Let the EMF of the cell be E and internal resistance be r. Using Ohm’s law:For 2Ω resistor: E = 0.9 (2 + r) For 7 resistor E = 0.3 (7 + r) Equating both: 0.9(2 + r) = 0.3 ...
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The resistance of the bulb is R = 220/100 = 484Ω. When operated at 110V, power consumed is P ′ = 110²/484 = 25W. Correct answer: (a) 25 W. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions ...
Current I is given by I = Q/t. Student A: Q = 300C. t = 60s, so I = 300/60 = 5A(correct). Student B: Q = (3.125 × 10¹⁹) × (1.6 × 10⁻¹⁹) = 5C, t = Is, so I ...
The internal resistance r of the cell is given by: r=R l1/l2−1 Given l1 =120 cm, l2 = 40 cm and R=1Ω: r = 1 ×120/40 −1 = 2Ω Answer: (d) 2Ω. Class 12th Science Physics NCERT MCQ Questions NCERT Books ...
The energy dissipated as heat in a resistor is given by: Q = P × t where power P is: P = V²/R Given: V =15V,R=50Ω, t = 1 minute = 60 seconds. P = 15²/50 = 225/50 = ...