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Ampere is equal to the one Coulomb charge per second. It is the SI unit of electric current and is named after the French Physicist and Mathematician Andre Marie Ampere. Ampere is denoted by the letter ‘A’. One ampere of current represents one coulomb of electrical charge, that is 6.24×1018 charge carriers, moving in one second.
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Key Terms: Ampere, Ammeter, Electric Current, Coulomb, Electric Charge, Voltage, Watts, Resistance, Volt, Current
What is Ampere?
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Ampere is the unit of electric current. It is equal to the flow of one Coulomb charge per second. One ampere can also be defined as the amount of current generated by one volt acting through one ohm of resistance in one second.
1 Ampere = 1 Coulomb / 1 Second
If the charge on particles traveling through an area receiving current increases, the Ampere value will also increase accordingly.
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Current Electricity Detailed Video Explanation:
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Table of Ampere Unit Prefixes
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In electrical and electronic engineering applications, electric current values can vary over a wide range. This value can be lower than 0.01 A or higher than 1000 A. Using multiples and submultiples of the standard units helps in expressing the current values in a comprehendible way.
The following table summarises most commonly used ampere unit prefixes, their symbols and conversions.
| Name | Symbol | Conversion | Example |
|---|---|---|---|
| Microampere(microamps) | μA | 1 μA = 10-6A | I = 40 μA = 40 × 10-6A |
| Ampere (amps) | A | – | I = 20 A |
| Milliampere(milliamps) | mA | 1 mA = 10-3A | I = 2 mA = 2 × 10-3 A |
| Kiloampere(kiloamps) | kA | 1 kA = 103A | I = 4 kA = 4 × 103 A |
Ampere Conversion
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For real-life applications, it is essential to use appropriate conversion units. The following are some examples of ampere conversions from one unit scale to another.
Conversion of Amps (A) to kiloamps (kA)
A thousand amps are equal to one kiloamp (kA).
1 kA = 1000 A or 1 × 103 A
For Example, 9 amperes of current can be converted to kiloamperes as follows:
9 A × 1000 = 9000 A or 9 kA
Conversion Amps (A) to milliamps (mA)
1000 amperes are equal to one milliampere.
1 mA = 1000 A
For example, 12 A is converted to milliampere as follows:
12 A/1000 = 0.012 A or 12 × 10-3 A = 12 mA
Conversion of Amps (A) to microamps (μA)
1 Ampere = 1000000 or 10-6 microamperes.
1 μA = 1000000 A
For example, 8 A can be converted to microamps as follows:
8 A / 1000000 = 0.000008 A or 8 × 10-6 A = 8 μA
Conversion of Watt, Volt and Ohm into Ampere
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The value of Electric Current can be calculated by using the values of voltage, power and resistance. Following is the conversion of volts, Watts and Ohms to Ampere.
Calculation of Amps with Watts and Volts
The circuit power is calculated using the following formula:
P (Watt) = V (Volt) × I (Ampere)
We may calculate the value of electric current by rearranging this equation as:
I (Ampere) = P (Watt) / V (Volt)
| Solved Example: Ques. Calculate the current flow in a circuit that consumes 50 W of power and has a supply voltage of 10 V? Ans. Using the equation, we can calculate the current as follows: I (A) = 50 W/10 V = 5 A |
Calculation of Amps with Volts and Ohms
The resistance in a circuit is calculated by the following formula:
V (Volt) = I (Ampere) × R (Ohm)
We may compute the value of electric current by rearranging the above equations as follows:
I (Ampere) = V (Volt) / R (Ohm)
| Solved Example: Ques. Calculate the current in a circuit having a voltage of 25 V with a resistance of 5 Ω? Ans. Using the above mentioned equation, we can calculate the value of current as follows: I = 25 V/5 Ω = 5 A |
Also Read: NCERT Solutions for Chapter 3 Current Electricity
Ampere Meter or Ammeter
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- The Ammeter or ampere meter is an electrical instrument which is used to measure electrical current in Amperes.
- By attaching an Ampere meter in series to the load, the electrical current on the load can be measured.
- Ammeter has no resistance due to which the measured circuit is unaffected.
- The ammeter cannot be connected in parallel to the load because It has low resistance. I
- On connecting the ammeter in parallel, it becomes a short circuit path and all the current flows through it at one time.
- This may lead to the burning of the meter due to the high value of current.
| Ques: When an ammeter is connected in parallel to load, what will happen? Ans: Due to its low resistance, the ammeter cannot be connected in parallel to the load. If it is connected in parallel it becomes a short circuit path that will allow all the current to flow through it which may lead to the burning of meter due to the high value of current. An ideal ammeter has zero impedance so that the power loss in the instrument is zero. But this ideal condition is not achievable practically. |
Types of Ammeters
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The ammeters are classified into two types on the basis of the construction design and the type of current.
Classification on the basis of Construction Design
- Electro-dynamometer ammeter
- Moving iron ammeter
- Permanent moving coil ammeter
- Rectifier type ammeter
Classification on the basis of Type of Current:
- DC ammeter
- AC ammeter
Permanent moving coil ammeters are the most common type of DC ammeter. Other ammeters are capable of measuring both AC and DC current.
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|---|---|---|
| Alternator Vs Generator | Electric Generator | Carbon Resistor |
| Meter Bridge | Drift Velocity | Current Density |
Things to Remember
- The ampere is a unit of electric current that is equal to one Coulomb per second of flow.
- Mathematically, 1 ampere is expressed as-1 Ampere = 1 Coulomb / 1 Second.
- The Electric Current can be measured by using the following formula: I = V/R
- Ammeter is used to measure the current flowing in a system.
- The Ammeter is linked in series with the circuit.
