Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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?

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.

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.