The currents i1 and i2 through the resistors R1 = 10 Ω

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 Multiple Choice QuestionsMultiple Choice Questions

401.

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


402.

The equivalent resistance across A and B is

  

  • 2 Ω

  • 3 Ω

  • 4 Ω

  • 5 Ω


403.

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


404.

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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405.

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

  • 9

  • 11.6

  • 14.5

  • 21.2


406.

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


407.

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


408.

Resistors P and Q are connected in the gaps of the meter bridge. The balancing point is obtained 13 m from the zero end. If a 6 Ω resistance is connected in series with P the balance point shifts to 23 m from the same end. P and Q are

  • 4, 2

  • 2, 4

  • both (a) and (b)

  • neither (a) nor (b)


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409.

The currents i1 and i2 through the resistors R1 = 10 Ω and R2 = 30 Ω in the circuit diagram with E1 = 3 V , E2 = 3 V and E3 = 2 V are respectively

  • 0.2 A, 0.1 A

  • 0.4 A, 0.2 A

  • 0.1 A, 0.2 A

  • 0.2 A, 0.4 A


A.

0.2 A, 0.1 A

In closed loop EFGDE

    i2R2 = E2

i2 × 30 = 3

        i2 = 0.1 A

In closed loop ABCEA

− i1R1 − E1 + E2 + E3 = 0

−i1 × 10 − 3 + 3 + 2  = 0

                              i1 = 0.2 A


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410.

A milliammeter of range O - 30 mA has internal resistance of 20Ω.The resistance to be connected in series to convert it into a voltmeter of maximum reading 3V is

  • 49 Ω

  • 80 Ω

  • 40 Ω

  • 30 Ω


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