

An Overview of Class 12 Physics Conversion Of Galvanometer Into Ammeter Of Desired Range Experiment
A galvanometer is a sensitive electrical instrument generally used to measure the magnitude and direction of current in a circuit. Due to its sensitivity, a galvanometer measures only small currents.
To measure large currents, a galvanometer can be converted into an ammeter. This conversion of a galvanometer into an ammeter can be done by connecting a shunt resistance parallel to the galvanometer. In this article, we will study the conversion of a galvanometer into an ammeter in detail.
Table of Contents:
Aim
Apparatus Required
Theory
Procedure
Observations
Result
Precautions
Lab Manual Questions
Viva Questions
Practical Based Questions
Aim
To convert the given galvanometer (of known resistance and figure of merit) into an ammeter of desired range and to verify the same.
Apparatus Required
A galvanometer
Key and battery
Sandpaper
A rheostat and connecting wires
An ammeter of 0-3 A range
Resistance box
Piece of sandpaper
Theory
An ammeter is a device which we can use to measure the strength of electric current through a circuit, and it is a modified form of the galvanometer. We should always connect this device in series in a circuit. To convert the galvanometer into the ammeter, we should connect a low resistance, also known as a shunt resistance, parallel to the galvanometer. The value of the shunt is so chosen that most of the current pass through the shunt. This prevents large currents from flowing through the galvanometer, which may damage it.

Conversion Of Galvanometer Into Ammeter
If ${I_g}$ is the current required to produce full-scale deflection in the galvanometer, then the remaining current will flow through the shunt resistance $S$, which is shown in the above figure.
Now, we know that the potential difference across $G$ and $S$ would be the same as they are parallel.
So, the potential difference across $G$ = Potential difference across shunt resistance $S$
${I_g}G = \left( {I - {I_g}} \right)S$ (From Ohm’s law $V = IR$)
So, the value for shunt resistance $S$ is,
$S = \dfrac{{{I_g}G}}{{I - {I_g}}}$
Where,
$S$ is the shunt resistance (low resistance)
$G$ is the resistance of the given galvanometer
${I_g}$ is the full-scale deflection current
$I$ is the range of the converted galvanometer
$I - {I_g}$ is the current flowing in shunt
The above equation gives the value of shunt resistance to be connected in parallel with the galvanometer to convert it into an ammeter of the desired range.
Procedure
First of all, we should count the total number of divisions on either side of zero of the galvanometer scale. Let it be $n$. Then, we need to calculate the current ${I_g}$ for full-scale deflection by using the formula ${I_g} = nk$
Now, use the formula,
$S = \dfrac{{{I_g}G}}{{I - {I_g}}}$ and calculate the value of shunt resistance for conversion into an ammeter.
The value of shunt resistance is minimal, such that the range is not available in the resistance box. Wires of copper, manganin etc., are used with suitable lengths and diameters to obtain the value of this small resistance.
Now, cut a length of the wire 2 cm more than the calculated value of $I$ such that there is 1 cm extra available at each end. Now, mark points on each end of the wire and connect them to the two terminals of the galvanometer. The wire should be such that the points are on the outside of the terminal screws. This galvanometer with shunt wire will now work as an ammeter in the range of $I$
Now, we need to make the electrical connections as shown in the above diagram.
Now, insert the key and adjust the rheostat to observe maximum and minimum deflection in the galvanometer.
We need to note the readings from the galvanometer scale and the corresponding ammeter reading and record the observations.
Observations
The resistance of the galvanometer is $G$ = _____
The figure of merit is $k$ = _____
Number of divisions in the galvanometer scale is $n$ =_____
Current for full scale deflection is ${I_g} = nk$
Range of conversion is $I$ =_____
Shunt resistance is $S = \dfrac{{{I_g}G}}{{I - {I_g}}}$
Least count of the galvanometer converted into an ammeter is $\dfrac{I}{n}$ = _____
Result
From this experiment, we have found that the actual value and measured value of currents are minimal. So, conversion is verified.
Precautions
We should calculate resistance accurately.
In this experiment, the length of the shunt wire must be correct.
For verification, we should use the same range conversion ammeter.
Lab Manual Questions
1. What is an ammeter?
Ans: It is an instrument designed to read electrical circuit currents. It is generally placed in series with the circuit.
