-deflection condition is $694.44\text{ mA}$ (or $\frac{6250}{9}\text{ mA}$).