- When an ammeter is linked in parallel, it creates a short circuit, allowing all of the current to flow through it, potentially causing the meter to burn due to the high current value.
Previous Year Questions
- A potentiometer wire AB having length L and resistance 12r is joined to… [JEE Main 2019]
- A resistance is shown in the figure. Its value and tolerance are given respectively by ...[JEE Main 2019]
- The resistance between any two vertices of the triangle is...[JEE Main 2019]
- A uniform wire of length l and radius r has a resistance of….
- In a potentiometer experiment, when three cells A,B, and C are connected in series…..[KEAM]
- When two resistances R1 and R2 are connected in series, they consume 12W power….[KEAM]
- Nichrome is used as electrical heating element because of its…[KEAM]
- In the figure shown below, the terminal voltage across E2 is…[KEAM]
- In a potentiometer of wire length ll, a cell of emf V is balanced at a...[KEAM]
- A resistance is shown in the figure. Its value and tolerance are given respectively by ...[JEE Main 2019]
- The resistance between any two vertices of the triangle is...[JEE Main 2019]
- In the given circuit the cells have zero internal resistance. The currents...[JEE Main 2019]
- In The Given Circuit Diagram…...[JEE Main 2019]
- In the given circuit, an ideal voltmeter connected across the 10Ω resistance...[JEE Main 2019]
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Sample Questions
Ques: What is the current in a circuit with a 25-volt voltage and a 5-ohm resistance? (1 Mark)
Ans: Given that- Voltage = 25 V and Resistance = 5 Ohm
As we know that, I (Ampere) = V (Volt) / R (Ohm)
Hence using this equation, we have
I = 25 V/5 Ω
I= 5 A
Ques: Name the device which is used to measure the Electric Current in Ampere. (1 Mark)
Ans: The value of Electric Current in Amperes can be measured by using an Ammeter or Ampere meter.
Ques: Can the Value of Ampere be Negative? (1 Mark)
Ans: Yes, the value of Ampere can be negative. It can be possible when the measured voltage is negative and the reference point is at very high value.
Ques: What is the current flow in a circuit with a 50 W power consumption and a 10 V supply voltage? (1 Mark)
Ans: Given that- Power = 50 W, Voltage = 10 V
The circuit's power is calculated using the following formula:
P (Watt) = V (Volt) × I (Ampere)
We may calculate the value of electric current by rearranging the previous equation:
I (Ampere) = P (Watt) / V (Volt)
We may calculate the current using the equation as follows:
I (A) = 50 W/10 V
I = 5 A
Ques: What happens if an ammeter is linked to a load in parallel? (2 Marks)
Ans: Due to its low resistance, the ammeter cannot be connected in parallel to the load. When linked in parallel, it forms a short circuit, enabling all current to flow through it at the same time. It potentially causes the metre to burn due to the high current value. The power loss in an ideal ammeter is zero since its impedance is zero. However, this ideal situation is not feasible in practice.
Ques: On what basis are the ammeters classified? Give their names. (2 Marks)
Ans: Ammeters are classified on the basis of The ammeters are classified into two types on the basis of the construction design and the type of current.
On the basis of construction design, they are of four type-
- Electro-dynamometer ammeter
- Moving iron ammeter
- Permanent moving coil ammeter
- Rectifier type ammeter
On the basis of type of Current, they are of two type-
- DC ammeter
- AC ammeter
Ques: An ammeter of resitance 1 ? can measure current upto 1.0 A
(i) What must be the value of the shunt resistance to enable the ammeter to measure upto 5.0 (A)?
(ii) What is the combined resistance of the ammeter and the shunt? (Delhi 2013, 2 Marks)
Ans: 
Ques: The reading of the (ideal) ammeter, in the circuit shown here, equals :
(i) I when key K1 is closed but key K2 is open.
(ii) I/2 when both keys K1 and K2 are closed.
Find the expression for the resistance of X in terms of the resistances of R and S. (Comptt. Delhi 2012, 2 Marks)
Ans: Finding the expression for the resistance X
(i) Current I when K2 is open and K1 is closed E
Ques: (a) The potential difference applied across a given resistor is altered so that the heat produced per second increases by a factor of 9. By what factor does the applied potential difference change?
(b) In the figure shown, an ammeter A and a resistor of 40 are connected to the terminals of the source. The emf of the source is 12 V having an internal resistance of 20. Calculate the voltmeter and ammeter readings. (Outside Delhi 2017, 2 Marks)
Ans: (a) If the potential difference is V then the heat produced in a resistor ‘R’ is :
(b) 
Ques: A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance R2 only
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42A, 1.05A, 1.4A and 4.2A, but not necessarily in that order. Identify the currents corresponding to the four cases mentioned above. (2 Marks)
Ans: 
Ques: What is ampere? (2 marks)
Ans: The ampere is a unit of measure of the rate of current or electron flow in an electrical conductor. It is the SI base unit of electric current. One ampere is equal to 6.241509074×10¹⁸ electrons worth of charge moving past a point in a second.
Ques: Although circular loop has symmetrical stucture, why ampere circuital law cannot be used in circular closed loop? (4 marks)
Ans: Ampere circuital law cannot be used in circular closed loop due to the following reasons:
- Ampere's Law is a simple tool in order to evaluate electrically induced magnetic field at a point.
- The magnetic field should be in a pattern which creates an Amperian Loop for Ampere's Law to be applicable.
- Amperian loop should have the magnetic field strength tangential to the loop or normal to the loop created by the magnetic field around it.
- This loop must also have a non zero or a uniform magnetic field in order to be applicable to Amperian Circuital Law.
- In case of a closed circular loop, the magnetic field induced due to the current flowing across the loop does not create such an Amperian loop.
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