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If it is known that a motor battery has an input voltage of 12V and a capacity of 6 Ah, how much power and resistor value is required to tur
Question
If it is known that a motor battery has an input voltage of 12V and a capacity of 6 Ah, how much power and resistor value is required to turn on 8 lamps with a parallel circuit, with the specifications of each lamp having a maximum voltage of 3V and an electric current of 140 mA? How long did all the lights go on until they off?
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Physics
5 years
2021-09-02T02:36:46+00:00
2021-09-02T02:36:46+00:00 1 Answers
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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,
= 
× 
∴ 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.