Electricity

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4 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- If the value of R is increased, the current through the wire will decrease which in turn decreases the potential difference across AB, and  hence potential gradient (k) across AB decreases.
Since, at neutral point, for given emf of cell, l increases as potential gradient (k) across AB has decreased because E' = kl
Thus, with the increase of l, which will result in increase in balance length. So, jockey J will shift towards B

New answer posted

4 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – power consumed through wire is P=V2/R =I2R . for transmitting the power there are two types either current is large or voltage is less or vice versa. But when current in the circuit is higher then power loss is also higher . so we transmit the power through high voltage circuits.

New answer posted

4 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Alloys are used for making standard resistance coil because they have low temperature coefficient of resistance with less temperature sensitivity. The alloys also have high resistivity and hence high resistance, because for given length and cross-section area of conductor .

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- For the selection of metal for wiring in home the main criterion are: the availability, conductivity and the cost of the metal. The Cu wires or Al wires are used  for wiring in the home. The main considerations are involved in this process are cost of metal and good conductivity of metal.

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4 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Metallic strips have negligible resistance and need not to be counted in the length l1, of the null point of potentiometer. That's why the thick metallic strips are used in potentiometer. It is for the convenience of experimenter as he measures only their lengths along the straight segments each of lengths 1 m.

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4 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- In a Wheatstone bridge the main advantage of null point method is that the resistance of galvanometer does not affect the balance point, there is no need to determine current in resistances and the internal resistance of a galvanometer. It is convenient and easy method for observer.
The R unknown can calculated applying Kirchhoff's rules to the circuit. We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer.

The necessary and sufficient condition for balanced Wheatst

...more

New answer posted

4 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- relaxation time =mean free path/rms velocity of electron

Also ρ = 1σ = mne2t relaxation time is inversely proportional to velocities

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4 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – In the circuit when an electron approaches a junction, in addition to the uniform E that faces it normally (which keep the drift velocity fixed), as drift velocity (vd) is directly proportional to Electric field (E). That's why there are accumulation of charges on the surface of wires at the junction.
These produce additional electric fields. These fields alter the direction of momentum. Thus, the motion of a charge across junction is not momentum conserving

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4 months ago

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- applying kirchhoff's junction rule I1 = I+I2

Applying kirchhoff's rule in outer loop containing 10V cell

10= IR+10I1……………. (1)

Applying kirchhoff's rule in outer loop containing 2V cell

2= 5I2-RI= 5 (I1-I)-RI

4= 10I1-10I-RI………… (2)

From 1 and 2

6=3RI+10I

2=I (R+10/3)

V= I (R+Reff)

After comparing V=2V, Reff= 10/3 ohm

Since effective internal resistance Reff of two cells 10/3 ohm, being the parallel combination 5 ohm and 10 ohm . the equivalent circuit is

 

New answer posted

4 months ago

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneVd  or Vd= i/neA

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