Answer to Question #235502 in Electric Circuits for maymay2902

Question

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.

Answer 1

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):

$\text{From t=0 s to t=8 s:} \\ \frac{dI}{dt} = \frac{I_2-I_1}{t_2 - t_1} =\frac{(4-0)A}{8\,s}=0.5\,A/s \\I_{a}=0.5t=\cfrac{dq_{a}}{dt} \implies q_{a}=\intop_{0}^{8} 0.5t \,dt \\ q_{a}=\large {[0.25 t^2]_{0}^{8} }=0.25(8^2-0^2)\,C=16\,C$

$\\ \text{From t=8 s to t=23 s:} \\ \frac{dI}{dt} = \frac{I_3-I_2}{t_3 - t_2} =\frac{(4-4)A}{15\,s}=0\,A/s \\I_{b}=4\,A=cte=\cfrac{dq_{b}}{dt} \implies q_{b}=\intop_{8}^{23} 4 \,dt \\ q_{b}=\large {[4t]_{8}^{23} }=4(23-8)\,C=60\,C$

$\\ \text{From t=23 s to t=43 s:} \\ \frac{dI}{dt} = \frac{I_4-I_3}{t_4 - t_3} =\frac{(3-4)A}{20\,s}=-0.05\,A/s \\I_{c}-3=-0.05(t-43) \implies I_{c} =\cfrac{dq_{c}}{dt}=5.15-0.05t \implies q_{c}=\intop_{23}^{43} (5.15-0.05t)dt \\ q_{c}=\large {[t(5.15-0.025t)]_{23}^{43} } \\ q_c=[(43(5.15-0.025\times43)) -(23(5.15-0.025\times23))]\,C \\ q_c=(175.225-105.225)\,C=70\,C$

$\Delta Q=q_a+q_b+q_c=(16+60+70)\,C=146\,C$

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

In conclusion, we find that

(a) the total charge transferred

in the elapsed time of 43 secs is $\mathbf{ \Delta Q}$=146 C,

and (b) the average current is <I>=3.395 A.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.