A galvanometer having a resistance of 20 ohm and 30 divisions on both sides has figure of merit 0.00

Q: What exactly is the "figure of merit" in a galvanometer?

The figure of merit is simply the amount of current required to produce a deflection of exactly one division on the galvanometer's scale. It's a measure of the instrument's sensitivity.

Q: Why do we connect the resistance in series and not in parallel to make a voltmeter?

A voltmeter is always connected in parallel to the component whose voltage is being measured. If the voltmeter had low resistance, it would draw a large current, changing the circuit's behavior. Connecting a high resistance in series ensures the voltmeter has a very high overall resistance, minimizing its impact on the circuit.

Q: What would happen if we connected the galvanometer directly to the 15V source without the series resistance?

The current flowing through the galvanometer would be $I = V/G = 15/20 = 0.75\text{ A}$. Since its maximum safe current is only $0.15\text{ A}$, the excessive current would produce too much heat and instantly burn out the delicate coil.

Q: The question says "30 divisions on both sides". Why don't we use 60 divisions?

The needle starts at the center (zero mark) and can deflect up to 30 divisions to the left OR 30 divisions to the right, depending on the direction of the current. The maximum deflection in any one direction is 30 divisions, so we use $n=30$.

Q: How is converting to an ammeter different from converting to a voltmeter?

To make an ammeter, you want the device to have very low resistance so it doesn't obstruct the current it's trying to measure. Therefore, you connect a very small resistance (called a shunt) in parallel with the galvanometer, rather than a large resistance in series.

Question

A galvanometer having a resistance of $20\Omega$ and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is

Options:

(A) $100\Omega$

(B) $125\Omega$

(C) $120\Omega$

(D) $80\Omega$

Sigma Percentile · Accepted Answer

### Understanding the Galvanometer

Imagine a galvanometer as a highly sensitive, delicate instrument. It is designed to detect even the faintest trickle of electric current. In our problem, the galvanometer has a resistance of $G = 20\Omega$ and a scale with $30$ divisions on either side of the zero mark.

But what exactly is the **figure of merit**? Think of it as the "price" of moving the needle by just one single division. Here, the figure of merit is $k = 0.005	ext{ A/div}$. This means it takes $0.005	ext{ A}$ of current to push the needle by one mark.

To find the absolute maximum current this delicate device can handle before the needle hits the end of the scale, we calculate the **full-scale deflection current** ($I_g$). We simply multiply the total number of divisions by the figure of merit:
$$I_g = n 	imes k = 30 	imes 0.005 = 0.15	ext{ A}$$

This $0.15	ext{ A}$ is the absolute limit. If we push any more current through it, we risk damaging the coil!

### The Physics of a Voltmeter

Now, we face a challenge. We want to use this delicate galvanometer to measure a hefty potential difference of up to $V = 15	ext{ V}$. If we connect it directly across a $15	ext{ V}$ source, the current would be $I = \frac{V}{G} = \frac{15}{20} = 0.75	ext{ A}$, which is five times its maximum limit! The coil would burn out instantly.

Furthermore, a good voltmeter must have a very high resistance. Why? Because when you connect a voltmeter in parallel across a component to measure its voltage, you don't want the voltmeter to draw a significant amount of current and alter the very circuit you are trying to measure.

To solve both problems—protecting the galvanometer and ensuring high resistance—we connect a large resistance $R$ in **series** with the galvanometer.

### The Master Equation

By connecting $R$ in series, the total resistance of our new "voltmeter" becomes $(G + R)$. According to Ohm's Law, the maximum voltage $V$ this combination can measure occurs when the maximum safe current $I_g$ flows through it:
$$V = I_g(G + R)$$

This is our master equation. It beautifully links the desired voltage range to the physical constraints of our galvanometer.

### Final Calculation

Let's plug in the values we've gathered. We want a maximum voltage $V = 15	ext{ V}$, our full-scale current is $I_g = 0.15	ext{ A}$, and the galvanometer's internal resistance is $G = 20\Omega$:
$$15 = 0.15(20 + R)$$

To isolate $R$, we first divide both sides by $0.15$:
$$20 + R = \frac{15}{0.15}$$

$$20 + R = 100$$

Finally, subtracting $20$ from $100$, we find the exact value of the series resistance required:
$$R = 100 - 20 = 80\Omega$$

By adding an $80\Omega$ resistor in series, we have successfully transformed our delicate galvanometer into a robust $15	ext{ V}$ voltmeter!

Animated Solution for Physics - Magnetic Effects of Current: A galvanometer having a resistance of $20Ω$ and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is

Select Answer:

Visualized Solution

\text{Galvanometer Parameters}

$I_{g} = n \times k$

$I_{g} = 30 \times 0.005$

$I_{g} = 0.15 A$

\text{Voltmeter Conversion}

$V = I_{g} (G + R)$

$15 = 0.15 (20 + R)$

$20 + R = \frac{15}{0.15}$

$20 + R = 100$

$R = 80Ω$

\text{Ammeter Conversion}

The Sigma Insight: Moving Coil Galvanometer

Understanding the Galvanometer

The Physics of a Voltmeter

The Master Equation

Final Calculation

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