PhysicsCompare the power consumed by a 2 Ω resistor in each of the following circuits:                                                     (OR)Find the current and resistance drawn by a bulb rated 40 𝑊; 220 𝑉, when it is connected to a 220 𝑉 supply. If the given bulb is replaced by a bulb if rating 25 𝑊; 220 𝑉, will there be any change in the value of current and resistance? Justify your answer and determine the change.

Compare the power consumed by a 2 Ω resistor in each of the following circuits:

                                                     (OR)

Find the current and resistance drawn by a bulb rated 40 𝑊; 220 𝑉, when it is connected to a 220 𝑉 supply. If the given bulb is replaced by a bulb if rating 25 𝑊; 220 𝑉, will there be any change in the value of current and resistance? Justify your answer and determine the change.

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    Solution:

    Given resistance V=6V, R1=1Ω and R2=2Ω to find l

    Total Resistance,

     Rs=R1+R2 Rs=1+2    Rs=3Ω

    According to Ohm's Law, V= IRS

    I=VRs   I=63  I=2A

    In a series combination, the current is same for all resistors, but the potential difference differs. 

    So, let the potential difference of 2 Ohm Resistor = V1 volts. 

    Let the power of 2 Ω resistors = P1W

    V1=IR2 V1=2×2 V1=4V

    Electric Power = V X I

    Power of 2 Ω Resistor = 𝑉1 × I

    P1=4×2P1  =8W

    For case (ii): Let us consider total current in circuit = 𝐼 

    In parallel combination V is same but I differ. 

    Let total resistance = Rp

    1Rp=1R1+1R21Rp=11+121Rp=2+121Rp=32Rp=23Ω

    According to Ohm's Law,

    V=IRp4=I×232=I3I=6A Now, I2=VR2I2=42I2=2A

    Let the power of R2=P2W

     P2=V×I2  P2=4×2    P2=8W

           

                                                (OR)

    Case 1: power of bulb, P = 40 watt 

    The voltage supplied across the bulb, 𝑉 = 220𝑉 

    We know, power supplied by bulb, 𝑃 = 𝑉𝐼

    or, 40 = 220 × 𝐼

    Current 𝐼 = 40/220 = 2/11 = 0. 18 𝐴𝑚𝑝 

    Again, using the formula, 𝑃 = 𝑉²/𝑅 to find resistance of bulb

    or, 𝑅 = 𝑉²/𝑃 

    𝑅 = (220)²/40 = 48400/40 = 1210 𝑜ℎ𝑚

    Case 2: if the given bulb is replaced by a bulb operating 25 𝑊; 220 V

    Current, 𝐼' = 𝑃'/𝑉 = 25/220 = 0. 1136 𝐴𝑚𝑝 

    resistance, 𝑅' = 𝑉²/𝑃' = (220)²/25 = 1936 𝑜ℎ𝑚. 

    When we change the bulb, current and resistance will be changed. 

    Change in current = 𝐼 − 𝐼' = 0. 18 − 0. 1136 = 0. 0664 𝐴𝑚𝑝 

    The current decreases when we use 25 watt bulb in place of 40 watt bulb. 

    Change in resistance = 1936 − 1210 = 726 𝑜ℎ𝑚

     The resistance increase when we use 25 watt bulb in place of 40 watt.

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