Question

This problem concerns the electric circuit shown in the figure below. A charged capacitor connected to an inductor causes a current to flow through the inductor until the capacitor is fully discharged. The current in the inductor, in turn, charges up the capacitor until the capacitor is fully charged again. If Q(t) is the charge on the capacitor at time t, and I is the current, then I=

dt
dQ
​
. If the circuit resistance is zero, then the charge Q and the current I in the circuit satisfy the differential equation L
dt
dI
​
+
C
Q
​
=0, where C is the capacitance and L is the inductance, so L
dt
2

d
2
Q
​
+
C
Q
​
=0. Then, just as as a spring can have a damping force which affects its motion, so can a circuit; this is introduced by the resistor, so that if the resistance of the resistor is R, L
dt
2

d
2
Q
​
+R
dt
dQ
​
+
C
1
​
Q=0. If L=1 henry, R=
3
2
​
ohm, and C=9 farads, find a formula for the charge when (a) Q(0)=0 and Q
′
(0)=6 Q(t)= (b) Q(0)=6 and Q
′
(0)=0 Q(t)=

Answer 1

The charge equation for the given electric circuit, with specific initial conditions and circuit parameters, results in the following solutions for charge Q at different starting points:

(a) Q(t) = 6 * cos(3t)

(b) Q(t) = 6 * e^(-1.5t)

Step-by-step explanation:

To solve the given differential equation for charge Q with initial conditions Q(0) and Q'(0), we consider the equation L*(d^2Q/dt^2) + R*(dQ/dt) + C*Q = 0. Given L = 1 henry, R = 3/2 ohms, and C = 9 farads, we first determine the characteristic equation: L*(d^2Q/dt^2) + R*(dQ/dt) + C*Q = 0 → d^2Q/dt^2 + (R/L)*(dQ/dt) + (1/(LC))*Q = 0. Substituting L, R, and C values, we get d^2Q/dt^2 + (3/2)*(dQ/dt) + (1/9)*Q = 0.

For part (a) with Q(0) = 0 and Q'(0) = 6, the roots of the characteristic equation are real and distinct, yielding the solution Q(t) = Ae^(r1*t) + Be^(r2*t), where r1 and r2 are the roots. Solving for r1 and r2 gives -1/6 and -3/2. Applying the initial conditions, Q(0) = A + B = 0 and Q'(0) = r1*A + r2*B = 6. Solving these equations gives A = 3 and B = -3. Therefore, Q(t) = 3e^(-t/6) - 3e^(-3t/2), which simplifies to Q(t) = 6 * cos(3t).

For part (b) with Q(0) = 6 and Q'(0) = 0, the roots of the characteristic equation are complex, giving the solution Q(t) = e^(at)*(A*cos(bt) + B*sin(bt)). Solving for a and b gives a = -R/(2L) = -1.5 and b = sqrt((4LC - R^2)/(4L^2)) = 3√2/2. Applying the initial conditions yields A = 6 and B = 0. Therefore, Q(t) = 6 * e^(-1.5t) * cos(3√2/2 * t).