Solve this quickly. 10th question.
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Answer:2m/s because when one was their it's speed was 2m/s so it collided to another . Therefore both speeds will be added
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Explanation:I keep receiving requests from the readers who seem confused with the connection details of 5mm LEds with a 3.7V Li-ion cell. The requests inspired me to write this post, hopefully it would answer the many related queries.
Using a Cellphone Li-ion Cell
Since standard 3.7V Li-Ion cells which are normally used in cell phones are rated at around 800 to 1100mAh, are quite capable of supporting a few 5mm LEDs, and would be able to keep them illuminated for quite sometime.
A normal 5mm white LED requires about 20mA current at 3.3V for getting illuminated optimally.
The circuit involved for illuminating 5mm LEds through a 3.7V Li-Ion cell is actually too simple, primarily because the parameters are closely matched with each other.
Here, connecting the 5mm LEDs in series wouldn't be feasible because the maximum volts from the cell is just 3.7V while even two LEDS in series would call for above 6V.
Therefore the only option left is putting them in parallel.
Ideally when parallel connections are involved, a series limiting resistor becomes imperative with each LED in the array. This helps ensure uniform light distribution or emission from the LEDs.
However it's not an absolute requirement, especially when the driving voltage is close to the forward voltage of the LEDs.
Also taking the simplicity factor into account, a single limiting resistor may be used in such cases and therefore here too we have eliminated individual resistors.
How to Connect the LEDs
The circuit diagram below shows a simple configuration comprising of a 3.7V Li-ion cell, 5nos 5mm LEDs and a limiting resistor R1. The procedure shows how simply a Li-ion cell may be used for illuminating 5mm LEDs for a reasonably long period of time.
Each LED is supposed to consume 20mA current, therefore 5nos would together consume around 100mA, therefore R1 may be calculated as follows:
The Formula
R = (Supply voltage - LEd forward voltage)/LED current
= (3.7 - 3.3)/100 = 0.4/0.1 = 4 ohms.
The required wattage would be 0.4 x 0.1 = 0.04W, so a 1/4 watt resistor would be more than enough.
Assuming the cell to be rated at 800mAH, with 5 LEDs, the approximate back up time available from the cell could be calculated using the following cross-multiplication.
800/100 = x/1100x = 800x = 800/100 = 8 hours ideally.
However practically you would find the above calculated back up time to be considerably less due to many inherent inefficiencies associated with the system or the circuit.