Find the charge q(t) flowing through an element if the current is
i(t) = 3e
−t + 20 cos(15t − π
6
⁄ )uA
q(0) = 1uC
i(t)=3e−t+20 cos(15t−π6)q=∫i(t)dt∫ 3e−t+20cos(15t−π6)dtApply the Sum Rule:∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫ 3e−tdt+∫ 20cos(15t−π6)dt=−3e−t+43sin(15t−π6)=−3e−t+43sin(15t−π6)+Cq(0)=1uC ⟹ c=−43q=−3e−t+43sin(15t−π6)−43i(t)= 3e^{-t}+20\:cos\left(15t-\frac{\pi }{6}\right)\\ q= \int i(t) dt\\ \int \:3e^{-t}+20\cos \left(15t-\frac{\pi }{6}\right)dt\\ \mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\ =\int \:3e^{-t}dt+\int \:20\cos \left(15t-\frac{\pi }{6}\right)dt\\ =-3e^{-t}+\frac{4}{3}\sin \left(15t-\frac{\pi }{6}\right)\\ =-3e^{-t}+\frac{4}{3}\sin \left(15t-\frac{\pi }{6}\right)+C\\ q(0) = 1uC \implies c=-\frac{4}{3}\\ q=-3e^{-t}+\frac{4}{3}\sin \left(15t-\frac{\pi }{6}\right)-\frac{4}{3}\\i(t)=3e−t+20cos(15t−6π)q=∫i(t)dt∫3e−t+20cos(15t−6π)dtApplytheSumRule:∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫3e−tdt+∫20cos(15t−6π)dt=−3e−t+34sin(15t−6π)=−3e−t+34sin(15t−6π)+Cq(0)=1uC⟹c=−34q=−3e−t+34sin(15t−6π)−34
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