Which of the following set up can be used to verify the from Class Physics Current Electricity

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CBSE

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NEET Physics : Current Electricity

Multiple Choice Questions

181.

Two identical cells of the same emf and same internal resistance give the same current through an external resistance 22, regardless of whether they are connected in series or parallel. The internal resistance of the cell is

  • 1 Ω

  • 2 Ω

  • 3 Ω

  • 4 Ω



182.

A uniform wire of resistance R and length L is cut into four equal parts, each of length L/4, which are then connected in parallel combination. The effective resistance of the combination will be 

  • R

  • 4R

  • R4

  • R16



183.

For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between

   

  • B and C

  • C and D

  • A and D

  • B1 and C1



184.

Which of the following set up can be used to verify the Ohm's law ?



185.

The equivalent resistance across A and B is

  

  • 2 Ω

  • 3 Ω

  • 4 Ω

  • 5 Ω



186.

In the given circuit the equivalent resistance between the points A and B in ohms is

  • 9

  • 11.6

  • 14.5

  • 21.2



187.

In the electric circuit shown each cell has an emf of 2 V and internal resistance of 1Ω. The external resistance is 2Ω. The value of the current I is (in amperes)

  • 2

  • 1.25 Ω

  • 0.4

  • 1.2



188.

Six equal resistance are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum between

  • P and Q

  • Q and R

  • P and Q

  • only two points



189.

Which of the following graphs represent variation of magnetic field B with distance r for a straight long wire carrying current ?



190.

The figure shows variation of photocurrent with anode potential for a photo- sensitive surface for three different radiations. Let Ia , lb and Ic be the intensities and va , vb and vc , be the frequencies for the curves a, b and c respectively. Then

  • va = vb and Ia ≠ Ib

  • va = vc and Ia = Ic

  • va = vb and Ia = Ib

  • vb = vc and Ib = Ic



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