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Electricity Test - 58

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Electricity Test - 58
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  • Question 1
    1 / -0
    You are given several identical resistance each of value R=10R=10 and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistance of 55 which can carry a current of 44 ampere. The minimum number of resistance's of the type RR that will be required for this job is
    Solution
    Resistance of each resistor R=10R = 10
    Equivalent resistance of each segment R=R+R=2RR' = R+R = 2R
    \therefore R=2×10=20R' = 2\times 10 = 20
    Such 44 segments are connected in parallel to each other.
    Thus equivalent resistance of the circuit Rp=R4R_p = \dfrac{R'}{4}
        \implies  Rp=204=5R_p = \dfrac{20}{4} = 5
    Maximum current flowing through each resistor is one ampere i.e. i =1Ai  =1 A
    Thus total current flowing through the circuit I=i+i+i+iI = i+i+i+i
    \therefore I=4i=4AI = 4i = 4 A
    Thus minimum 88 resistors are required to satisfy the above conditions.

  • Question 2
    1 / -0
    A $$30\ V,\ 90\ W$$ lamp is to be operated on a 120 V DC120\ V\ DC line. For proper glow, a resistor of ..... should be connected in series with the lamp. 
    Solution
    Resistance of the lamp R=V2PR = \dfrac{V^2}{P}
    \therefore R=30290R = \dfrac{30^2}{90}
        \implies R=10ΩR = 10 \Omega
    Current flowing through the lamp I=PVI = \dfrac{P}{V}
    \therefore I=9030=3AI = \dfrac{90}{30} =3 A
    DC line voltage Vd=120V_d = 120 V
    Resistance of the line must be R=VdIR' = \dfrac{V_d}{I}
    We get R=1203=40ΩR' = \dfrac{120}{3} = 40 \Omega
    Let R1R_{1} be the external resistance to be connected in series with the lamp for proper glow.
    \therefore R+R1=RR + R_1 = R'
    OR 10+R1=4010 + R_1 = 40
        \implies R1=30ΩR_1 = 30 \Omega
  • Question 3
    1 / -0
    1Wh1 Wh (Watt hour) is equal to :
    Solution
    Watt is a unit of power such that  1W=1Js1W =1\dfrac{J}{s}
    Also we know  1h=36001h = 3600 ss
    \therefore  1Wh=1Js×36001Wh = 1\dfrac{J}{s} \times 3600 s=3600s = 3600 JJ
  • Question 4
    1 / -0
    Which of the following shows energy changes that takes place when an incandescent bulb glows using a battery?
    Solution
    Energy stored in the battery in the form of chemical energy. Once the bulb is switched on the chemical energy changes to electrical, followed by heating of the filament then the bulb glows bright and gives light energy.
  • Question 5
    1 / -0
    A 30V-90W lamp is operated on a 120 V DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is
    Solution
    Resistance of lamp
    R0=V2P=(30)290=10ΩR_0 = \dfrac{V^2}{P} = \dfrac{(30)^2}{90} = 10 \Omega
    Current in the lamp
    I=VR0=3010=3AI = \dfrac{V}{R_0} = \dfrac{30}{10} = 3A
    As the lamp is operated on 120V DC, then resistance becomes
    R=Vi=1203=40ΩR' = \dfrac{V'}{i} = \dfrac{120}{3} = 40 \Omega
    For proper glow, a resistance R is joined in series with the bul
    R=R+R0R' = R + R_0
    Rα=RR0=4010=30Ω\Rightarrow R^{\alpha} = R' - R_0 = 40 - 10 = 30 \Omega
  • Question 6
    1 / -0
    Two resistors 400 Ω400\ \Omega and 800 Ω800\ \Omega are connected in series with a 6 V battery. the potential difference measured by voltmeter across 400 Ω400\ \Omega resistor is:
    Solution
    Total resistance of the circuit (series combination)  Rs=R1+R2R_s = R_1+R_2
    \therefore Rs=400+800 =1200ΩR_s = 400+800  =1200\Omega

    Current, I=ERs=61200I=\dfrac{E}{R_s}=\dfrac{6}{1200}

    I=0.005 AI=0.005\ A

    Potential difference across R1R_1,  V=IR1V' = IR_1
    \therefore V=0.005×400V' = 0.005 \times 400
    V=2 VV' =2\ V

  • Question 7
    1 / -0
    In the following circuit, the 1 Ω\Omega resistor dissipates power P. If the resistor is replaced by 9 Ω\Omega, the power dissipated in it is

    Solution
    Equivalent resistance in the circuit =(3+1)=4Ω= (3+1)=4\Omega

    By Ohm's law,
    10 V=(4 Ω)(i Ampere)10 \ V=(4 \ \Omega)(i \ Ampere)
    i=2.5 Ai=2.5 \ A

    Power through a resistance is given by P=I2RP=I^2R 
    Power through 1Ω,1 \Omega, 
    P1Ω=i2×1=6.25 WP_{1 \Omega}=i^2\times 1=6.25 \ W
    Given, P1Ω=PP_{1\Omega}=P 

    When 1Ω1\Omega is replaced by 9Ω9\Omega, equivalent resistance =(3+9)=12Ω=(3+9)=12\Omega
    By Ohm's law, 
    10 V=(12 Ω)(i Ampere)10 \ V=(12 \ \Omega)(i' \ Ampere)
    Current i=1012 Ai'=\dfrac{10}{12} \ A

    Power dissipated through replaced resistor, P9Ω=(1012)2×9=6.25=P1Ω=PP_{9\Omega}=(\dfrac{10}{12})^2\times 9=6.25=P_{1\Omega}=P

    Option A is correct.
  • Question 8
    1 / -0
    Which of the following are the properties of fuse wire? 
    Solution
    All the given options hold true for fuse wire. A fuse wire should have these characteristics:
    • low melting point, 
    • high conductivity and 
    • least deterioration due to oxidation

    It is connected in series with main supply.
  • Question 9
    1 / -0
    The ratio of resistance of two copper wire of the same length and of same cross-sectional area, when connected in series to that when connected in parallel, is
    Solution
    Let RR be the resistance of copper wire. In the first condition,
    Rseries=R+R=2R....(i)R_{series} = R + R = 2R .... (i)
    In the second condition,
    1Rparallel=1R+1R=1Rparallel=1+1R\dfrac {1}{R_{parallel}} = \dfrac {1}{R} + \dfrac {1}{R} = \dfrac {1}{R_{parallel}} = \dfrac {1 + 1}{R}
    Rparallel=R2.....(ii)R_{parallel} =\dfrac {R}{2} ..... (ii)

    On dividing Eq. (i) by Eq. (ii), we get

    RseriesRparallel=2RR2=41\dfrac {R_{series}}{R_{parallel}} = \dfrac {2R}{\dfrac {R}{2}} = \dfrac {4}{1}.
  • Question 10
    1 / -0
    An electric circuit contains an operating heating element and a lit lamp. Which statement best explains why the lamp remains lit when the heating element is removed from the circuit?
    Solution
    When the components of an electrical circuit are connected in parallel, same potential difference appears across each of the element and hence, removal of an element does not affect the working of other elements. Hence, A is the correct option.
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