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:
Discussion Forum Latest Questions
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 ...
For four identical resistors in parallel, the effective resistance is: R/4 = 0.25 ⇒ R = 1Ω In series, total resistance is: R series= 4R = 4 × 1 = 4Ω Thus, the correct answer is (a) 4 Ω.
Resistance of a wire is given by R = ρ L/A . When stretched, L ′ = 1.5L and = A/(1.5)², so new resistance: R ′ = ρ 1.5L/A/(1.5)² = 2.25R To make R′ = 4R, solving gives fraction 1/6. ...