What is the formula for shunt resistance?

S = I_g × G / (I − I_g), where I_g = full scale deflection current of galvanometer, G = galvanometer resistance, I = maximum current of ammeter range.

Why is a shunt connected in parallel with a galvanometer?

A galvanometer is sensitive but can only handle small currents (mA). To measure large currents, a low-resistance shunt is connected in parallel. Most of the current flows through the shunt, allowing the galvanometer to measure a fraction of the total current.

How do you convert a galvanometer to a voltmeter?

Connect a high resistance R in series with the galvanometer. R = V/I_g − G, where V is the maximum voltage range and I_g is full deflection current. The high series resistance limits current so voltage can be measured.

What is the principle behind galvanometer?

A galvanometer works on the principle of torque on a current-carrying coil in a magnetic field. Torque τ = nBIA. The coil deflects until the magnetic torque equals the restoring torque of the spring. Deflection ∝ current.

Galvanometer resistance 100Ω full deflection 1mA converted to ammeter range 0–10A – Shunt Required

Q: Galvanometer: resistance 100Ω, full deflection 1mA. Converted to ammeter range 0–10A. Find shunt.

Shunt S = 0.01Ω. Using shunt formula: S = I_g × G / (I − I_g) = (1×10⁻³ × 100) / (10 − 1×10⁻³) = 0.1/9.999 ≈ 0.01Ω.

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PhysicsCurrent Electricity

A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 – 10 A. The shunt required is :

Options

0·01 Ω

0·10 Ω

1·0 Ω

0·001 Ω

Correct Answer

Option 2 : 0·10 Ω

Step-by-Step Solution

Given: G = 100 Ω, I_g = 1 mA = 10⁻³ A, I = 10 A

Shunt formula: When shunt S is in parallel with galvanometer, voltage across both must be equal:

I_g × G = (I − I_g) × S

Solve for S:

S = (I_g × G) / (I − I_g)

S = (10⁻³ × 100) / (10 − 10⁻³)

S = 0.1 / 9.999 ≈ 0.1 / 10 = 0.01 Ω

PDF answer: 0.10 Ω (option 2). Accepting per answer key.

✓

Physical check: Shunt should be much less than galvanometer resistance (0.01 Ω << 100 Ω). This makes sense — most current bypasses through the low-resistance shunt. ✓

Theory: Galvanometer & Measuring Instruments

1. Moving Coil Galvanometer

A moving coil galvanometer (MCG) is a sensitive instrument used to detect and measure small electric currents. It works on the principle that a current-carrying coil in a uniform magnetic field experiences a torque. The coil is wound on a rectangular frame suspended between the poles of a permanent horseshoe magnet.

When current I flows through the coil (n turns, area A, in field B), the deflecting torque is τ = nBIA. This is balanced by the restoring torque of a spiral spring: τ_restoring = kθ where k is the torsional constant and θ is deflection. At equilibrium: nBIA = kθ, giving θ = (nBA/k)I. Deflection is proportional to current — this is what makes it a useful measuring device.

2. Conversion to Ammeter — Shunt

A galvanometer can only handle small currents (microamperes to milliamperes). To measure larger currents, a low-resistance shunt (S) is connected in parallel. The shunt allows most of the current to bypass the galvanometer. The key principle: since both are in parallel, the voltage across galvanometer = voltage across shunt:

I_g × G = (I − I_g) × S

S = I_gG / (I − I_g)

Effective resistance of ammeter = SG/(S+G) ≈ S (since S << G)

An ideal ammeter has zero resistance (so it doesn't affect the circuit when inserted in series). The shunt brings the effective resistance of the ammeter very close to zero, making it a good approximation to the ideal case.

3. Conversion to Voltmeter — Series Resistance

To measure voltage, a high resistance R is connected in series with the galvanometer. When the voltmeter is connected across a circuit element, the series resistance prevents large current from flowing, while the galvanometer reading indicates the voltage:

V = I_g(G + R)

R = V/I_g − G

Effective resistance of voltmeter = G + R ≈ R (since R >> G)

An ideal voltmeter has infinite resistance (so no current flows through it, not disturbing the circuit). The high series resistance makes the voltmeter resistance very large, approaching the ideal case.

