Answer to Question #216802 in Differential Equations for Masande

Question

In the L-C Circuit   L= 1 HENRY , C 1/16 farad and E(t) =60 volt

The differential equation q’’ +16q’  = 60 represent the capacitor charger at anytime t

Q(0)=0 and q’(0) =0 = i(0) =0 find the charge q on the capacitor at anytime t

Answer 1

Solve:

we know i(current)=$\frac{dq}{dt}$

Thus,

$\frac{di}{dt}+16i=60$

$$\frac{di}{dt}=60-16i \newline \frac{di}{60-16i}=dt \newline integrating\ both\ side \newline \int_0^i\frac{di}{60-16i}=\int_0^t dt \newline \ln(60-16i)-ln(60)=t \newline \ln\frac{(60-16i)}{60}=t \newline 60-16i=60 \epsilon^t \newline i=\frac{1}{16}(60-60 \epsilon^t) \newline i= \frac{dq}{dt}=\frac{60}{16}(1-\epsilon^t) \newline \int_0^qdq=\frac{15}{4}\int_0^t(1-\epsilon^t)dt \newline q=\frac{15}{4}(t-(\epsilon^t-1))\newline q=\frac{15}{4}(t+1-\epsilon^t)$$

Thus charge is given by $q=\frac{15}{4}(t+1-\epsilon^t)$