2. What will happen if an ammeter is connected in parallel to the circuit?
Ans: If we connect an ammeter in parallel to the circuit, then it will measure only a part of the current flowing through it and not the total current.
3. Why should we connect an ammeter in series in this experiment?
Ans: An ammeter is connected in series in this experiment so that the whole current to be measured in the circuit can pass through the ammeter.
Viva Questions
1. State various types of galvanometers.
Ans: Following are the various types of galvanometer:
Tangent galvanometer,
Moving coil galvanometer,
Ballistic galvanometer,
Astatic galvanometer
2. On which principle does an ammeter work?
Ans: It works on the principle of the magnetic effect of electric current.
3. State two advantages of a moving coil galvanometer.
Ans: The moving coil galvanometer provides high sensitivity, and we get more accurate readings due to its high sensitivity. The second advantage is that strong magnetic fields do not easily affect it.
4. On which factors does the sensitivity of the galvanometer depend?
Ans: The sensitivity of the galvanometer depends on the area of the coil, the number of turns in the coil, the strength of the magnetic field and the magnitude of a couple per unit twist.
5. What should we measure with the help of a Clamp-on ammeter?
Ans: We can measure large alternating currents with the help of a Clamp-on ammeter.
6. Write the SI unit of electric current.
Ans: The SI unit of electric current is Ampere and it is denoted by the symbol A.
7. State different types of the ammeter.
Ans: Different ammeters are rectifier-type ammeters, moving iron ammeters, electrodynamometer ammeters and hot-wire ammeters.
8. State a few disadvantages of the galvanometer.
Ans: Following are some disadvantages of the galvanometer:
It measures only direct current
It is not suitable for high current values
Restoring torque can not be changed easily.
Practical-Based Questions
Which instrument can be used to measure resistance?
Voltmeter
Voltameter
Ammeter
Ohmmeter
Ans: Option D - Ohmmeter
What should we connect in parallel in an electric circuit?
Voltmeter
Volatmeter
Fuse
Ammeter
Ans: Option A - Voltmeter
The main function of a shunt in an ammeter is to
Bypass the current
Increase the resistance of the ammeter
Increase the sensitivity of the ammeter
Increase the specific resistance of the ammeter
Ans: Option A - Bypass the current
Which type of electric current are we using when we turn on our TV at home?
DC
Conductive
AC
Both
Ans: Option C - AC
The symbol for voltage is
W
VO
V
Amp
Ans: Option C - V
What physical quantity does the ammeter measure?
Voltage
Current
Power
Resistance
Ans: Option B - Current
We should connect a voltmeter in _____ across the cell.
Parallel
Series
Either series or parallel
None of the above
Ans: Option A - parallel
Resistance of an ideal ammeter is
Zero
Infinite
Very low
Very high
Ans: Option A - Zero
We should connect an ammeter in _____ across the cell.
Series
Triangular
Either series or triangular
Parallel
Ans: Option A - Series
Resistance of an ideal voltmeter
Very low
Zero
Infinity
Very high
Ans: Option C - Infinity
Conclusion
From the above experiment,
We have found that the actual value and measured value of the potential differences are very small and conversion is perfect.
We can also conclude that the galvanometer can be converted into an ammeter of desired range by connecting a shunt resistance in parallel with the galvanometer this shunt prevents a galvanometer from being damaged due to large currents and it is also used to increase the range of ammeter.
FAQs on Class 12 Physics Conversion Of Galvanometer Into Ammeter Of Desired Range Experiment
1. What is the fundamental principle for converting a galvanometer into an ammeter for the CBSE Class 12 board exam?
The fundamental principle is to allow only a small fraction of the main circuit current to pass through the galvanometer, while the rest is bypassed. This is achieved by connecting a low-resistance wire, called a shunt resistor (S), in parallel with the galvanometer coil. The combination then acts as an ammeter, which is always connected in series in the circuit.
2. Derive the expression for the shunt resistance required to convert a galvanometer into an ammeter of a specific range. This is a frequently asked 3-mark question.
