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Test: Current Electricity - 1 - JEE MCQ


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30 Questions MCQ Test - Test: Current Electricity - 1

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Test: Current Electricity - 1 - Question 1

According to Kirchhoff’s Loop Rule

Detailed Solution for Test: Current Electricity - 1 - 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 - 1 - 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 - 1 - 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 - 1 - Question 3

Resistance of a conductor is

Detailed Solution for Test: Current Electricity - 1 - Question 3

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

Test: Current Electricity - 1 - 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 - 1 - 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 - 1 - Question 5

Flow of charges in direction of electrons is called 

Detailed Solution for Test: Current Electricity - 1 - 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 - 1 - 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 - 1 - 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 - 1 - 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 - 1 - 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 - 1 - Question 8

Resistance of a conductor is

Detailed Solution for Test: Current Electricity - 1 - Question 8

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

Test: Current Electricity - 1 - Question 9

According to Ohm's law

Detailed Solution for Test: Current Electricity - 1 - 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 - 1 - Question 10

Potentiometer is

Detailed Solution for Test: Current Electricity - 1 - Question 10

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

Test: Current Electricity - 1 - Question 11

The unit of current is

Detailed Solution for Test: Current Electricity - 1 - 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 - 1 - 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 - 1 - 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 - 1 - Question 13

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

Detailed Solution for Test: Current Electricity - 1 - 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 - 1 - Question 14

Electromotive force is

Detailed Solution for Test: Current Electricity - 1 - 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 - 1 - Question 15

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

Detailed Solution for Test: Current Electricity - 1 - 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.

Test: Current Electricity - 1 - Question 16

Unit of Resistivity is

Detailed Solution for Test: Current Electricity - 1 - Question 16

Test: Current Electricity - 1 - Question 17

Potentiometer measures the potential difference more accurately than a voltmeter, because

Detailed Solution for Test: Current Electricity - 1 - Question 17

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 - 1 - Question 18

Meter Bridge is based on the principle of

Detailed Solution for Test: Current Electricity - 1 - Question 18

Meter bridge is the application of wheatstone’s bridge. A Wheatstone bridge is an electrical network which is used to calculate the value of unknown resistance.

Test: Current Electricity - 1 - Question 19

A piece of copper and another of germanium are cooled from room temperature to 80K. The resistance

Detailed Solution for Test: Current Electricity - 1 - Question 19
  • Copper is a conductor and we know that for conductors, resistance is directly proprtional to temperature. Therefore on decreasing temperature resistance also decreases.
  • Whereas, germanium is a semiconductor and for semiconductors, resistance is inversely proportional to temperature.
  • So on decreasing temperature resistance increases.
Test: Current Electricity - 1 - Question 20

Unit of power is

Detailed Solution for Test: Current Electricity - 1 - Question 20

Power is rate of doing work.P=W/t.The unit of power is J/s which is also called watt.

Test: Current Electricity - 1 - Question 21

In a potentiometer experiment, for measuring internal resistance of a cell, the balance point has been obtained on the fourth wire. The balance point can be shifted to fifth wire by

Detailed Solution for Test: Current Electricity - 1 - Question 21

If the current due to the auxillary battery is decreased, the potential gradient will be decreases, so the balancing length increases.Thus null point will move to fifth wire.

Test: Current Electricity - 1 - Question 22

Meter Bridge is used to

Detailed Solution for Test: Current Electricity - 1 - Question 22

With a known resistance in one of the gaps, the meter bridge is used to determine the value of unknown resistance by the formula. 

Test: Current Electricity - 1 - Question 23

According to Kirchhoff’s Junction Rule

Detailed Solution for Test: Current Electricity - 1 - Question 23

Kirchhoff’s junction rule is based on the principle of conservation of charge. For steady currents, there can be no accumulation of charge in a junction. The currents entering a junction bring charge to the junction and the currents leaving a junction carry away the charge brought to the junction.

Test: Current Electricity - 1 - Question 24

Ammeter is always used

Detailed Solution for Test: Current Electricity - 1 - Question 24

Ammeter is a device used to measure current. Since it has to allow the complete current flowing in the circuit through it, it has to be connected in series. For this reason, ammeters have very low values of resistances so that they do not add to the value of resistance connected in the circuit.

Test: Current Electricity - 1 - Question 25

The sensitivity of the potentiometer can be increased by:

Detailed Solution for Test: Current Electricity - 1 - Question 25
  • A potentiometer is considered to be sensitive if the potential gradient dV/dl is low.
  • Such a potentiometer can measure very small changes in potential difference.
  • Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity.
  • Increasing the emf of the primary cell and by decreasing the length, potential gradient increases.
Test: Current Electricity - 1 - Question 26

Direction of the conventional current

Detailed Solution for Test: Current Electricity - 1 - Question 26

Current flows in a conductor due to the flow of negatively charged electrons. However, the direction of conventional current is taken to be opposite to the direction of flow of electrons. It can therefore be considered as the direction in which positive charges move.

Test: Current Electricity - 1 - Question 27

Kirchhoff’s second law is a consequence of

Detailed Solution for Test: Current Electricity - 1 - Question 27

Kirchhoff's second law, also known as the Kirchhoff's voltage law (KVL) states that the sum of all voltages around a closed loop in any circuit must be equal to zero. This again is a consequence of conservation of energy.

Test: Current Electricity - 1 - Question 28

A battery is connected with a potentiometer wire. The internal resistance of the battery is negligible. If the length of the potentiometer wire of the same material and radius is doubled then

Detailed Solution for Test: Current Electricity - 1 - Question 28

Potential gradient Is given by V/l. If V and l are constant, the potential gradient also remains constant. The change in the radius will cause a change in the current. This does not change the potential gradient.

Test: Current Electricity - 1 - Question 29

In a meter bridge experiment a balance point is obtained at a distance of 60 cm from the left end when unknown resistance R is in a left gap and 8 ohms resistor is connected in the right gap. When the position of R and 8 ohm resistor is interchanged the balance point will be at distance of

Detailed Solution for Test: Current Electricity - 1 - Question 29

60/40 = l/100-l, = R/8

so R=12

now changed position 8 is proportional to l and 12 to (100-l)

8/12 = l/100-l

l = 40ohm

Test: Current Electricity - 1 - Question 30

A Wheatstone bridge is balanced for four resistors R1, R2, R3 and R4 with a Lech lanche cell between A and C and a galvanometer between B and D. The positions of the cell and the galvanometer are interchanged. The balance will

Detailed Solution for Test: Current Electricity - 1 - Question 30

The balance point will not change . For balancing of wheatstone bridge four arms of resistors are responsible. Interchange the position of galvanometer and cell does not affect balancing of wheatstone bridge.
 

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