Let each cell have emf E and internal resistance r. For aiding: (12E + 2E) / R = 3A → 14E/R = 3. For opposing: (12E – 2E) / R = 2A → 10E/R = 2. Solving, x = 2 ...
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For two parallel 1.5V cells, total external resistance is 10Ω. Using V = E – Ir and solving with V = 1.2V, we get r = 5Ω. Answer: (c) 5Ω.
The heat produced in a conductor is proportional to the square of the current (H ∝ I²). If the current is tripled, the heat increases by 3² = 9 times. Thus, the temperature rise will be 4°C × 9 = ...
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 ...
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: