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Electricity and Magnetism
Ammeter
An ammeter is a low-resistance measuring instrument used to determine the magnitude of electric current flowing in a circuit branch. It is invariably connected in series with the load component to ensure the entire branch current passes through it.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Conversion of a moving coil galvanometer into an ammeter using a low shunt resistance, calculation of shunt value, effective resistance of an ammeter, and error analysis due to meter resistance.
PGT extension
Design considerations for multi-range ammeters, calibration of ammeters, extension of range using shunts in AC circuits, temperature effects, and ballistic ammeter principles.
Prerequisites
Ohm's Law and equivalent resistance calculations
Torque on a current-carrying loop in a uniform magnetic field
Moving coil galvanometer construction and working
Principle and Basic Construction
An ammeter is fundamentally built by connecting a very low resistance, known as a shunt resistor (S), in parallel with a sensitive moving coil galvanometer (G).
The primary design objective is to minimize the equivalent resistance of the instrument so that its insertion into a circuit does not alter the actual current being measured.
The deflecting torque in the galvanometer is directly proportional to the current passing through its moving coil, leading to a linear scale under a radial magnetic field.
Shunt Resistance and Current Division
A major fraction of the total circuit current (I) is bypassed through the parallel shunt resistor (S), while only a safe fraction (I_g) flows through the galvanometer coil.
The value of the shunt resistance depends strictly on the maximum current to be measured and the full-scale deflection current of the galvanometer.
Smaller shunt resistance values allow the ammeter to measure larger ranges of electric current.
Equivalent Resistance of an Ammeter
The effective resistance (R_A) of an ideal ammeter is exactly zero, ensuring zero potential drop across the instrument.
For a practical ammeter consisting of galvanometer resistance G and shunt S, the equivalent resistance is given by the parallel combination formula: $R_A = (G \times S) / (G + S)$.
Since S is much smaller than G, the effective resistance R_A is slightly less than S, making it exceptionally small compared to typical load resistances.
Multi-Range Ammeter Design
To convert a single galvanometer into a multi-range ammeter, a selector switch is used to connect different shunt resistances in parallel with the galvanometer.
Universal or Ayrton shunts are often deployed to avoid opening the circuit when changing ranges, preventing high induced voltage spikes.
Precision in multi-range instruments requires non-inductive wire-wound resistors to minimize phase errors when dealing with alternating currents.
Derivations
Derivation of Shunt Resistance for Ammeter Conversion
Conditions: Galvanometer of resistance G, full-scale deflection current I_g, maximum current to be measured I.
Let the total current entering the parallel combination be I.
The current through the galvanometer is I_g, leaving the current through the shunt resistor as $I_s = I - I_g$.
Since the galvanometer and shunt are connected in parallel, the potential difference across both must be equal: $V_g = V_s$.
P = Power loss, I = Current, $R_A = Ammeter resistance$ Watts (W)
Steady DC flow
Fraction of Current through Galvanometer
$$I_g = I \left(\frac{S}{G + S}\right)$$
$I_g = Galvanometer current$, I = Total current Amperes (A)
Current divider rule application
Graphs and experiments
Calibration Curve of an Ammeter
Draw: A plot of standard/actual current (I_standard) on the x-axis versus ammeter reading (I_observed) on the y-axis.
Interpret: A straight line making a 45-degree angle with the axis indicates a perfectly calibrated ammeter. Deviations reveal systematic calibration errors.
Exam use: Tested conceptually to determine zero error, calibration correction, and least count adjustments.
Torque vs Current Characteristic
Draw: Linear graph plotting deflecting torque (\tau) against current (I) passing through the ammeter.
Interpret: A linear straight line passing through the origin, proving that the scale of a moving-coil based ammeter is uniformly graduated under a radial magnetic field.
Exam use: Questions focus on scale uniformity and spring restoring torque balancing.
Effect of Ammeter Resistance on Circuit Current (Experiment Setup)
Draw: Circuit diagram showing a variable DC source, standard resistor, and ammeter in series, with a high-resistance voltmeter across the load.
Interpret: Demonstrates that a practical ammeter introduces a small voltage drop and reduces the main circuit current compared to the ideal theoretical value.
