The charge cycle shown in Figure P2.10 is an example of a two-rate charge. The current is held constant at 70 mA for 1 h. Then it is switched to 60 mA for the next 1 h. Find: a. The total charge transferred to the battery. b. The total energy transferred to the battery. Hint: Recall that energy w is the integral of power, or P = dw/dt.
Let
What will be an ideal response?
Known quantities:
See Figure P2.10
Find:
a) The total charge transferred to the battery.
b) The energy transferred to the battery.
Analysis:
a. 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.
0.070A?3600s=252C
0.060A?3600s=216C
252C+216C=468C
b. P=V•I, therefore, an equation for power can be found by multiplying the two graphs together.
First separate the voltage graph into two equations:
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