Two 100 W, 200 V bulbs are connected in series across 200 V supply. Total power consumed is
A. 200 W
B. 100 W
C. 50 W
D. 25 W
Answer: Option C
Solution (By JKSSB Mock Tests)
Each bulb resistance R = V²/P = 40000/100 = 400 Ω. In series, total R = 800 Ω. Total power P_total = V²/(total R) = 200²/800 = 40000/800 = 50 W. So each bulb gets 25 W. Series reduces power.
Explanation:
R ∝ L/A. New L'=2L, A'=A/2 => R' = ρ(2L)/(A/2) = 4ρL/A = 4R. Alternatively, volume constant? Not given, here both changed independently. So 4R.
Explanation:
An ideal solenoid consists of tightly packed coils. When current passes through it, the magnetic field lines inside the solenoid are parallel straight lines. This indicates that the magnetic field is uniform in magnitude and direction at all points strictly inside the central region of the solenoid. The field strength is given by B = mu_0 * n * I.
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