A galvanometer has a resistance of 50 \Omega and gives a full-scale deflection for a current of 2 mA. Calculate the value of the shunt resistance required to convert it into an ammeter of range 0 to 5 A.
Identify given parameters: G = 50 \Omega, $I_g = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$, I = 5 \text{ A}.
Use the shunt formula: $S = \frac{I_g G}{I - I_g}$.
Calculate S: S = \frac{0.1}{4.998} \approx 0.02 \Omega.
Answer: 0.02 \Omega
PGT extension
A moving coil galvanometer of resistance 99 \Omega requires 10 mA for full-scale deflection. If a shunt is connected to increase its range 10 times (n = 10), find the new resistance of the ammeter and the fraction of total current passing through the galvanometer.
Identify given parameters: G = 99 \Omega, $I_g = 10 \text{ mA}$, $n = 10 (\text{so } I = 10 I_g)$.
Calculate the required shunt resistance using $S = \frac{G}{n - 1}: S = \frac{99}{10 - 1} = \frac{99}{9} = 11 \Omega$.
An ammeter with an internal resistance of 0.05 \Omega reads up to 2 A. What additional shunt resistance must be connected across this ammeter to extend its measuring range up to 10 A?
Treat the existing ammeter as the galvanometer unit with internal resistance $G_new = 0.05 \Omega$ and current limit $I_g = 2 \text{ A}$.
Target maximum current I = 10 \text{ A}.
Apply the shunt formula: $S_{add} = \frac{I_g G_{new}}{I - I_g}$.
Assumes the shunt and galvanometer resistances remain constant with temperature variations, which is limited by Joule heating during high current passage.
Moving-coil based ammeters are strictly restricted to Direct Current (DC) measurements unless used with rectifiers, because alternating current would produce zero net average deflection due to rapid torque reversal.
Common misconceptions
Common error: An ideal ammeter has infinite resistance to block current and measure it accurately.
Correct: An ideal ammeter has zero internal resistance.
Infinite resistance would drastically alter the circuit current being measured, violating the purpose of the measurement.
Common error: An ammeter should be connected in parallel across the component.
Correct: An ammeter must always be connected in series with the branch.
Parallel connection routes nearly all circuit current through the extremely low shunt resistance, causing a short circuit and potential damage to the meter.
Common error: A standard DC moving-coil ammeter can directly measure 50 Hz domestic alternating current accurately.
Correct: A standard DC moving-coil ammeter shows zero deflection on alternating current.
The coil cannot keep up with high-frequency alternating torque reversals, averaging the deflection to zero.
Quick revision
An ammeter is a low-resistance device placed in series to measure current.
Conversion of galvanometer to ammeter requires a parallel shunt resistance $S = (I_g G)/(I - I_g)$.
An ideal ammeter possesses zero internal resistance.
Effective resistance of an ammeter is always less than the smallest individual resistance in the parallel network.
Must remember
An ammeter is constructed by connecting a very low resistance shunt ($S$) in parallel with a moving coil galvanometer ($G$).
An ideal ammeter has zero internal resistance ($R_A = 0$) so that its insertion does not alter the circuit current.
A practical ammeter's equivalent resistance is given by $R_A = \frac{GS}{G + S}$, which is always less than the shunt resistance $S$.
The multiplying power of a shunt is given by $n = \frac{I}{I_g} = 1 + \frac{G}{S}$, allowing range extension.
Shunt resistance required for a given multiplying factor $n$ is calculated using $S = \frac{G}{n - 1}$.
An ammeter must always be connected in series with the circuit component whose current is to be measured.
Standard DC moving-coil ammeters read zero average deflection when subjected to high-frequency alternating current due to rapid torque reversals.
Multi-range ammeters utilize universal or Ayrton shunts to prevent opening the circuit when switching ranges, avoiding high induced voltage spikes.
Glossary
Shunt Resistor ($S$)A very low resistance connected in parallel with a galvanometer to bypass the major fraction of current.
Multiplying Power ($n$)The ratio of maximum measurable current to the full-scale deflection current of the galvanometer ($I / I_g$).
Effective Resistance ($R_A$)The total equivalent internal resistance of the ammeter combining the galvanometer and shunt resistance.
Full-Scale Deflection Current ($I_g$)The maximum current that can safely pass through the moving coil galvanometer without damaging it.
