PhysicsStudy the circuit shown in which three identical bulbs B1, B2 and B3 are connected in parallel with a battery of 4.5 V. Each bulb has a wattage of 1.5 W. The total resistance of the circuit is

Study the circuit shown in which three identical bulbs B1, B2 and B3 are connected in parallel with a battery of 4.5 V. Each bulb has a wattage of 1.5 W. The total resistance of the circuit is


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  1. A
    4.5 ohm
  2. B
    5.5 ohm
  3. C
    6.5 ohm
  4. D
    None  

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    Solution:

    Study the circuit shown in which three identical bulbs B1, B2 and B3 are connected in parallel with a battery of 4.5 V. The total resistance of the circuit is 4.5 ohm.
    It is given that each bulb has a wattage of 1.5 W. But we know the relation between power, voltage and resistance as Power P= V2R.
    Given that, Voltage V=4.5 V, Power P=1.5 W
    By substituting voltage and power, we get resistance in the circuit. It is given by the formula,
    Resistance R= V2P
          4.521.5
          13.5 ohm
    To find out current in the circuit, we know the relation, Power P=IV
    Substituting the values of V and R in the above equation, we get Current I=1 A.
    As the bulbs B1, B2 and B3 are connected in parallel, the voltage across all the components remains the same but the current is distributed equally among the three bulbs. So, I1=13 A
    We know that, P=I2R
          1.5=I2R
          R=1.5132
          R=13.5 ohm In a parallel circuit, current remains the same, potential difference and resistance are distributed.
    So, R1=R2=R3=R/3
    Total resistance in the circuit is 13.53=4.5 ohm.
    Thus, total resistance in the circuit is 4.5 ohm.
     
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