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.
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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 = ...
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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:
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.