Collegedunia Team Content Curator
Content Curator
Ampere is the SI unit of electric current flowing through a conductor. Ampere is named after the French Physicist and Mathematician Andre-Marie Ampere who defined one ampere of current as the flow of one coulomb of electrical charge, i.e. 6.24×1018 charge in one second. Ampere is discussed in Current Electricity chapter from the latest CBSE Class 12 Physics Syllabus.
Ampere is denoted by the letter ‘A’ and can be mathematically written as
1 A= 1C/s
Let us understand an ampere of current with a visual representation:
Current flowing through a Conductor
Ampere is measured using the ammeter which indicated how much current is flowing through the circuit. The value of ampere can be positive or negative. The following article lists some of the important questions related to electric current and its measurement in Ampere units.
Important Questions on Ampere
Ques. What is the value of 1 Ampere of current? (2 marks)
Ans. One Ampere current is defined as one coulomb of electrical charge that flows through a conductor in one second. This means 6.24×1018 charge carriers move in one second through a conductor when current flows through it. ohm.
Ques. Define Ampere in terms of the volt and resistance offered by a conducting material. (2 marks)
Ans. One Ampere is defined as the amount of current produced by the force of one volt acting through a resistance of one
Ques. What is the symbol of current? (2 marks)
Ans. Current flowing through the conductor is denoted by the letter ‘I’. As per Ohm’s law, the current flowing through a conductor is directly proportional to the voltage V and resistance R. Mathematically,
V (voltage) = Current (I) x Resistance (R)
Thus, I = V/R
Ques. Convert 9 mA and 10 kA to respective Ampere units. (2 marks)
Ans. 1 mA= 10-3 A
Thus 9 mA= 9 x 10-3 A
= 0.009 A
Similarly 1 kA = 10 3 A.
Thus, 10 kA = 10 x 103
= 10000 A.
Read More:
| Important Topics from Class 12 Chapter 3 Current Electricity | ||
|---|---|---|
| Ohm’s Law | Cells in Series & Parallel | Resistor Applications |
| Current Density Formula | Internal Resistance | Carbon Resistor |
| Electric Charge | Unit of Measurement | Potentiometer |
| Static Electricity | Network Analysis | Resistance and Length Formula |
| Resistance | Power and Resistance | Voltage Divider Formula |
Ques. How to calculate Amperes from Watts and Volts? (2 marks)
Ans. Electric current can be calculated using voltage, power, and resistance as per the following relation:
P (Watt) = V (Volt) × I (Ampere)
Therefore,
I (Ampere) = P (Watt) / V (Volt)
Ques. How to calculate Amperes with volts and Ohms? (2 marks)
Ans. From Ohm’s Law we know that
V (Volt) = I (Ampere) × R (Ohm)
Therefore,
I (Ampere) = V (Volt) / R (Ohm)
Ques. List down the important features of an ammeter. (3 marks)
Ans. Some of the important features of an ammeter are:
- It is used to measure the electric current in a circuit in Amperes.
- The ammeter is attached in series to the load to measure the electric current.
- Ammeter is not connected in parallel in a circuit due to its low resistance value.
- The ammeter connected in parallel becomes a short circuit path and all the current starts to flow at once burning the ammeter due to the excessive value of the current.
Check Out:
Ques. State the different types of Ammeter. (3 marks)
Ans. The different ammeter types are
- Electro-dynamometer ammeter
- Moving iron ammeter
- Permanent moving coil ammeter
- Rectifier type ammeter
Ques. The potential difference and the resistance in a given electric circuit are 20 V and 2 Ω respectively. What is the current flowing through the circuit? (2 marks)
Ans. Given: V = 20 V
R = 2 Ω
From Ohm’s Law V= IR
Therefore, I = V/R
Thus, I = 20/2
=10 A
Ques. What is the potential difference across the circuit with an electric current of 40 A and resistance if the wire being 15 Ω? (2 marks)
Ans. Given: R = 15 Ω
I = 40 A
Thus, from Ohm’s Law V=IR
Substituting we get,
V= 15 x 40
= 600 V
Ques. A parallel combination of 3 resistors takes a current of 7.5 A from a 30 V battery. If the two resistances are 10 Ω and 12 Ω, find the third resistance. (3 Marks)
Ans. Let the resistances of three resistors be R1, R2, and R3.
Given
- R1 = 10 Ω
- R2 = 12 Ω
- I = 7.5 A
- V = 30 V
The total resistance of the circuit is given by Ohm’s law
R = V/I = 30/7.5 = 4 Ω
Since all three resistors are connected in parallel, therefore equivalent resistance of the circuit is given by
1/R = 1/R1 + 1/R2 + 1/R3
Substituting the value of R, R1, and R2 in above equation, we get
1/4 = 1/10 + 1/12 + 1/R3
⇒ 1/R3 = 1/4 – 1/10 – 1/12
⇒ 1/R3 = 1/15
⇒ R3 = 15 Ω
Ques. In copper, there are 1028 electrons in a unit cubic meter, all of which contribute to a current of 2 A in the wire of copper of 1 x 10-6 m2 cross-sectional area. What is the electric field in the wire? Given, the resistivity of the material of wire ρ = 1.6 x 10-8 Ωm. (5 Marks)
Ans. Given
- The number density of electrons in the wire, n = 1028 m3
- Current flowing through the wire, I = 2 A
- Area of the cross-section of the wire, A = 1 x 10-6 m2
- Resistivity of the material of wire ρ = 1.6 x 10-8 Ωm
The current density is given by
J = I/A = 2 /1 x 10-6 = 2 x 106 Am2
Also, the current density in terms of electric field E, is given by
J = σE
Where σ is the conductivity of the wire and it is reciprocal of resistivity i.e. σ = 1/ρ
⇒ E = J x ρ
Substituting the value of J and ρ, we get
E = (2 x 106) x (1.6 x 10-8) = 3.2 x 10-2 Vm-1
Hence, the electric field in the wire is 3.2 x 10-2 Vm-1
Check Out:
Comments