Resistance of a bulb is R = V²/P . For 25 W: R₁ = 220²/25 = 1936Ω and for 100 W: R₂ = 220²/100 = 484Ω. Total resistance Rtotal = 1936 + 484 = 2420Ω. Current I = 440/2420 ...
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The resistance of each lamp is R = V/P = 100/50 = 200Ω. Let n be the number of lamps. The total resistance in parallel is: 1/Req = n/200 Including battery resistance: Rtotal = 200/n + 10 For full power:
Given: Supply voltage = 120V, Lead wire resistance = 6Ω, Bulb power = 60W, Heater power = 240W. using R = V²/P, Bulb resistance = 240, Heater resistance = 60Ω. Current before: 0.5A, Current after: 2.5A. Voltage drop increase: 12V.
The fuse current (I) is proportional to the square root of the cross-sectional area. Since area is proportional to the square of the radius, doubling the radius increases I by √4 = 2, giving = 5 × 2¹.5 = 14.7 ...