When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 ohm, it shows

Q: Why do we connect the resistance in series and not parallel?

A voltmeter is connected in parallel across the component whose voltage we want to measure. To ensure it doesn't draw a significant current from the main circuit (which would alter the actual voltage being measured), a voltmeter must have a very high resistance. Connecting a large resistor in series with the galvanometer achieves this high overall resistance.

Q: What happens if we don't use a series resistance at all?

A galvanometer is extremely sensitive and has a low resistance. If connected directly across a high voltage source, a massive current would flow through it, far exceeding its full-scale deflection limit ($I_g$), and the delicate coil would instantly burn out.

Q: Why is $I_g$ used in the formula for maximum voltage?

The full-scale deflection current ($I_g$) is the maximum current the galvanometer can safely handle. When the voltmeter is measuring its maximum rated voltage (in this case, 10 V), we want exactly $I_g$ to flow through it. This ensures the needle points to the maximum mark on the scale without damaging the instrument.

Q: How do I handle the milliampere (mA) unit in the calculation?

It is crucial to convert all units to standard SI units before performing calculations. Milli means $10^{-3}$, so you must convert $5\text{ mA}$ to $5 \times 10^{-3}\text{ A}$. Forgetting this conversion is a very common source of error!

Q: Is the resistance of an ideal voltmeter zero or infinite?

An ideal voltmeter has infinite resistance so that it draws absolutely zero current from the circuit it is measuring. While we can't achieve infinite resistance in reality, adding a very large series resistance $R_s$ makes the real voltmeter approximate an ideal one.

Question

When a current of $5	ext{ mA}$ is passed through a galvanometer having a coil of resistance $15\,\Omega$, 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 $0	ext{-}10	ext{ V}$ is

Options:

(A) $2.045 	imes 10^3\,\Omega$

(B) $2.535 	imes 10^3\,\Omega$

(C) $4.005 	imes 10^3\,\Omega$

(D) $1.985 	imes 10^3\,\Omega$

Sigma Percentile · Accepted Answer

Converting a delicate galvanometer into a robust voltmeter is a classic problem that beautifully illustrates the practical application of Ohm's Law. Let's embark on this journey to understand the physics and the math behind it.

### Analyzing the Setup

A galvanometer is a highly sensitive instrument designed to detect very small currents. In our problem, the galvanometer has an internal coil resistance of $G = 15\,\Omega$ and achieves full-scale deflection with a mere $I_g = 5	ext{ mA}$ (which is $5 	imes 10^{-3}	ext{ A}$).

If we were to connect this galvanometer directly across a $10	ext{ V}$ source, the current would be $I = \frac{V}{G} = \frac{10}{15} \approx 0.67	ext{ A}$. This is massively larger than its $5	ext{ mA}$ limit and would instantly burn out the delicate coil!

To prevent this and allow the device to measure up to $10	ext{ V}$, we must restrict the current. We achieve this by connecting a large resistance, $R_s$, in **series** with the galvanometer.

### The Master Equation

When the series resistance $R_s$ is added, the total resistance of our new 'voltmeter' becomes $(G + R_s)$.

According to Ohm's Law, the maximum voltage $V$ that this setup can safely measure corresponds to the maximum safe current $I_g$ flowing through the total resistance. This gives us our governing equation:

$$V = I_g(G + R_s)$$

Our goal is to find $R_s$. Let's rearrange the equation to isolate it:

$$G + R_s = \frac{V}{I_g}$$

$$R_s = \frac{V}{I_g} - G$$

### Final Calculation

Now, we substitute the given values into our rearranged formula. **Always remember to convert milliamperes to Amperes to maintain SI unit consistency!**

$$R_s = \frac{10}{5 	imes 10^{-3}} - 15$$

Let's tackle the fraction first. Bringing the $10^{-3}$ from the denominator to the numerator changes its sign, making it $10^3$ or $1000$:

$$\frac{10}{5 	imes 10^{-3}} = \frac{10}{5} 	imes 10^3 = 2 	imes 1000 = 2000\,\Omega$$

This $2000\,\Omega$ represents the *total* resistance required. To find the *additional* series resistance $R_s$, we subtract the galvanometer's own resistance:

$$R_s = 2000 - 15 = 1985\,\Omega$$

To match the options provided, we express this in scientific notation:

**$$R_s = 1.985 	imes 10^3\,\Omega$$**

This elegant calculation shows how a simple resistor can completely transform the capability of an electrical instrument.

Select Answer:

Visualized Solution

Visualizing the Voltmeter Setup

Applying Ohm's Law

Isolating the Series Resistance

Calculating the Resistance

Final Answer

The Way Forward

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Final Calculation

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