Unit of Current Important Questions

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Unit of Current is Ampere. It is denoted by the symbol A. It is also denoted by the shorted form ‘amps’. One ampere is equal to 6.241509074 × 1018 electrons worth of charge moving past a point in a second. The charge of an electron e is 1.602 176634 ×10-19 Coulomb. Unit of current is an SI unit and thus used universally for all current measurements. This topic is under Class 12 Chapter 3 Current Electricity. 

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Important Questions on Unit of Current

Ques 1. Define Ampere. (2 marks)

Ans. Ampere is the flow of electric charge at the rate of 1 coulomb per second. Ampere is denoted by the symbol A.

Ques 2. What are the uses of Ammeter? (2 marks)

Ans. Ammeter is a device used to measure electric current. It can measure both alternating current and direct current. It is connected in series inorder to measure the current.

Ques 3. Calculate the resistance of a light bulb if its voltage is 10V and carries a current 5A. (3 marks)

Ans: By Ohm’s law,

V = IR

R = V/I

R =10/5

R = 2 Ohms

Ques 4. A battery connected to a 0.8 amp light bulb can effectively store 5,500 C of charge when fully charged. How long will the battery last before going dead if the light is left on? (3 marks)

Ans. We know that

I = Q/t

t = Q/I

= 5500/0.8

= 6875 sec

= 1.91 hours

Ques 5. A light bulb draws 300mA of current when its voltage is 240V. Calculate the resistance. (3 marks)

Ans. I = 300mA = 300/1000 A

= 0.3 A

V = 240V

By Ohm’s law,

V = IR

R = V/I

R = 240/0.3

R = 800 Ohms

Ques 6. Calculate the current passing through an electrical device of resistance 10 ohms and voltage 100V. (3 marks)

Ans. R = 10 ohms

V = 100V

By Ohm’s law,

V = IR

I = V/R

= 100/10

= 10 Ampere

Ques 7. What is electric current? (2 marks)

Ans. The rate of flow of charges through a conductor with a defined potential difference is called electric current. If Q is the charge flowing through a conductor in a time interval of t, then current I is given by,

I = Q/t

Ques 8. If current flowing through a conductor for 6 minutes id 0.6A, find the charge. (3 marks)

Ans. We know that,

I = Q/t

Q = ?

I = 0.6A

t = 6 min = 360 sec

Q = Ixt

= 0.6 x 360

= 216 C

Ques 9. If 2A current flows for 5 minutes, the amount of charge flowing through the conductor is? (3 marks)

Ans. We know that,

I = Q/t

I = 2A

t = 5 min = 300 sec

Q = Ixt

= 2 x 300

= 600C

Ques 10. If 600C charge flows for 5 minutes, calculate the current flowing through the conductor.  (3 marks)

Ans. We know that,

I = Q/t

I = ?

Q = 600C

t = 5 min = 300 sec

I = Q/t

= 600/300

= 2A

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CBSE CLASS XII Related Questions

  • 1.

    A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.


      • 2.

        (I) The real image of an object placed between $f$ and $2f$ from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? Explain.

        (II) Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.


          • 3.

            The number of free electrons per unit volume ($\text{m}^3$) in a wire made of copper (resistivity $1.72 \times 10^{-8}\,\Omega\,\text{m}$) is $8.5 \times 10^{28}$. Calculate the average number of collisions made by an electron in one second when an electric field is maintained in the wire.


              • 4.

                A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

                  • 80 Wb
                  • 16 Wb
                  • \(8 \times 10^{-2}\) Wb
                  • \(8 \times 10^{-3}\) Wb

                • 5.
                  Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

                    • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})\)
                    • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})\)
                    • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
                    • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})\)

                  • 6.

                    Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

                      CBSE CLASS XII Previous Year Papers

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