daily energy requirement?

Question

Please help?
i need help with this question please. estimate the daily energy requirement of a LED if it is designed to operate for up to 12 hours during the night?

the internal battery supplies a e.m.f of 1.2 V and the LED draws a current of 25 mA when it is illuminated.

Answer 1

If I understand the question correctly... LED outdoor lighting?

Give yourself a bit of extra -- 30mAh x 12 h = 360mAh.

Your charging (solar cell, right?) and battery system should be capable of capturing and delivering at least that much power in order to keep you lit for 12 hours. The trick will be assuring that your charging source works well enough on cloudy days. You might only get an hour or less of bright sun on a day, so plan for a considerably larger charging source. If it has a capacity of 1Ah, you'll have a better chance of getting a full charge during the other 12 hours on a less than optimal day.

Answer 2

You asked about energy. Energy is expressed as power times time. This can be watt-hours, Joules (watt-seconds) or any unit that applies to energy, not power or charge (i.e., not watts and not maH, milliamp-hours.) So the simple answer is 1.2 V * 0.025 A * 12 hours = 0.36 watt-hours. If this is a rechargeable battery system and you're asking about the recharge energy needed, 0.36 w-h has to be multiplied by some factor greater than 1 to obtain the charging energy, since the charge-discharge cycle is not 100% efficient. For instance, NiCads typically are charged at "10 hour rate" current (i.e., 1/10 * their amp-hour capacity) not for 10 but 14 hours. Add to that factor (1.4) the fact that charging voltage is higher than discharge voltage, and you can end up with needed charge energy twice the discharge energy.
But the simple answer may not be right for a couple of reasons. One is that most LEDs have a higher voltage drop than 1.2 V. The ref. cites 1.7 to 4.6 V depending on type and color. So a 1.2 V battery without some sort of power conversion won't fire most LEDs. Second, without power conversion you would need to provide a battery with a higher voltage than the LED requires and use a voltage-dropping series resistor to limit the current to the desired value. Without the limiting the LED would overheat and possibly blow out.
Since you say there is an internal battery, this suggests a manufactured item. In that case the LED might have a voltage requirement that is below typical, or there might be a dc-to-dc converter to increase the voltage applied to the LED. In the first case the simple answer still applies. However if there is a voltage-increasing converter then the battery sees a larger current drain than 25 ma and one would have to know the converter efficiency to state the actual power or energy used.

Answer 3

You haven't provided enough information to develop an answer; but the method is easy enough to master. Basically it's an power balance (energy over time). On one side you you what the current drawn by the LED is. You need to find out what voltage (or millivolts - 1/1000 of a volt) the LED device operates at. Most likely that info is on the device or can be obtained from the manufacturer. That gives you the power consumed.

Then on the battery side, you know the voltage for the battery. The missing info is: what current does the battery produce in milli-amperes?

Once you find that info, then you can easily determine the number of seconds the device will operate for using arithmetic. It isn't difficult, just make sure you use consistant unit and that you know what the definitions of volt and amperes are.

Good luck.

Answer 4

Both of the first two answers are thinking to hard. 25mA is the current draw of the LED. 25mA x 12h = 300mAh. That is the power used by the LED in one day assuming 12 hours a day. All the remaining information is extraneous.

Answer 5

Volts * Amps = Power (in Watts)

(1.2 Volt) * (0.025 Amp) = 0.030 Watt

This would be power consumed in one hour.
For 12 hour operation...
(12 hour) * (0.030 Watt) = 0.36 Watt

Therefore will consume slightly more than 1/3 watt over half-day operation.

Answer 6

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