If three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel, the equivalent resistance of the combination is:
A. 0.5 Ω
B. 1 Ω
C. 2 Ω
D. 11 Ω
Answer: Option B
Solution (By JKSSB Mock Tests)
For resistors in parallel, the reciprocal of the equivalent resistance (1/Rp) is the sum of the reciprocals of individual resistances. 1/Rp = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6. Finding a common denominator (6): 1/Rp = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore, Rp = 1 Ω.
Explanation:
The relationship between kinetic energy (K) and momentum (p) is K = p² / 2m. If K is increased by 300%, the new kinetic energy K' = K + 3K = 4K. Since p = √(2mK), the new momentum p' = √(2m × 4K) = 2√(2mK) = 2p. The momentum has doubled. Percentage increase = [(2p - p)/p] × 100 = 100%.
Explanation:
The formula for capacitors in series is exactly the opposite of resistors. 1/C_eq = 1/C1 + 1/C2 + 1/C3. Here, all are 3 uF. So, 1/C_eq = 1/3 + 1/3 + 1/3 = 3/3 = 1. Therefore, C_eq = 1 microfarad. When identical capacitors are in series, you simply divide the capacitance by the number of capacitors (3/3 = 1).
No comments yet. Be the first to start the discussion!