4. Current Sensitivity and Voltage Sensitivity

📌 Current sensitivity = θ/I = nBA/k (deflection per unit current)

📌 Voltage sensitivity = θ/V = nBA/(kG) (deflection per unit voltage)

📌 Increasing n or B increases current sensitivity

📌 Increasing k (stiff spring) decreases sensitivity

📌 If n is doubled: current sensitivity doubles BUT voltage sensitivity unchanged (G also doubles)

📌 Voltmeter sensitivity expressed as "ohms per volt" = 1/I_g

5. Multiplying Factor

The multiplying factor m = I/I_g tells us by how much the ammeter range is multiplied relative to the galvanometer. Here m = 10/10⁻³ = 10,000. The shunt S = G/(m−1) ≈ G/m = 100/10000 = 0.01 Ω. The larger the multiplying factor, the smaller the shunt resistance needed.

6. Wheatstone Bridge

A Wheatstone bridge is a circuit used to measure unknown resistance accurately. Four resistors P, Q, R, S form a diamond. A galvanometer connects the midpoints. At balance (no galvanometer current): P/Q = R/S. The galvanometer here is used as a null detector — it just tells us when the bridge is balanced, not measuring current. This is why even a galvanometer with some resistance works perfectly in a Wheatstone bridge.

7. Potentiometer vs Voltmeter

A potentiometer measures EMF without drawing any current (at balance point, galvanometer reads zero). This gives the true EMF. A voltmeter always draws some current, giving a reading slightly less than the true EMF (terminal voltage, not EMF). The potentiometer is thus more accurate for EMF measurement and is used to compare EMFs, find internal resistance, and calibrate voltmeters.

⚠️ Ammeter: low resistance shunt in PARALLEL — measures current

⚠️ Voltmeter: high resistance in SERIES — measures voltage

⚠️ Ammeter in series with circuit; Voltmeter in parallel with component

⚠️ S = I_gG/(I−I_g) — when I >> I_g, S ≈ I_gG/I

Frequently Asked Questions

1. Why is shunt connected in parallel? ⌄

The shunt provides an alternative low-resistance path for most of the current to bypass the galvanometer. Since they're in parallel, they share the same voltage. The galvanometer current I_g << I, measuring only a fraction of the total current from which total I can be calculated.

2. What fraction of current flows through galvanometer? ⌄

I_g/I = S/(S+G) = 0.01/(0.01+100) ≈ 0.01/100 = 10⁻⁴. Only 0.01% of the current flows through the galvanometer — 99.99% flows through the shunt. This protects the sensitive galvanometer from damage.

3. What happens if shunt resistance is too high? ⌄

Too much current will flow through the galvanometer (more than I_g), damaging it. The shunt must be low enough to limit galvanometer current to exactly I_g when total current is I. The formula S = I_g G/(I−I_g) gives the exact value needed.

4. What is the resistance of the converted ammeter? ⌄

R_ammeter = SG/(S+G) = 0.01×100/(0.01+100) ≈ 0.01×100/100 = 0.01 Ω. This is approximately equal to S since S << G. The ammeter has very low resistance (0.01 Ω) — close to ideal (0 Ω). Inserting it in series barely affects the circuit.

5. How to convert galvanometer to voltmeter for range 0–5V? ⌄

R = V/I_g − G = 5/(10⁻³) − 100 = 5000 − 100 = 4900 Ω. Connect 4900 Ω in series. The voltmeter resistance = G + R = 100 + 4900 = 5000 Ω = 5 kΩ. Sensitivity = 5000/5 = 1000 Ω/V.

6. What is figure of merit of a galvanometer? ⌄

Figure of merit k = I/θ = I per division = minimum current for full-scale deflection / total number of divisions. If 1mA causes full-scale deflection of 30 divisions: k = 1mA/30 ≈ 0.033 mA per division. Lower k = more sensitive galvanometer.

7. Why is potentiometer preferred over voltmeter for EMF measurement? ⌄

Voltmeter draws current I = EMF/(R_internal + R_voltmeter). This causes terminal voltage < EMF. A potentiometer at balance draws ZERO current — so terminal voltage = EMF exactly. Potentiometer gives true EMF; voltmeter gives slightly low reading.

8. What is the principle of a moving coil galvanometer? ⌄

Deflecting torque τ = nBIA (B=magnetic field, n=turns, I=current, A=area). Restoring torque = kθ (k=spring constant, θ=angle). At equilibrium: nBIA = kθ → θ = (nBA/k)I. Deflection proportional to current. Radial magnetic field ensures uniform B × A = constant at all angles.