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Given:
- (P) Power= 100W
- (V) Potential Difference = 230V
- (I) Current = ?
- (R) Resistance = ?
a) Electricity flowing per second = Electric Current.....
Putting all values.....
I = 100/230
I = 10/23
I = 1/2.3 A ( ampere)
b) Resistance
Putting all the values......
R = 230÷1/2.3
R = 230 × 23/10
R = 23 × 23
R = 529Ω
c) New Power ( Pn) = 25w
Reduce in the power is 75%.......
So to do so....... reduce the electric current consumption of bulb by 75%.....
Rather I am confused in this answer......
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