As shown in the figure, the resistance of a galvanometer G can be found by the half-deflection metho

Q: Why does the current through the galvanometer decrease when the key is closed?

When the key is closed, the resistor $R_2$ provides an alternative, parallel path for the current. This 'shunts' some of the current away from the galvanometer, reducing the current flowing through it.

Q: Why do we use the current divider rule here?

The total current $i$ splits at the junction into two parallel branches: one through the galvanometer $G$ and one through $R_2$. The current divider rule is the most direct way to calculate the exact fraction of current going through each branch.

Q: Is the total current $i$ the same in both cases (key open and closed)?

No! Closing the key adds a parallel resistor, which decreases the overall equivalent resistance of the circuit. This causes the battery to supply a larger total current $i$ compared to the initial state.

Q: Why is $R_1$ typically a large value in this experiment?

$R_1$ is placed in series to limit the total current drawn from the battery. Galvanometers are highly sensitive instruments; without a large $R_1$, a high current could easily damage the coil.

Q: Can we use this method if the battery has internal resistance?

Yes, but the internal resistance would simply add to $R_1$ in all our equations. Since the problem doesn't mention it, we assume the battery is ideal with zero internal resistance.

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:

Sigma Percentile · Accepted Answer

## The Magic of the Half-Deflection Method

Welcome to a classic exploration of electrical instruments! The half-deflection method is an elegant experimental technique used to determine the internal resistance of a galvanometer without relying on an external ammeter. Let's decode the physics and mathematics behind this clever circuit.

### Analyzing the Setup

Our circuit consists of a $10	ext{V}$ battery, a series resistor $R_1$, a galvanometer with resistance $G = 6\ \Omega$, and a shunt resistor $R_2 = 4\ \Omega$ connected in parallel with the galvanometer through a key $K$.

To understand the system, we must analyze it in two distinct states: when the key $K$ is open, and when it is closed.

### The Master Equation: Key Open vs. Key Closed

**State 1: Key $K$ is Open**
When the key is open, the shunt resistor $R_2$ is disconnected. The current has only one path to flow—straight through $R_1$ and the galvanometer $G$. The total resistance of the circuit is simply the sum of these two components. By Ohm's law, the initial current $I_g$ flowing through the galvanometer is:

$$I_g = \frac{10}{R_1 + G} = \frac{10}{R_1 + 6}$$

**State 2: Key $K$ is Closed**
When we close the key, the current suddenly has a choice! It splits at the junction, dividing between the galvanometer and the shunt resistor $R_2$. The equivalent resistance of this parallel combination is:

$$R_p = \frac{G \cdot R_2}{G + R_2} = \frac{6 	imes 4}{6 + 4} = 2.4\ \Omega$$

Because the overall resistance of the circuit has decreased, the battery supplies a new, larger total current $i$:

$$i = \frac{10}{R_1 + R_p} = \frac{10}{R_1 + 2.4}$$

This total current $i$ reaches the parallel junction and divides. We need to find the new current flowing specifically through the galvanometer, which we will call $I_g'$. Using the **current divider rule**, we multiply the total current by the resistance of the *opposite* branch, divided by the sum of the parallel branches:

$$I_g' = i 	imes \frac{R_2}{G + R_2} = i 	imes \frac{4}{10} = \frac{4}{R_1 + 2.4}$$

### Applying the Half-Deflection Condition

Here is where the magic happens. The problem states that closing the key reduces the galvanometer's deflection to exactly half of its original value. Since deflection is directly proportional to current, this means the new current $I_g'$ is exactly half of the initial current $I_g$:

$$I_g' = \frac{I_g}{2}$$

Substituting our expressions into this condition gives us a beautiful algebraic equation:

$$\frac{4}{R_1 + 2.4} = \frac{1}{2} \left( \frac{10}{R_1 + 6} ight)$$

Simplifying the right side:

$$\frac{4}{R_1 + 2.4} = \frac{5}{R_1 + 6}$$

Now, we cross-multiply to solve for $R_1$:

$$4(R_1 + 6) = 5(R_1 + 2.4)$$

$$4R_1 + 24 = 5R_1 + 12$$

$$R_1 = 12\ \Omega$$

### Final Calculation

We are almost at the finish line! The question asks for the current flowing through $R_1$ in this half-deflection condition (when the key is closed). This is simply our total current $i$.

Substituting $R_1 = 12\ \Omega$ back into our equation for $i$:

$$i = \frac{10}{12 + 2.4} = \frac{10}{14.4}\ 	ext{A}$$

To convert this value into milliamperes (mA), we multiply by $1000$:

$$i = \frac{10}{14.4} 	imes 1000 = 694.44\ 	ext{mA}$$

And there we have it! The current through $R_1$ is **$694.44\ 	ext{mA}$**. The half-deflection method is a brilliant testament to how simple circuit laws can be manipulated to measure unknown properties with high precision.

Enter Numerical Value:

Visualized Solution

\text{Half-Deflection Method}

\text{Key } K \text{ is OPEN}

\text{Key } K \text{ is CLOSED}

\text{Current Division}

\text{Half-Deflection Condition}

\text{Solving for } R_1

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Applying the Half-Deflection Condition

Final Calculation

Similar Questions

A moving coil galvanometer of resistance $100Ω$ is used as an ammeter using a resistance $0.1Ω$ . The maximum deflection current in the galvanometer is $100 μ A$ . Find the current in the circuit, so that the ammeter shows maximum deflection.

A galvanometer having a coil resistance $100Ω$ gives a full scale deflection when a current of $1 mA$ is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of $10 V$ ?

In the circuit, the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance, the value of the resistor $R$ will be

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 Ω$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

When a current of $5 mA$ is passed through a galvanometer having a coil of resistance $15 Ω$ , it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range $0 - 10 V$ is

A potentiometer wire $A B$ having length $L$ and resistance $12 r$ is joined to a cell $D$ of EMF $ε$ and internal resistance $r$ . A cell $C$ having emf $\frac{ε}{2}$ and internal resistance $3 r$ is connected. The length $A J$ at which the galvanometer as shown in figure shows no deflection is

A galvanometer gives full scale deflection with $0.006 A$ current. By connecting it to a $4990 Ω$ resistance, it can be converted into a voltmeter of range $0 - 30 V$ . If connected to a $\frac{2 n}{249} Ω$ resistance, it becomes an ammeter of range $0 - 1.5 A$ . The value of $n$ is

Shown in the figure adjacent is a meter-bridge set up with null deflection in the galvanometer. The value of the unknown resistor $R$ is