CBSE Class 12 - Physics - Electric Charges And Fields - Very Short Q and A

Electric Charges And Fields

Very Short Q and A

Charging By Induction
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Q1: Define frictional electricity.

Answer: It is the electricity developed on bodies when they rubbed against each other.


Q2: What is the smallest amount of charge?

Answer: Charge on an electron or proton is ± 1.62 × 10^-19 C


Q3: State the two basic properties of electric charge?

Answer:  i. Electric charge is quantised.
ii. Electric charge is always conserved.



Q4: Define Dielectric constant.

Answer: The Dielectric constant of a medium is the ratio of the force between two charges placed at a certain distance apart in vacuum(or air) to the force between the same two charges at the same distance apart in that medium.


Q5: What is induction?

Answer: It is a process of charging a body without making physical contact.

CBSE Class 12 - Physics- Marking Scheme- Sample Question Paper (2016)

CBSE Class 12 - Physics- Marking Scheme- Sample Question Paper (2016) 

CBSE Class 12 - Physics - Sample Question Paper (2015)

CBSE Class 12 - Physics- Sample Question Paper (2015) 

CBSE Class 12 - Physics - CH3 - Current Electricity (NCERT Solutions)

Current Electricity

NCERT Solution 

Q1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?

Answer: Emf of the battery, E = 12 V
    Internal resistance of the battery, r = 0.4 Ω
    Maximum current drawn from the battery = I

   According to Ohm’s law,
          E = Ir
          I = E/r  = 12/0.4 = 30 A
   Thus, the maximum current drawn from the given battery is 30 A.


Q2: A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer: Given,
                Emf of the battery, E = 10 V
                Internal resistance of the battery, r = 3 Ω
                Current in the circuit, I = 0.5 A
                Resistance of the resistor = R
       Applying Ohm’s law,
                I = E/(R + r)
           R + r = E/I  = 10/0.5 = 20 Ω
           ∴ 20 - 3 = 17 Ω

       Terminal voltage of the resistor = V
       According to Ohm’s law,
                    V = IR = 0.5 × 17 = 8.5 V
       ∴ the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V.

Q3: a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
       b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer:
    (a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series.
         Total resistance of the series combination is the algebraic sum of individual resistances.
         ⇒ Total resistance = 1 + 2 + 3 = 6 Ω