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Electricity Test - 46

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Electricity Test - 46
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  • Question 1
    1 / -0
    If the specific resistance of a wire of length ll and radius rr is kk then resistance is
    Solution
    Given :   Specific resistance    ρ=k\rho = k

    Cross-section area of the wire       A=πr2A = \pi r^2

    Resistance of wire    is given by    R=ρlA=klπr2R = \rho \dfrac{l}{A} = \dfrac{kl}{\pi r^2}
  • Question 2
    1 / -0
    Which one of the following statements is not correct?
    Solution
    Switches, fuses and circuit breakers must be placed in the live wire, not in neutral wire. And all the other statements given are correct.
    Hence option C
  • Question 3
    1 / -0
    n identical resistors each of resistance r when connected in parallel, have a total resistance R. When these resistors are connected in series, then effective resistance in terms of R is:
    Solution
    1R=1r+1r+........ n\displaystyle \frac{1}{R} = \frac{1}{r} + \frac{1}{r} + ........ n terms =nr= \displaystyle \frac{n}{r}

     R=rn   r=Rn\therefore \displaystyle R = \frac{r}{n}       \Rightarrow r = Rn
    When they are connected in series
    Req=r+r+......nR_{eq} = r + r + ...... n terms
    =nr=n×Rn=n2R= nr = n \times Rn = n^2 R
  • Question 4
    1 / -0
    What is the effective resistance between points P and Q?

    Solution
    Hint: Resistances are connected in parallel\textbf{Hint: Resistances are connected in parallel}
    Step 1: Find the the type of connection for each resistance
    After moving from first 2r resistance, it has two paths one is with one more 2r resistance and other is pass through the conductor. So, these two 2r resistances are in parallel connection. Similarly, there is one more way that goes from above and connects directly to r which is other option with these 2r path. So, it is also in parallel connection. Hence, all resistance are parallel to each other. 
    Step2: Find the net resistance of the circuit
    Assume that net resistance is R_{net}. So, from parallel resistance formula, 
    1Rnet=12r+12r+1r=1r(12+12+1)=2r\dfrac{1}{R_{net}} = \dfrac{1}{2r} + \dfrac{1}{2r} + \dfrac{1}{r} = \dfrac{1}{r} (\dfrac{1}{2} + \dfrac{1}{2} + 1) = \dfrac{2}{r}

    Rnet=r2\Rightarrow R_{net} = \dfrac{r}{2}
    Answer:Answer:
    Hence, option A is the correct answer. 
  • Question 5
    1 / -0
    A man has five resistors each of value 15Ω\frac {1}{5}\Omega . What is the minimum resistance he can obtain by connecting them ?
    Solution
    Minimum resistance is obtained when all the resistance are connected in parallel.
    Effective resistance in parallel connection      11Rmin=1R+1R+....1\dfrac{1}{R_{min}} = \dfrac{1}{R}+\dfrac{1}{R}+.... 5 terms
        \implies  Minium resistance Rmin=R5=1/55=125ΩR_{min} = \dfrac{R}{5} =\dfrac{1/5}{5} =\dfrac{1}{25}\Omega
  • Question 6
    1 / -0
    A 4Ω\Omega resistance is bent through 180o^o at its midpoint and the two halves are twisted together. Then the resistance is
    Solution
    As the 4Ω4\Omega resistance is folded at its midpoint, thus the resistance of each segment which are now connected in parallel is 2Ω2\Omega.
    Equivalent resistance between A and B        1RAB=12+12\dfrac{1}{R_{AB}} = \dfrac{1}{2}+\dfrac{1}{2}
        \implies      RAB=1ΩR_{AB} = 1\Omega

  • Question 7
    1 / -0
    What is the effective resistance between points P and Q?

    Solution
    Hint: Here all the resistors are connected in parallel\textbf{Hint: Here all the resistors are connected in parallel}
    Step1: Find the type of connection for each resistance
    Here, there are 3 resistance of 12 Ω, 15 Ω and 20 Ω. From point P to Q, there are 3 paths in which all resistance are different. (i.e. current can pass from any resistance). So, all the resistance are in parallel connection with each other. 
    Step2: Find the net resistance of the circuit
    Assume that net resistance of the circuit is R_{net}. So, from the formula of parallel resistance connection, 
    1Rnet=112+115+120\dfrac{1}{R_{net}} = \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{20}

    LCM of 12, 15, 20 is 60. So, 
    1Rnet=5 + 4 + 360=1260=15\dfrac{1}{R_{net}} = \dfrac{5\ +\ 4\ +\ 3}{60} = \dfrac{12}{60} = \dfrac{1}{5}

    Rnet=5 Ω\Rightarrow R_{net} = 5\ Ω
    Answer:Answer:
    Hence, option A is the correct answer. 
  • Question 8
    1 / -0
    The smallest resistance which can be obtained with ten 0.1 ohm resistors is
    Solution
    When the resistances are connected in parallel, the effective resistance becomes less than the smallest individual resistance. Thus smallest resistance is obtained when ten 0.1Ω0.1\Omega resistors are connected in parallel.
    \therefore  Effective resistance      Rp=r10=0.110=0.01ΩR_p = \dfrac{r}{10} =\dfrac{0.1}{10} = 0.01\Omega
  • Question 9
    1 / -0
    We have n resistors each of resistance R. The ratio of the combination for maximum and minimum values is
    Solution
    The maximum resistance is obtained when all the nn resistors are connected in series.
    Thus maximum resistance    Rmax=nRR_{max} = nR
    Minimum resistance is obtained when all the nn resistors are connected parallel to each other.
    \therefore Minimum resistance     Rmin=RnR_{min} = \dfrac{R}{n}
        \implies   RmaxRmin=nRRn=n2\dfrac{R_{max}}{R_{min}} = \dfrac{nR}{\frac{R}{n}} = n^2
  • Question 10
    1 / -0
    A wire had a resistance of 12 Ω12\ \Omega. It is bent in the form of a circle. The effective resistance between two points on any diameter is:
    Solution
    Resistance of the wire, R=12 ΩR=12\ \Omega
    As the length of each segment is half of the complete wire, thus the resistance of each segment, R=122=6ΩR' =\dfrac{12}{2} =6\Omega
    Effective resistance between P and Q (RR' in parallel), ReqR_{eq}
    1Req=1R1+1R2\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

    1Req=1R+1R\dfrac{1}{R_{eq}}=\dfrac{1}{R'}+\dfrac{1}{R'}

    1Req=2R\dfrac{1}{R_{eq}}=\dfrac{2}{R'}

    Req=62R_{eq}= \dfrac62

    Req=3 ΩR_{eq} =3\ \Omega

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