Ayrton ShuntA universal shunt arrangement used in multi-range ammeters to change ranges without opening the main circuit.
Ideal Ammeter vs. Practical Ammeter
Parameter
Ideal Ammeter
Practical Ammeter
Internal Resistance ($R_A$)
Zero ($R_A = 0$)
Extremely small ($R_A = \frac{GS}{G+S} > 0$)
Potential Drop
Zero potential drop across terminals
Very small potential drop ($V_A = I R_A$)
Circuit Impact
Does not alter circuit current at all
Causes a negligible, acceptable perturbation in circuit current
Practice Quiz
Choose an option to reveal the explanation.
1. How is a galvanometer converted into an ammeter?
Explanation: An ammeter is constructed by placing a very low resistance shunt ($S$) in parallel with a sensitive moving coil galvanometer ($G$) to bypass excess current.
2. What is the effective resistance ($R_A$) of an ideal ammeter?
Explanation: An ideal ammeter must not alter the current in the circuit, which requires its internal resistance to be zero so there is no potential drop across it.
3. If a galvanometer of resistance $G$ has a shunt $S$ connected across it, what is the equivalent resistance of the ammeter?
Explanation: Since the shunt and galvanometer are in parallel, their equivalent resistance is given by the parallel combination formula $R_A = \frac{GS}{G + S}$.
4. What is the formula for the shunt resistance $S$ required to measure a maximum current $I$ using a galvanometer of resistance $G$ with a full-scale deflection current $I_g$?
Explanation: Applying parallel circuit voltage equality: $I_g G = (I - I_g) S$, which yields $S = \frac{I_g G}{I - I_g}$.
5. What does the multiplying power ($n$) of a shunt represent?
Explanation: The multiplying power of a shunt is defined as $n = \frac{I}{I_g} = 1 + \frac{G}{S}$, indicating how many times the range of the ammeter has increased.
6. Why must an ammeter always be connected in series in a circuit?
Explanation: An ammeter measures electric current and therefore must carry the entire current of the branch, requiring a series connection.
7. What happens if an ammeter is mistakenly connected in parallel across a voltage source or component?
Explanation: Connecting a very low resistance ammeter in parallel creates an effective short circuit, drawing excessive current that can burn out the galvanometer coil.
8. What is the fraction of total current $I$ that passes through the galvanometer in an ammeter?
Explanation: Using the current divider rule, the current through the galvanometer branch is $I_g = I \left(\frac{S}{G + S}\right)$.
9. If the multiplying factor $n$ of an ammeter is $10$ and the galvanometer resistance is $G$, what is the required shunt resistance $S$?
Explanation: Using the formula $S = \frac{G}{n - 1}$, substituting $n = 10$ gives $S = \frac{G}{10 - 1} = \frac{G}{9}$.
10. Why are non-inductive wire-wound resistors used in precision multi-range ammeters for AC applications?
Explanation: In AC circuits, standard wire-wound resistors can introduce unwanted inductive reactance, causing phase errors. Non-inductive winding cancels out self-inductance.
11. What is the purpose of using an Ayrton (universal) shunt in a multi-range ammeter?
Explanation: Ayrton shunts allow range selection without breaking the circuit, preventing high induced voltage spikes that could damage sensitive coils.
12. What is the reading of a standard moving-coil DC ammeter when connected in a 50 Hz domestic AC circuit?
Explanation: The coil of a moving-coil ammeter cannot keep up with high-frequency alternating torque reversals, averaging the torque and deflection to zero.
13. If a galvanometer has resistance $G$ and shunt $S$, how does the effective resistance $R_A$ compare to $S$?
Explanation: Since $R_A = \frac{GS}{G+S} = S \left(\frac{G}{G+S}\right)$, and $\frac{G}{G+S} < 1$, the effective resistance $R_A$ is always slightly less than the shunt resistance $S$.
14. Which parameter determines the maximum current that can be measured by an ammeter for a fixed shunt?
Explanation: The maximum measurable current $I = I_g \left(1 + \frac{G}{S}\right)$ depends directly on the full-scale deflection current $I_g$ of the internal galvanometer.
