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Electric Current Test - 11

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Electric Current Test - 11
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  • Question 1
    1 / -0

    A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is 0.5 A. The terminal voltage of the battery, when the circuit is closed, is

    Solution

    V = E - Ir = 10 - 0.5 x 3 = 8.5 V

  • Question 2
    1 / -0

    A galvanometer of resistance 22.8 Ω measures 1 A. How much shunt should be used, so that it can be used to measure 20 A?

    Solution

    To measure 20 A, 19 A passes through shunt (in parallel) and 1 A through ammeter.

    Rshunt/Rgalvanometer = Igalvanometer/Ishunt = 1/19

    Rshunt = 1/19 x Rgalvanometer = 1.2 Ω

  • Question 3
    1 / -0

    In a potentiometer, the null point is received at 7th wire. If now we have to change the null point at 9th wire, what should we do?

    Solution

    The working of potentiometer is based on the fact that the fall of potential across any portion of the wire is directly proportional to the length of that portion provided the wire is of uniform area of cross-section and a constant current is flowing through it.
    To shift the balance point on higher length, the potential gradient of the wire is to be decreased. The same can be obtained by decreasing the current of the main circuit, which is possible by increasing the resistance in series of potentiometer wire.

  • Question 4
    1 / -0

    An electric meter of internal resistance 20Ω gives a full scale deflection when 1 mA current flows through it. The maximum current, that can be measured using three parallel resistors of resistance 12Ω each, in mA is

    Solution

    Voltage Source = I x R = 1 mA x 20 ohms = 20 mV
    Equivalent Resistance of three equal resistors = 4 ohms
    Therefore, Current drawn will be = 20 mV divided by 4 ohms = 5 mA

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