Moving Charges and Magnetism

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  1. The resistance of an ammeter is 13 Ω and its scale is graduated for a current upto 100 amps. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 amperes by this meter. The value of shunt-resistance is








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    ​We know ​ ​ ​

    I
    		 = 1 +
    G
    ISS

    750
    		 = 1 +
    13
    100S

    S ⇒ 2Ω

    Correct Option: A

    ​We know ​ ​ ​

    I
    		 = 1 +
    G
    ISS

    750
    		 = 1 +
    13
    100S

    S ⇒ 2Ω

  1. A galvanometer acting as a voltmeter will have








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    A galvanometer can be converted into a voltmeter by connecting the high ressistance in series with the galvanometer so that only a small amount of current passes through it.

    Correct Option: C

    A galvanometer can be converted into a voltmeter by connecting the high ressistance in series with the galvanometer so that only a small amount of current passes through it.

  1. A galvanometer of 50 ohm resistance has 25 divisions. A current of 4 × 10–4 ampere gives a deflection of one per division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of








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    Rg = 50Ω, Ig = 25 × 4 × 10–AΩ = 10–2 A​​Range of V = 25 volts ​
    V = Ig(Re + Rg) ​

    ∴ Re =
    V
    		 - Rg = 2450 Ω
    Ig

    Correct Option: A

    Rg = 50Ω, Ig = 25 × 4 × 10–AΩ = 10–2 A​​Range of V = 25 volts ​
    V = Ig(Re + Rg) ​

    ∴ Re =
    V
    		 - Rg = 2450 Ω
    Ig

  1. A galvanometer having a resistance of  8 ohms is shunted by a wire of resistance 2 ohms. If the total current is 1 amp, the part of it passing through the shunt will be








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    Is = I ×
    G
    		 = 1 ×
    8
    		 =
    8
    		 = 0.8 amp
    S + G2 + 810

    Correct Option: B

    Is = I ×
    G
    		 = 1 ×
    8
    		 =
    8
    		 = 0.8 amp
    S + G2 + 810

  1. A galvanometer of resistance 20 Ω gives full scale deflection with a current of 0.004 A. To convert it into an ammeter of range 1 A, the required shunt resistance should be








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    Maximum current which can pass through galvanometer,  Ig = 0.004A ​

    Let R be the resistance of shunt. ​
    We know potential drop across AB ​= Potential drop across CD ​
    R (I – Ig) = Ig (20) 
    ​⇒ R (1 – 0.004) = 0.004 × 20 ⇒ R = 0.08 Ω

    Correct Option: C

    Maximum current which can pass through galvanometer,  Ig = 0.004A ​

    Let R be the resistance of shunt. ​
    We know potential drop across AB ​= Potential drop across CD ​
    R (I – Ig) = Ig (20) 
    ​⇒ R (1 – 0.004) = 0.004 × 20 ⇒ R = 0.08 Ω