To derive the formula for shunt resistance, let:
- G = Resistance of the galvanometer
- Ig = Current required for full-scale deflection in the galvanometer
- I = The maximum current to be measured by the ammeter (the desired range)
- S = Resistance of the shunt
Since the galvanometer and shunt are connected in parallel, the potential difference across them is the same:
Potential Difference across Galvanometer = Potential Difference across Shunt
Ig × G = (I - Ig) × S
Therefore, the required shunt resistance is S = (Ig × G) / (I - Ig). This formula is crucial for solving numerical problems in the board exam.
3. A galvanometer has a resistance of 15 Ω and shows full-scale deflection for a current of 4 mA. How would you convert it into an ammeter of range 0 to 6 A?
This is a typical numerical question that can be expected for 2 or 3 marks. Here are the steps to solve it as per the CBSE 2025-26 pattern:
Given:
- Galvanometer Resistance (G) = 15 Ω
- Current for full-scale deflection (Ig) = 4 mA = 4 × 10-3 A
- Desired Ammeter Range (I) = 6 A
Formula: S = (Ig × G) / (I - Ig)
Calculation:
S = (4 × 10-3 A × 15 Ω) / (6 A - 0.004 A)
S = (0.06) / (5.996)
S ≈ 0.01 Ω
Conclusion: To convert the galvanometer into an ammeter of range 0-6 A, a very low resistance shunt of approximately 0.01 Ω must be connected in parallel with the galvanometer.
4. Why must an ammeter have a very low effective resistance? Explain how this is achieved in a converted galvanometer.
It is crucial for an ammeter to have very low effective resistance for the following reasons:
- Circuit Integrity: An ammeter is connected in series to measure the current in a circuit. If it had high resistance, it would significantly increase the total resistance of the circuit, thereby decreasing the current it is intended to measure. This would lead to an inaccurate reading.
- Accuracy: A low resistance ensures that the ammeter does not alter the original state of the circuit, providing a more accurate measurement.
This is achieved by connecting a low-resistance shunt (S) in parallel with the high-resistance galvanometer (G). The effective resistance (Reff) of this parallel combination is even lower than the shunt resistance itself (Reff = (G × S) / (G + S)), ensuring the overall device has minimal impact on the circuit.
5. What would happen if the shunt resistor was accidentally connected in series with the galvanometer instead of parallel? Is the device still an ammeter?
If the shunt resistor (S) were connected in series, the device would fail as an ammeter. Here's the impact:
- Extremely High Resistance: The total resistance of the instrument would become very high (Rtotal = G + S).
- Circuit Alteration: When connected in series in a circuit, this high-resistance instrument would drastically reduce the flow of current, essentially acting as an open circuit or a very high-resistance voltmeter.
- Incorrect Function: It would not measure the true current. Instead, it would measure a much smaller current or none at all.
Therefore, the device would no longer function as an ammeter. The parallel connection is essential for bypassing the majority of the current and keeping the overall resistance low.
6. If two ammeters, one with a range of 0-1A (ammeter) and another with a range of 0-10mA (milliammeter), are created from the same type of galvanometer, which one will have a higher resistance and why?
The milliammeter (0-10mA range) will have a higher resistance. This is a common conceptual question based on the shunt formula: S = (Ig × G) / (I - Ig).
Reasoning:
- The value of shunt resistance (S) is inversely proportional to the range of the ammeter (I).
- For the milliammeter, the range 'I' is very small (10mA). This makes the denominator (I - Ig) smaller, resulting in a larger shunt resistance (S).
- For the ammeter, the range 'I' is larger (1A). This makes the denominator larger, resulting in a smaller shunt resistance (S).
7. How does a real ammeter, created from a galvanometer, differ from an 'ideal' ammeter? What is the key implication of this for practical exams?
The key difference lies in their internal resistance:
- An ideal ammeter is a theoretical concept and has zero internal resistance. It measures the current in a circuit without altering it in any way.
- A real ammeter (like a converted galvanometer) has a small but non-zero internal resistance. This is because the parallel combination of the galvanometer and shunt, while very low, is not zero.
Implication for Practical Exams: For practical purposes, this small internal resistance means the ammeter will always slightly decrease the current it is trying to measure. This introduces a small systematic error in the readings. While this error is usually negligible in Class 12 experiments, understanding its source is an important concept for viva voce questions, demonstrating a deeper understanding of the instrument's limitations.

















