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Current Electricity Test - 80

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Current Electricity Test - 80
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  • Question 1
    1 / -0
    Calculate the potential difference between, points A and B current flowing in 10Ω10\Omega resistor in the part of the network below:-                                 

    Solution

  • Question 2
    1 / -0
    A part of circuit with current is shown. The value of II is :

    Solution

    According to KCL the total current that enters ie equal to the total current living from the node. Hence the current in the shown part is zero.

  • Question 3
    1 / -0

    The potential difference across the terminals of a battery is 10V10V when there is a current of 3A3A in the battery from the negative to the positive terminal. When the current is 2A2A in the reverse direction, the potential difference becomes 15V15V. The internal resistance of the battery is?
    Solution

  • Question 4
    1 / -0
    In the circuit shown, the current in the 1 Ω1\ \Omega resistor is :

    Solution

  • Question 5
    1 / -0
    The equivalent resistance between points AA and BB is

    Solution

  • Question 6
    1 / -0
    In the circuit shown below, the cell is ideal, with emf=2Vemf=2V. The resistance of the coil of the galvanometer GG is 1Ω1 \Omega. The steady current through the galvanometer is

    Solution

  • Question 7
    1 / -0
    If the current in the toroidal solenoid increases uniformly from zero to 6.0A6.0 A in 3.0μs3.0\mu s.Self inductance of the toroidal solenoid is 40μH40 \mu H. The magnitude of self-induced emf is
    Solution

  • Question 8
    1 / -0
    In the determination of the internal resistance of a cell using a potentiometer, when the cell is shunted by a resistance "R""R" and connected in the secondary circuit, the balance length is found to be L1L_1. On doubling the shunt resistance, the balance length is found to increase to L2L_2. The value of the internal resistance is:
    Solution

  • Question 9
    1 / -0
    In the adjoining figure the equivalent resistance between AA and BB is

    Solution
    We are given circuit, we have to calculate equivalent between A and B. 
    Clearly in 2nd fig.
    2, 2 and 2 are in series
    So, R1=2+2+2R_1 = 2 + 2 + 2
    =6Ω= 6\Omega
    Now, 6Ω6\Omega and 6Ω6\Omega are in parallel
    R2=6×612=3Ω\therefore R_2 = \dfrac{6\times 6}{12} = 3\Omega
    Now, Equivalent Resistance (RABR_{AB}) = 1 + 3 + 1
    =5Ω= 5\Omega
    Hence option A is correct. 

  • Question 10
    1 / -0
    A horizontal straight conductor of length ll placed along east-west direction falls under gravity from height hh at a place where horizontal component of magnetic field is HH and vertical component of magnetic field is VV. Then :
    Solution

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