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Sagot :
Answer:
Part A
The power to turn on the lamp, ∑P = 3.36 W
Part B
The Resistor required is approximately 8.04 Ohms
Part C
The time for all the lights to go out is approximately 21.43 hours
Explanation:
The input voltage of the motor battery , V = 12 V
The capacity of the battery, Q = 6 Ah
The number of lamps in parallel = 8 lamps
The maximum voltage of each lamp, = 3 V
The electric current in each lamp = 140 mA
The energy available in a battery, E = Q × V
For the battery, we have;
E = 6 Ah × 12 V = 72 Wh
The energy available in a battery, E = 72 Wh
Part A
The power used by the lamps, [tex]P_i[/tex] = [tex]I_i[/tex] × [tex]V_i[/tex]
∴ The total power used by the lamp, ∑P = 8 × 0.14 A × 3 V = 3.36 W
The power to turn on the lamp, ∑P = 3.36 W
Part B
The resistance required, is given as follows;
Resistor required = (Battery voltage - Lamp voltage)/(The sum of bulb current)
∴ Resistor required = (12 V - 3 V)/(8 × 0.14 A)
The Resistor required = 8.03571429 Ohms
The Resistor required ≈ 8.04 Ohms
Part C
The time for all the lights to go out = The time for the lamps to use all the power available in the battery
The time for all the lights to go out, t = E/∑P
∴ t = 72 Wh/(3.36 W) = 21.4285714 h
∴ The time for all the lights to go out, t ≈ 21.43 h
The time for all the lights to go out = The time for the lamps to use all the power available in the battery = t ≈ 21.43 h
∴ The time for all the lights to go out ≈ 21.43 hours.
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