Question

As shown in the figure, the resistance of a galvanometer $G$ can be found by the half-deflection method. Here the resistance $R_2$ 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 $R_2 = 4\ \Omega$ and thus the galvanometer resistance is found to be $6\ \Omega$. In this half-deflection condition the current (in mA) through the resistor $R_1$ is:

Hint:

Remember that in the half-deflection method, $R_1$ is typically very large compared to $G$, but if values are small like in this problem, you must use the exact formula $R_1 = \frac{G R_2}{G - R_2}$ if applicable.
Always distinguish between the total current and the galvanometer current in parallel circuits.
Check the final units carefully as the question asks for mA.

Answer 1

Step 1: Understanding the Question:
The question describes the standard experimental setup for the half-deflection method used to measure galvanometer resistance.
In this circuit, a battery of e.m.f. $V$ (here $10\text{ V}$) is connected in series with a high resistance $R_1$ and the galvanometer $G$.
A shunt resistor $R_2$ is connected in parallel with the galvanometer via a key $K$.
When the key $K$ is open, the current flows only through $R_1$ and $G$.
When the key $K$ is closed, $R_2$ is adjusted such that the galvanometer reading drops to half its original value.
Step 2: Key Formula or Approach:
For the half-deflection method, if $R_1$ is much larger than $G$, the resistance $G$ is approximately equal to $R_2$.
However, more precisely, for half-deflection, the following relation holds:
\[ G = \frac{R_1 R_2}{R_1 - R_2} \]
The current through $R_1$ is the total current in the circuit when the key $K$ is closed:
\[ I_{R_1} = \frac{V}{R_1 + R_{parallel}} \text{ where } R_{parallel} = \frac{G R_2}{G + R_2} \]
Step 3: Detailed Explanation:

Calculating $R_1$:
Using the half-deflection condition $G = \frac{R_1 R_2}{R_1 - R_2}$ with $G = 6\ \Omega$ and $R_2 = 4\ \Omega$.
\[ 6 = \frac{4 R_1}{R_1 - 4} \implies 6R_1 - 24 = 4R_1 \implies 2R_1 = 24 \implies R_1 = 12\ \Omega. \]

Calculating total resistance when key $K$ is closed:
The galvanometer and $R_2$ are in parallel. Their equivalent resistance is:
\[ R_p = \frac{G \times R_2}{G + R_2} = \frac{6 \times 4}{6 + 4} = \frac{24}{10} = 2.4\ \Omega. \]
The total resistance of the circuit is $R_{total} = R_1 + R_p = 12 + 2.4 = 14.4\ \Omega$.

Calculating current through $R_1$:
The current through $R_1$ is the total source current since it is in series with the battery and the parallel combination.
\[ I_{R_1} = \frac{V}{R_{total}} = \frac{10}{14.4} \text{ A}. \]
\[ I_{R_1} \approx 0.694444 \text{ A}. \]
Converting to milliamperes (mA):
\[ I_{R_1} = 0.694444 \times 1000 = 694.44 \text{ mA}. \]

Step 4: Final Answer:
The resistance of the series resistor $R_1$ is determined to be $12\ \Omega$.
When the shunt is connected, the total circuit resistance becomes $14.4\ \Omega$.
The resulting current through $R_1$ is $694.44\text{ mA}$.