Answer to Question #271368 in Differential Equations for Hemanth

Question

An inductor of 2 henries, resistor of 16 ohms and capacitor of 0.02 farads are connected in series with a battery of

e.m.f E = 100sin33t. At t=0, the charge on the capacitor and current in the circuit are zero. Find the charge and

current at time t.

Answer 1

Solution;

Given;

$L=2H$

$R=16\Omega$

$C=0.02F$

$E=100sin33t$

At t=0,Q=0 and I=0

Also from E;

$V_0=100V$

$w=33$

From Kirchoff's Loop Law;

$L\frac{dI}{dt}+IR+\frac{Q}{C}=V_0sinwt$

Since the capacitor is intially uncharged,$I=\frac{dQ}{dt}$ , Substituting;

$L\frac{d^2Q}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=V_0sinwt$

Which is a second order differential equation.

One possible solution of the above differential equation is;

$Q(t)=Q_0cos(wt-\phi)$

$Q_0=\frac{V_0}{w\sqrt{R^2+(X_L-X_C)^2}}$

$tan\phi=\frac{X_L-X_C}{R}$

$X_L=wL$

$X_C=\frac{1}{wC}$

By direct substitution of values;

$X_L=33×2=66$

$X_C=\frac{1}{33×0.02}=1.515$

$tan\phi=\frac{66-1.515}{16}=4.03$

$Q_0=\frac{100}{33\sqrt{16^2+(66-1.515)^2}}=0.0456$ $\phi=tan^{-1}4.03=76.07$

Hence , charge at any time t;

$Q(t)=0.0456cos(33t-76.07)$

But ;

$I=+\frac{dQ}{dt}$

Hence;

$I(t)=1.505sin(33t-76.07)$