4. (20 points) The battery information is listed in the following figure. (a) Pl

Question

4. (20 points) The battery information is listed in the following figure. (a) Please detemine the current battery state-of-charge. b) The current battery capacity is found to be 95mAh Assume the current drain of discharge remains the same, how long does it take to discharge the battery to almost empty (i e.. only 2.5% of maximum. charge is left)? (c) The curent battery capacity is found to be 95mAh. Assume the current drain of discharge remains the same and the charge rate is twice the current drain of discharge, how long does it take to charge the battery to almost full (ie-only 97.5% of maximum charge is attained)? Battery Info 88.88F(31.60C) CurrentCapacity 95mAh 1295mAh DesignCapacity: Battery Charge CycleCount Voltage: 1420mAh 455 3833mV 3528mV 717mA Drain:

Explanation / Answer

a.) In this we have been asked the current battery state of charge, In actual as we all know that charge is product of time and current. So in the question itself it has been cleared that current capacity is 95mAh which is nothing but a product of current in milliamperes and time in hours.

b.) We have been given as the current capacity, which is nothing but charge equal to 95mAh and there cycles counted equal to 455(as given) So in general frequency is 50 cycles/sec so we can calculate the time as 455/50 = 9.1 seconds

Now time for 2.5% of maximum capacity(1295mAh) is 32.375 and now the time for this is calculated as

95mAh = 9.1 seconds

32.375mAh = ?

this gives ? = 3.1 seconds.

c.) Now we have been given that charging rate is doubled, so the number of cycles gets reduced to reach upto 95mAh that is 455/2 = 227.5. Now time in seconds for 227.5 cycles count is 227.5/50 = 4.55 seconds.

97.5% of 1295mAh will be 1262.625 mAh. Now it can be done in similar manner as b was done

95mAh = 4.55 seconds

1262.625mAh = ?

this gives ? = 60.47 seconds

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