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Electricity Test - 32

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Electricity Test - 32
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  • Question 1
    1 / -0
    Which of the following is a conductor ?
    Solution

  • Question 2
    1 / -0
    A (75 - j40 ) Ω\Omega load is the connected to a coaxial line of z0=75Ωz _ { 0 } = 75 \Omega The load matching on the line can be accomplished by connecting
    Solution
    A (75 - j40 ) ΩΩ load is the connected to a coaxial line of z0=75Ωz0=75Ω The load matching on the line can be accomplished by connecting a short circuit stub at the load.
    so the option is A.
  • Question 3
    1 / -0
    A cell of emf 6 V6\ V and resistance 0.5 ohm0.5\ ohm is short circuited. The current in the cell is 
    Solution
    i=Er=60.5=12 ampi=\dfrac Er=\dfrac{6}{0.5}=12\ amp
  • Question 4
    1 / -0
    The current in a car headlamp is 3.03.0A when connected to a 1212V battery. What is the resistance of the lamp when it is lit?
    Solution
    Given that,
    Current in a car headlamp ,I=3.0 AI=3.0\ A
    Voltage applied across it , V=12 VV= 12\ V

    Resistance of the lamp  ,R=VI=123=4.0 ΩR=\dfrac VI= \dfrac{12}{3}= 4.0\ \Omega
  • Question 5
    1 / -0
    If four resistances are connected as shown in the fig. between A and B, the effective resistance is:

    Solution
    Simplified, or equivalent circuit diagram is shown in the figures.

    From figure, effective resistance:

    Req=4×42×4 R_{eq} = \dfrac{4 \times4}{2 \times 4} =2Ω = 2 \Omega

  • Question 6
    1 / -0
    The resistivity of a conductor with a current density  of 2.5Am2.5 Am2^{-2}  when an electric field 5×108Vm15\times 10^{-8}Vm^{-1} is applied to it, is
    Solution

    Step 1:\textbf{Step 1:}  Relation between Electric field(E), Resistivity and Current density(J)\textbf{Relation between Electric field(E), Resistivity and Current density(J)}

    Suppose length of conductor =l = l
    We know that, Potential difference, V=E.lV=E.l
    $$\textbf{&}$$ Current Density J=I/AJ=I/A
    So, I=JA I = J A
    and   R=ρlA R=\dfrac{\rho l}{A}

    From V=IR V = IR
    E.l=JA×ρlA\Rightarrow E.l = J A \times \dfrac{\rho l}{A}
    E=J(ρ)\Rightarrow E = J(\rho)
    ρ=EJ\therefore \rho = \dfrac{E}{J}

    Step 2: Putting the values in above equation\textbf{Step 2: Putting the values in above equation}
    ρ=EJ=5×108Vm12.5Am2\Rightarrow \rho =\dfrac{E}{J}= \dfrac{5 \times 10^{-8}Vm^{-1}}{2.5Am^2}   =2×108Ωm = 2 \times 10^{-8} \Omega m

    Hence, Option B is correct
  • Question 7
    1 / -0
    A wire of resistance 20Ω20\Omega  is bent in the form of a square. The resistance between the ends of diagonal is:

    Solution
    Since the wire is bend in form of the square and we know that sides of square are equal so each side will have same resistance i.e 5 ohm. Now if we make diagonal two faces in each side of the diagonal will have two sides and resistance will be 10 ohm.
    R=20ΩR=20 \Omega
    So, R4=5Ω\dfrac{R}{4}=5 \Omega
    \Rightarrow R1=10Ω R_{1} = 10 \OmegaR2=10ΩR_{2} = 10\Omega
    So, resistance between the ends of a diagonal = RD=10×102×10Ω=5Ω R_{D} = \dfrac{10 \times10}{2 \times10}\Omega = 5\Omega

  • Question 8
    1 / -0
    The resistance of a conductor is 1.08 Ω\ \Omega . To reduce it to 1 Ω\ \Omega , the resistance that must be connected is:
    Solution
    To reduce the resistance value we should connect another resistance in parallel.
    So, R=R1R2R1+R2 R = \dfrac{R_{1}R_{2}}{R_{1}+R_{2}}
    or, 1=1.08×R21.08+R2 1 = \dfrac{1.08 \times R_{2}}{1.08 + R_{2}}
    or, 1.08+R2=1.08R2 1.08 + R_{2} = 1.08 R_{2}
    or, 1.08=0.08R2 1.08 = 0.08 R_{2}
    \therefore R2=1088=13.5  Ω R_{2} = \dfrac{108}{8} = 13.5\  \Omega
  • Question 9
    1 / -0
    Three resistances each of 3Ω3\Omega  are connected as shown in figure. The resultant resistance between A and F is:

    Solution
    To draw the circuit bring point B to D and Point C to point E. From the diagram, inner two resistances are in parallel,  and its equivalent is also parallel to the third resistance.
    RAF=R2×R3R2\therefore R_{AF} = \dfrac{\dfrac{R}{2} \times R}{\dfrac{3R}{2}}
                =R3 = \dfrac{R}{3}
    RAF=1Ω\Rightarrow R_{AF} = 1 \Omega

  • Question 10
    1 / -0
    The least resistance that one can have from six resistors each of 0.1 ohm resistance is:
    Solution
    Least resistance is achieved when all the resistors are connected in parallel.
     Req=R6=0.16=0.0167 Ω \therefore  R_{eq} = \dfrac{R}{6} = \dfrac{0.1}{6} = 0.0167\ \Omega
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