In an electric circuit, let Q(t)be the charge on a capacitor at time t.Then the current in the circuit is given by $I(t)=\frac{d Q}{d t}$ (the rate of change of the charge).The charge and the current satisfy the differential equation $L \frac{d I}{d t}+\frac{Q}{C}=0$ .If L = 34 henry, and C = 10 farad, find a formula for Q(t)with initial conditions Q(0)= 0 and I(0)= 2.Round constants to three decimal places.
A)Q(t)= 36.878sin(0.054t)
B)Q(t)= 36.878cos(0.054t)
C)Q(t)= 0.054sin(0.054t)
D)Q(t)= 36.878cos(0.054t)+ 0.054sin(34t)
E)Q(t)= 10cos(34t)+ 10sin(34t)

Question

Quizplus · Accepted Answer

The differential equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is $r^2+\frac{1}{LC}=0$, which has roots $r=\pm i\sqrt{\frac{1}{LC}}$. Therefore, the general solution to the differential equation is $Q(t)=c_1\cos\left(\sqrt{\frac{1}{LC}}t\right)+c_2\sin\left(\sqrt{\frac{1}{LC}}t\right)$. Using the initial conditions, we get $Q(0)=c_1=0$ and $I(0)=\frac{dQ}{dt}(0)=c_2\sqrt{\frac{1}{LC}}=2$. Therefore, $Q(t)=2\sqrt{\frac{LC}{10}}\sin\left(\sqrt{\frac{1}{LC}}t\right)=36.878\sin(0.054t)$.

In an Electric Circuit, Let Q(t)be the Charge on a Capacitor