Answer to Question #235502 in Electric Circuits for maymay2902

Question #235502

The current in a conductor varies as follows : during the first 8 secs there is a linear change from zero to 4 amp; during the next 15 sec the current is constant at 4 amp, during the third period of 20 secs the current decreases linearly to 3 amp. Determine (a) the total charge transferred in the elapsed time of 43 secs, (b) the average current.


1
Expert's answer
2021-09-10T12:17:42-0400

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:dIdt=I2I1t2t1=(40)A8s=0.5A/sIa=0.5t=dqadt    qa=080.5tdtqa=[0.25t2]08=0.25(8202)C=16C\text{From t=0 s to t=8 s:} \\ \frac{dI}{dt} = \frac{I_2-I_1}{t_2 - t_1} =\frac{(4-0)A}{8\,s}=0.5\,A/s \\I_{a}=0.5t=\cfrac{dq_{a}}{dt} \implies q_{a}=\intop_{0}^{8} 0.5t \,dt \\ q_{a}=\large {[0.25 t^2]_{0}^{8} }=0.25(8^2-0^2)\,C=16\,C


From t=8 s to t=23 s:dIdt=I3I2t3t2=(44)A15s=0A/sIb=4A=cte=dqbdt    qb=8234dtqb=[4t]823=4(238)C=60C\\ \text{From t=8 s to t=23 s:} \\ \frac{dI}{dt} = \frac{I_3-I_2}{t_3 - t_2} =\frac{(4-4)A}{15\,s}=0\,A/s \\I_{b}=4\,A=cte=\cfrac{dq_{b}}{dt} \implies q_{b}=\intop_{8}^{23} 4 \,dt \\ q_{b}=\large {[4t]_{8}^{23} }=4(23-8)\,C=60\,C


From t=23 s to t=43 s:dIdt=I4I3t4t3=(34)A20s=0.05A/sIc3=0.05(t43)    Ic=dqcdt=5.150.05t    qc=2343(5.150.05t)dtqc=[t(5.150.025t)]2343qc=[(43(5.150.025×43))(23(5.150.025×23))]Cqc=(175.225105.225)C=70C\\ \text{From t=23 s to t=43 s:} \\ \frac{dI}{dt} = \frac{I_4-I_3}{t_4 - t_3} =\frac{(3-4)A}{20\,s}=-0.05\,A/s \\I_{c}-3=-0.05(t-43) \implies I_{c} =\cfrac{dq_{c}}{dt}=5.15-0.05t \implies q_{c}=\intop_{23}^{43} (5.15-0.05t)dt \\ q_{c}=\large {[t(5.15-0.025t)]_{23}^{43} } \\ q_c=[(43(5.15-0.025\times43)) -(23(5.15-0.025\times23))]\,C \\ q_c=(175.225-105.225)\,C=70\,C


ΔQ=qa+qb+qc=(16+60+70)C=146C\Delta Q=q_a+q_b+q_c=(16+60+70)\,C=146\,C


Then, we can calculate the average current as: <I>=ΔQΔt=146C43s=3.395A<I>=\frac{\Delta Q}{\Delta t}=\frac{146\,C}{43\,s}=3.395\,A


In conclusion, we find that

(a) the total charge transferred

in the elapsed time of 43 secs is ΔQ\mathbf{ \Delta Q}=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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