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Electricity Test - 73

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Electricity Test - 73
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  • Question 1
    1 / -0
    Two wires A and B of same metal are connected in parallel. Wire A has length $$l$$ and radius $$r$$ and wire B has length $$2l$$ and radius $$2r$$. then the ratio of total resistance of parallel combination and the resistance of wire A is
    Solution

  • Question 2
    1 / -0
    Five resistances of R $$\Omega$$ were taken. First three resistances are connected in parallel combination and rest two are connected in series combination, then the equivalent resistance is :
    Solution
    $$R_{equi} = \frac{R}{3} + R + R = \frac{7R}{3} \ \Omega$$
  • Question 3
    1 / -0
    1μA =________  mA1μA =________  m
    Solution
    One microampere is $$1μA = 10^{-6}A$$ 
    On rewriting it,
     $$1 \mu A=(10^{-3}) \times (10^{-3})A$$
    Since $$1 mA=10^{-3}A$$
    So $$1 \mu A= 10^{-3}mA$$
  • Question 4
    1 / -0
    The resistance of one conducting wire is $$10 $$. How much electric current will flow by connecting it with a battery of $$1.5 V$$ ?
    Solution
    From Ohm's law,
    $$V =  IR$$
    $$ \Rightarrow I = \dfrac{V}{R}= \dfrac{1.5}{10}= 0.15A = 150 mA$$
  • Question 5
    1 / -0
    If the five equal pieces of a resistance wire having $$5 $$ resistance each is connected in parallel, then their equivalent resistance will be
    Solution
    Number of resistance given $$n=5$$
    Resistance of each resistor $$R=5\Omega$$
    All resistors are connected in parallel 

    $$\dfrac 1{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_5}+\dfrac{1}{R_5}=\dfrac{5}{R}$$

    $$\dfrac 1{R_{eq}}=\dfrac{5}{5}=1\ \Omega$$
  • Question 6
    1 / -0
    Materials which allow current to pass through them easily are called?
    Solution
    Materials which allow current to pass through them easily are called as conductor. The most common conductors are metals: copper, silver, gold, platinum and aluminum etc.
    So, the correct option is $$(B)$$
  • Question 7
    1 / -0
    The amperage of the fuse wire used in a circuit that works on $$230 \ V$$ is $$2.2 \ A$$. If so. the power of the device is
    Solution
    The amperage of the fuse wire used in a circuit that works on $$V = 230 \ V$$ is $$i = 2.2 \ A$$.
     
    Power, $$P = Vi = 230 \times 2.2 = 506 \ W$$

    Hence, the power of the device is between $$500 \ W$$ and $$510 \ W$$.

    The correct option is C.
  • Question 8
    1 / -0
    If $$500\Omega$$ of resistance is made by adding five $$100\Omega$$ resistance of tolerance $$4\%$$, then the tolerance of the combination is .
    Solution
    Individual resisistors are having resistance:
    $$R=100\pm 4$$
    On combining 5 similar resistors we get:
    $$5R=500\pm 20$$
    So, tolerance of the combination is also $$4\%$$
  • Question 9
    1 / -0
    Answer the question:
    Two $$5.0\Omega$$ resistors are connected as shown in the diagram.
    What is the total resistance of the combination ?

    Solution
    As we can see that both resistance are connected in parallel.
    For parallel combination:
    $$R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$$

    $$R_{eq}=\dfrac{5\times 5}{5+5}=2.5\Omega$$

    Hence option A is correct, equivalent resistance is less than 5 ohm
  • Question 10
    1 / -0
    In which direction conventional current flow in an electric circuit: 
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