A moving coil galvanometer has 50 turns and each turn has an area 2 * 10^-4 m^2. The magnetic field

Q: Why do we connect the shunt in parallel?

To measure large currents, most of the current must bypass the sensitive galvanometer. A low-resistance shunt in parallel provides this alternate path.

Q: What is the torsional constant $C$?

It is the restoring torque per unit angle of twist of the suspension wire. It opposes the magnetic torque to bring the coil to equilibrium.

Q: Why is the shunt resistance very small?

A small shunt resistance ensures that the overall resistance of the ammeter is very low, so it doesn't alter the current in the circuit it is measuring.

Q: What happens if we pass 1 A directly through the galvanometer?

The galvanometer coil is very sensitive and designed for small currents (like 0.1 A here). Passing 1 A directly would burn the coil or damage the pointer.

Q: How does the number of turns affect the galvanometer?

More turns increase the magnetic torque for a given current, making the galvanometer more sensitive.

Question

A moving coil galvanometer has 50 turns and each turn has an area $2 	imes 10^{-4} 	ext{ m}^2$. The magnetic field produced by the magnet inside the galvanometer is $0.02 	ext{ T}$. The torsional constant of the suspension wire is $10^{-4} 	ext{ N-m rad}^{-1}$. When a current flows through the galvanometer, a full scale deflection occurs, if the coil rotates by $0.2 	ext{ rad}$. The resistance of the coil of the galvanometer is $50 \Omega$. This galvanometer is to be converted into an ammeter capable of measuring current in the range $0 - 1.0 	ext{ A}$. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance in ohms, is ............. .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a highly sensitive moving coil galvanometer. It's a delicate instrument designed to measure very tiny currents. The problem gives us a detailed anatomy of this device: it has $50$ turns, an area of $2 	imes 10^{-4} 	ext{ m}^2$, and sits in a magnetic field of $0.02 	ext{ T}$. The suspension wire, which acts like a tiny spring, has a torsional constant of $10^{-4} 	ext{ N-m rad}^{-1}$.

When a current flows through the coil, it experiences a magnetic torque that makes it rotate. The coil stops rotating when this magnetic torque is perfectly balanced by the restoring torque of the suspension wire. We are told that a full-scale deflection corresponds to a rotation of $0.2 	ext{ rad}$.

### The Master Equation

To find the maximum current $i_g$ the galvanometer can handle (the full-scale deflection current), we equate the magnetic torque to the restoring torque:

$$N i_g A B = C 	heta$$

Here, $N$ is the number of turns, $A$ is the area, $B$ is the magnetic field, $C$ is the torsional constant, and $	heta$ is the angle of deflection.

Let's substitute the given values into our master equation:

$$50 	imes i_g 	imes (2 	imes 10^{-4}) 	imes 0.02 = 10^{-4} 	imes 0.2$$

Solving for $i_g$, we find:

$$i_g = \frac{10^{-4} 	imes 0.2}{50 	imes 2 	imes 10^{-4} 	imes 0.02} = 0.1 	ext{ A}$$

So, our delicate galvanometer maxes out at just $0.1 	ext{ A}$. But we need it to measure up to $1.0 	ext{ A}$! If we pass $1.0 	ext{ A}$ directly through it, the coil would likely burn out.

### Converting to an Ammeter

To solve this, we provide an alternate, low-resistance path for the excess current. This is called a **shunt resistance** ($S$), and it is connected in parallel with the galvanometer.

When a total current $i = 1.0 	ext{ A}$ enters the setup, it splits. The galvanometer takes its maximum safe current, $i_g = 0.1 	ext{ A}$, and the rest of the current, $(i - i_g) = 0.9 	ext{ A}$, bypasses through the shunt.

Because the galvanometer and the shunt are in parallel, the potential difference across them must be identical:

$$V_{ab} = i_g R_g = (i - i_g) S$$

### Final Calculation

We know the resistance of the galvanometer coil is $R_g = 50 \Omega$. Let's plug in all our knowns:

$$0.1 	imes 50 = (1.0 - 0.1) 	imes S$$

$$5 = 0.9 	imes S$$

Finally, solving for the shunt resistance $S$:

$$S = \frac{5}{0.9} = \frac{50}{9} \approx 5.55 \Omega$$

**The required shunt resistance is $5.55 \Omega$.** By adding this small resistor in parallel, we've successfully upgraded our sensitive galvanometer into a robust ammeter!

Enter Numerical Value:

Visualized Solution

$Given Parameters$

$Torque Balance Equation$

$Full Scale Deflection Current$

$Ammeter Conversion Circuit$

$Parallel Voltage Equality$

$Substituting Values$

$Final Shunt Resistance$

The Sigma Insight: Moving Coil Galvanometer

Analyzing the Setup

The Master Equation

Converting to an Ammeter

Final Calculation

Similar Questions

A galvanometer having a coil resistance of $100 Ω$ gives a full scale deflection when a current of $1 mA$ 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 $10 A$ , is

The resistance of a galvanometer is $50 Ω$ and the maximum current which can be passed through it is $0.002 A$ . What resistance must be connected to it in order to convert it into an ammeter of range $0 - 0.5 A$ ?

Consider a galvanometer shunted with $5 Ω$ resistance and $2%$ of current passes through it. What is the resistance of the given galvanometer ?

For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be

A galvanometer coil has 500 turns and each turn has an average area of $3 \times 1 0^{- 4} m^{2}$ . If a torque of $1.5 N-m$ is required to keep this coil parallel to a magnetic field when a current of $0.5 A$ is flowing through it, the strength of the field (in T) is ........... .

A moving coil galvanometer has resistance $50 Ω$ and it indicates full deflection at $4 mA$ current. A voltmeter is made using this galvanometer and a $5 k Ω$ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to

A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 V, the resistance in Ohms needed to be connected in series with the coil will be

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Parameters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque Balance Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Full Scale Deflection Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ammeter Conversion Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Voltage Equality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Shunt Resistance