A galvanometer having a coil resistance of 100 ohm gives a full scale deflection when a current of 1

Q: Why do we connect the shunt in parallel?

To bypass the excess current so the galvanometer doesn't burn out, and to keep the overall resistance of the ammeter very low.

Q: Why is the shunt resistance so small?

An ideal ammeter should have zero resistance so it doesn't alter the current in the circuit it's measuring. A small shunt achieves this.

Q: Can we use the exact value $I - I_g$ instead of approximating?

Yes, $10 - 0.001 = 9.999$. Then $R_s = 0.1 / 9.999 = 0.010001\ \Omega$, which is practically $0.01\ \Omega$. Approximating saves time.

Q: What happens if we connect a high resistance in parallel?

Most of the current would still try to pass through the galvanometer, potentially damaging it, and it wouldn't measure large currents.

Q: How do we convert it to a voltmeter instead?

By connecting a very high resistance in series with the galvanometer.

Question

A galvanometer having a coil resistance of $100\ \Omega$ gives a full scale deflection when a current of $1	ext{ mA}$ is passed through it. The value of the resistance which can convert this galvanometer into ammeter giving a full scale deflection for a current of $10	ext{ A}$, is

Options:

(A) $0.01\ \Omega$

(B) $2\ \Omega$

(C) $0.1\ \Omega$

(D) $3\ \Omega$

Sigma Percentile · Accepted Answer

### The Challenge: Measuring a Massive Current

Imagine you have a delicate instrument, a galvanometer, that is incredibly sensitive. It gives a full-scale deflection with just a tiny current of $1	ext{ mA}$ ($10^{-3}	ext{ A}$). Its internal coil resistance is $100\ \Omega$.

Now, you are faced with a challenge: you need to measure a massive current of $10	ext{ A}$ using this very same delicate instrument. If you pass $10	ext{ A}$ directly through it, the coil will instantly burn out! So, how do we solve this?

### The Master Logic: Parallel Circuits to the Rescue

To protect the galvanometer and measure the large current, we provide an alternate, low-resistance path for the excess current to flow. We connect a very small resistance, called a **shunt resistance** ($R_s$), in parallel with the galvanometer.

Because they are in parallel, the fundamental law of parallel circuits applies: **The potential difference across the galvanometer must equal the potential difference across the shunt.**

### The Mathematical Setup

Let the total current entering the setup be $I = 10	ext{ A}$. 
The maximum safe current that can flow through the galvanometer is $I_g = 10^{-3}	ext{ A}$. 
The remaining current must flow through the shunt. According to Kirchhoff's Current Law, the current through the shunt is $(I - I_g)$.

Equating the potential differences:
$$V_G = V_S$$
$$I_g 	imes R_g = (I - I_g) 	imes R_s$$

### The Art of Approximation

Now, let's substitute our known values into the equation:
$$10^{-3} 	imes 100 = (10 - 10^{-3}) 	imes R_s$$

Here is where we can be smart and save time. Look at the term $(10 - 10^{-3})$. Subtracting $0.001$ from $10$ gives $9.999$. For all practical purposes in this calculation, $9.999$ is approximately $10$.

So, we can safely approximate $(I - I_g) \approx I = 10	ext{ A}$.

### The Final Calculation

Using our approximation, the equation simplifies beautifully:
$$0.1 = 10 	imes R_s$$

Solving for $R_s$:
$$R_s = \frac{0.1}{10} = 0.01\ \Omega$$

And there we have it! By simply connecting a tiny $0.01\ \Omega$ resistor in parallel, we have successfully transformed our delicate galvanometer into a robust ammeter capable of measuring up to $10	ext{ A}$.

Select Answer:

Visualized Solution

$Circuit Setup$

$Parallel Combination Logic$

$Current Distribution$

$Substitution and Approximation$

$Solving for R_{s}$

$Final Answer$

$The Way Forward$

The Sigma Insight: Moving Coil Galvanometer

The Challenge

The Master Logic

The Mathematical Setup

The Art of Approximation

The Final Calculation

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