The charge cycle shown in Figure P2.4 is an example of a three-rate charge. The current is held constant at 30 mA for 6 h. Then it is switched to 20 mA for the next 3 h. Find: a. The total charge transferred to the battery. b. The energy transferred to the battery.
Hint: Recall that energy w is the integral of power, or P = dw/dt.
Known quantities:
See Figure P2.4
Find:
The total charge transferred to the battery.
The energy transferred to the battery.
Analysis:
Current is equal to Coulombs/Second, therefore given the current and a duration of that current, the transferred charge can be calculated by the following equation:
A?t=C
The two durations should be calculated independently and then added together.
0.030A?21600s=648C
0.020A?10800s=216C
648C+216C=864C
P=V•I, therefore, an equation for power can be found by multiplying the two graphs together.
First separate the voltage graph into three equations:
0 h ? 3 h :V=9.26•10^(-6) t+0.5
3 h ? 6 h :V=5.55?10^(-5) t
6 h ? 9 h :V=1.11?10^(-4) t-1.9
Next, multiply the first two equations by 0.03A and the third by 0.02A.
0 h ? 3 h :P=2.77?10^(-7) t+0.015
3 h ? 6 h :P=1.66?10^(-6) t
6 h ? 9 h :P=2.22?10^(-6) t-0.038
Finally, since Energy is equal to the integral of power, take the integral of each of the equations for their specified times and add them together.
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