A galvanometer of resistance G is converted into a voltmeter of range 0-1 V by connecting a resistan

Q: Why does the full-scale deflection current $i_g$ remain constant?

The full-scale deflection current is an intrinsic property of the galvanometer coil itself. It depends on the number of turns, magnetic field, and spring constant, which don't change when we add external resistors.

Q: Why do we connect the resistance in series to make a voltmeter?

A voltmeter needs to have a very high resistance so that it draws negligible current from the main circuit it's measuring. Connecting a high resistance in series achieves this.

Q: What if we wanted to convert it into an ammeter instead?

To convert a galvanometer into an ammeter, you would connect a very small resistance (called a shunt) in parallel with it, so most of the current bypasses the sensitive galvanometer coil.

Q: Can I just double $R_1$ to double the voltage range?

No, because the total initial resistance was $G + R_1$, not just $R_1$. To double the voltage, you must double the *total* resistance, which means adding an extra $G + R_1$.

Q: What happens if I connect the extra resistance in parallel?

Connecting a resistance in parallel would decrease the overall equivalent resistance. This would actually decrease the voltage range of the voltmeter, which is the opposite of what we want!

Question

A galvanometer of resistance $G$ is converted into a voltmeter of range $0-1	ext{ V}$ by connecting a resistance $R_1$ in series with it. The additional resistance that should be connected in series with $R_1$ to increase the range of the voltmeter to $0-2	ext{ V}$ will be

Options:

(A) $R_1 + G$

(B) $R_1$

(C) $G$

(D) $R_1 - G$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you have a galvanometer. It's a sensitive device that measures small currents. But what if you want to measure voltage? You can convert it into a voltmeter by adding a high resistance in series.

In our problem, we start with a galvanometer of resistance **$G$**. To make it measure up to $1	ext{ V}$, we connect a resistance **$R_1$** in series. Let's say the current required for full-scale deflection is **$i_g$**.

### The Master Equation
According to Ohm's law, the total voltage $V$ across a series circuit is the product of the current $I$ and the equivalent resistance $R_{	ext{eq}}$. 
Here, the equivalent resistance is simply the sum of the individual resistances:
$$R_{	ext{eq}} = G + R_1$$

Substituting our values into Ohm's law, we get our first master equation:
$$1 = i_g(G + R_1)$$

### Extending the Range
Now, we want to upgrade our voltmeter. We want it to measure up to $2	ext{ V}$. To do this, we need to increase the total resistance. We add an additional unknown resistance, let's call it **$x$**, in series with $R_1$.

The current for full-scale deflection remains $i_g$ because the galvanometer itself hasn't changed. Applying Ohm's law to this new, upgraded circuit gives us our second equation:
$$2 = i_g(G + R_1 + x)$$

### Final Calculation
We now have a system of two equations. The easiest way to solve for $x$ is to divide the first equation by the second. This elegantly cancels out the $i_g$ term, which we don't know anyway!

$$\frac{1}{2} = \frac{i_g(G + R_1)}{i_g(G + R_1 + x)}$$
$$\frac{1}{2} = \frac{G + R_1}{G + R_1 + x}$$

Now, it's just a matter of simple algebra. Cross-multiplying gives:
$$G + R_1 + x = 2(G + R_1)$$

Expanding the right side:
$$G + R_1 + x = 2G + 2R_1$$

Finally, isolating $x$:
$$x = 2G + 2R_1 - G - R_1$$
$$x = G + R_1$$

And there we have it! To double the range of our voltmeter, we need to add an additional resistance equal to the total initial resistance of the circuit.

Select Answer:

Visualized Solution

Initial Setup

Ohm's Law

First Equation

Extended Range Setup

Second Equation

Solving for $x$

Algebraic Manipulation

Final Answer

The Way Forward

The Sigma Insight: Electrical Instruments

Analyzing the Setup Imagine you have a galvanometer

The Master Equation According to Ohm's law, the total voltage $V$ across a series circuit is the product of the current $I$ and the equivalent resistance $R_{eq}$

Extending the Range Now, we want to upgrade our voltmeter

Final Calculation We now have a system of two equations

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