We have to find the equation that describes I=f(t) and then integrate to find the total charge as the sum of the charge for the three pars of the process (a, b, and c):
From t=0 s to t=8 s:dtdI=t2−t1I2−I1=8s(4−0)A=0.5A/sIa=0.5t=dtdqa⟹qa=∫080.5tdtqa=[0.25t2]08=0.25(82−02)C=16C
From t=8 s to t=23 s:dtdI=t3−t2I3−I2=15s(4−4)A=0A/sIb=4A=cte=dtdqb⟹qb=∫8234dtqb=[4t]823=4(23−8)C=60C
From t=23 s to t=43 s:dtdI=t4−t3I4−I3=20s(3−4)A=−0.05A/sIc−3=−0.05(t−43)⟹Ic=dtdqc=5.15−0.05t⟹qc=∫2343(5.15−0.05t)dtqc=[t(5.15−0.025t)]2343qc=[(43(5.15−0.025×43))−(23(5.15−0.025×23))]Cqc=(175.225−105.225)C=70C
ΔQ=qa+qb+qc=(16+60+70)C=146C
Then, we can calculate the average current as: <I>=ΔtΔQ=43s146C=3.395A
In conclusion, we find that
(a) the total charge transferred
in the elapsed time of 43 secs is ΔQ=146 C,
and (b) the average current is <I>=3.395 A.
Reference:
- Sears, F. W., & Zemansky, M. W. (1973). University physics.
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