15. What is the unit of the multiplying power ($n$) of an ammeter shunt?
Explanation: Multiplying power $n = \frac{I}{I_g}$ is the ratio of two currents, making it a dimensionless quantity.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A galvanometer has a resistance of $99\;\Omega$ and a full-scale deflection current of $10\;\text{mA}$. What shunt resistance is required to convert it into an ammeter reading up to $1\;\text{A}$?
Explanation: Using $S = \frac{I_g G}{I - I_g}$, where $G = 99\;\Omega$, $I_g = 0.01\;\text{A}$, and $I = 1\;\text{A}$: $S = \frac{0.01 \times 99}{1 - 0.01} = \frac{0.99}{0.99} = 1\;\Omega$? Wait, $\frac{0.99}{0.99} = 1\;\Omega$. Let's check: $S = \frac{0.01 \times 99}{0.99} = 1\;\Omega$. Let's re-verify options. Option A is $1\;\Omega$.
2. A galvanometer with resistance $G$ is shunted by a resistance $S$. The fraction of the total current passing through the galvanometer is:
Explanation: By the current divider rule, the current through the galvanometer branch is $I_g = I \left(\frac{S}{G + S}\right)$, so the fraction is $\frac{I_g}{I} = \frac{S}{G + S}$.
3. If the range of an ammeter of resistance $R_A$ is increased $n$ times, the new shunt resistance to be connected is:
Explanation: When converting or extending a galvanometer (or ammeter of resistance $G$) by factor $n$, the shunt required is $S = \frac{G}{n - 1}$.
4. A milliammeter of resistance $20\;\Omega$ has a full-scale deflection of $25\;\text{mA}$. To convert it into an ammeter reading up to $2.5\;\text{A}$, the shunt resistance required is:
6. A galvanometer requires $0.01\;\text{A}$ for full-scale deflection and has a resistance of $10\;\Omega$. The multiplying power of a shunt of $0.1\;\Omega$ is:
Attempt these questions under a ten-minute limit using a separate timer.
1. Why is the resistance of an ammeter designed to be extremely small?
Explanation: An ammeter is inserted in series; keeping its resistance exceptionally small ensures that the total resistance of the circuit remains virtually unchanged.
2. If the shunt resistance $S$ in an ammeter is decreased, what happens to the range of the ammeter?
Explanation: A smaller shunt resistance bypasses more current away from the galvanometer, allowing a larger total current $I$ to be measured safely, thus increasing the range.
3. A galvanometer of resistance $G$ is converted into an ammeter. If $3\%$ of the total circuit current passes through the galvanometer, the shunt resistance $S$ in terms of $G$ is:
4. What is the power dissipation in an ammeter of equivalent resistance $R_A$ carrying a total current $I$?
Explanation: Power dissipation in any resistive component carrying current $I$ is given by Joule's law: $P = I^2 R_A$.
5. In a multi-range ammeter design, what is the primary advantage of a universal shunt over individual shunts for each range?
Explanation: Universal (Ayrton) shunts keep the circuit closed during switching, preventing dangerous transient high-voltage induction spikes.
6. A galvanometer has resistance $G$ and full-scale current $I_g$. The potential difference across the ammeter when measuring current $I$ is:
Explanation: The potential drop across the ammeter equals the product of the total current $I$ and the effective ammeter resistance $R_A = \frac{GS}{G + S}$.
7. If a galvanometer has a resistance of $36\;\Omega$ and the shunt resistance is $4\;\Omega$, what fraction of the total current flows through the galvanometer?
8. Which of the following errors is most likely introduced if an ammeter has a non-negligible internal resistance when measuring current in a high-resistance circuit?
Explanation: A non-negligible ammeter resistance adds to the total circuit resistance, reducing the current being measured—a classic loading error.
9. What happens to the scale of a moving-coil galvanometer when converted into an ammeter using a radial magnetic field?
Explanation: A radial magnetic field ensures that the deflecting torque is directly proportional to the current, resulting in a linear scale.
10. If a shunt of resistance $S$ is connected across a galvanometer of resistance $G$, the ratio of current through the shunt ($I_s$) to the current through the galvanometer ($I_g$) is:
Explanation: Since the potential difference is the same across both branches, $I_s S = I_g G$, which gives $\frac{I_s}{I_g} = \frac{G}{S}$.