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Test: Current Electricity (15 Nov) - JEE MCQ


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Test: Current Electricity (15 Nov) - Question 1

According to Kirchhoff’s Loop Rule

Detailed Solution for Test: Current Electricity (15 Nov) - Question 1

Kirchhoff’s loop rule is based on the principle of conservation of energy. The work done in transporting a charge in a closed loop is zero. The algebraic sum ( since potential differences can be both positive and negative) of potential differences around any closed loop is always zero.

Test: Current Electricity (15 Nov) - Question 2

The Wheatstone bridge is balanced for four resistors R1,R2,R3 and R4 with a cell of emf 1.46 V. The cell is now replaced by another cell of emf 1.08 V. To obtain the balance again

Detailed Solution for Test: Current Electricity (15 Nov) - Question 2

The balance point of the Wheatstone’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.
 

Explanation:


  • Initial Balanced Wheatstone Bridge: In the initial balanced Wheatstone bridge configuration, the emf of the cell is 1.46 V and all four resistors R1, R2, R3, and R4 are set to specific values to achieve balance.

  • Replacement of Cell: When the cell is replaced by another cell with an emf of 1.08 V, the balance of the Wheatstone bridge is disrupted.

  • Requirement for Rebalancing: In order to rebalance the Wheatstone bridge with the new cell of emf 1.08 V, no resistance needs to be changed.

  • Reasoning: The balance of the Wheatstone bridge is determined by the ratio of the resistances in the bridge arms and not by the absolute values of the resistances. As long as the ratio of the resistances remains the same, the balance will be maintained regardless of the emf of the cell.

  • Conclusion: Therefore, in this scenario, no resistance needs to be changed to obtain the balance again with the new cell of emf 1.08 V.


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Test: Current Electricity (15 Nov) - Question 3

Resistance of a conductor is

Detailed Solution for Test: Current Electricity (15 Nov) - Question 3

According to Ohm’s law, V = IR. Therefore R= V/I

Test: Current Electricity (15 Nov) - Question 4

Two cells of emf 1.25V , 0.75V and each of internal resistance 1Ω are connected in parallel. The effective emf will be

Detailed Solution for Test: Current Electricity (15 Nov) - Question 4

We want the Voltage difference VAB.
let A and B be open and not connected to any thing.
There is a current that flows from cell of larger emf to the cell of small emf.
Call that current as  I Amperes.

1.25 V - 0.75 V - I * R2 - I * R1 = 0

I = 0.5 / (R1+R2)
VAB =  -0.75 - I * R2 =  - 0.75 - 0.5 * R2 / (R1+R2)

= - (0.75 R1 + 1.25 R2) / (R1+R2)

= - ().75 * 1 + 1.25 * 1) / (1 + 1)  volts

=  - 1 volts

Test: Current Electricity (15 Nov) - Question 5

Flow of charges in direction of electrons is called 

Detailed Solution for Test: Current Electricity (15 Nov) - Question 5
  1. Current: This is a general term for the flow of electric charge. It can refer to any type of charge flow, not specifically electrons.

  2. Electronic current: This specifically refers to the flow of electrons through a conductor.

  3. Conventional current: This is the flow of positive charge and is considered to flow from the positive side to the negative side of a power source. Historically, this was defined before the discovery of electrons, so it is opposite to the direction of electron flow.

  4. Photonic current: This is not a standard term in electrical engineering or physics. It might refer to currents related to photons in some contexts, but it is not related to the flow of electrons.

So, if you are referring to the flow of electrons specifically, electronic current is the accurate term.

Test: Current Electricity (15 Nov) - Question 6

In the measurement of resistance by a metre bridge, the current is necessarily reversed through the bridge wire to eliminate

Detailed Solution for Test: Current Electricity (15 Nov) - Question 6

The connecting metal strips include a small resistance in the circuit called the end resistance.On reversing the current, the end resistance tends to get cancelled out.

Test: Current Electricity (15 Nov) - Question 7

 If the length of the filament of a heater is reduced by 10% the power of the heater will

Detailed Solution for Test: Current Electricity (15 Nov) - Question 7

Power P= V2/R​
If length reduced 10% then new resistance of filament will be R′.
R′=R−10% of R
R′=0.9R
Now new power of heater is P2​
P2​= V2​​/R′ = V2/0.9R​​=1.1P
% increase power=11%

Test: Current Electricity (15 Nov) - Question 8

Resistance of a conductor is

Detailed Solution for Test: Current Electricity (15 Nov) - Question 8

According to Ohm’s law, V=I R. Therefore R=V/I

Test: Current Electricity (15 Nov) - Question 9

According to Ohm's law

Detailed Solution for Test: Current Electricity (15 Nov) - Question 9

Ohm’s law states I is proportional to V.This holds good at steady temperatures and for the flow of constant current.

Test: Current Electricity (15 Nov) - Question 10

Potentiometer is

Detailed Solution for Test: Current Electricity (15 Nov) - Question 10

Potentiometer is a long wire of uniform cross section made of manganin.

Test: Current Electricity (15 Nov) - Question 11

The unit of current is

Detailed Solution for Test: Current Electricity (15 Nov) - Question 11

1. Newton is SI Unit of Force
2. Joule is SI Unit of Work Done
3. Coulomb is SI Unit of Charge
Hence, the unit of current is Ampere.

Test: Current Electricity (15 Nov) - Question 12

A Wheatstone bridge ABCD is balanced with a galvanometer between the points B and D. At balance the resistance between the points B and D is :

Detailed Solution for Test: Current Electricity (15 Nov) - Question 12

No current passes through the Galvanometer in the bridge balance condition.Therefore, the resistance between B and D is infinite.

Test: Current Electricity (15 Nov) - Question 13

The instrument for the accurate measurement of the e.m.f of a cell is

Detailed Solution for Test: Current Electricity (15 Nov) - Question 13
  • Both potentiometer and voltmeter are devices to measure potential difference.
  • EMF is the terminal p.d between the electrodes of a cell in open circuit, i.e., when no current is drawn from it.
  • Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection.
  • So, accurate measurement of p.d is done using a potentiometer.
Test: Current Electricity (15 Nov) - Question 14

Electromotive force is

Detailed Solution for Test: Current Electricity (15 Nov) - Question 14
  • An electric field exists in the electrolyte between the positive and negative terminals of the battery.
  • In the external circuit, the current flows from the positive electrode to the negative electrode.
  • To maintain continuity, in the electrolyte, the current (positive charges) flow from the negative electrode (lower potential) to the positive electrode (higher potential).
  • Work done by the source in taking unit positive charge from lower to higher potential is called electromotive force.
Test: Current Electricity (15 Nov) - Question 15

Drift is the random motion of the charged particles within a conductor,

Detailed Solution for Test: Current Electricity (15 Nov) - Question 15

The electrons in a conductor have random velocities and when an electric field is applied, they suffer repeated collisions and in the process move with a small average velocity, opposite to the direction of the field. This is equivalent to positive charge flowing in the direction of the field.

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