Answer to Question #273058 in Differential Equations for Chintu

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.

J. A spring with a mass of 2 kg has natural length m. A force of 25.6 N

Answer 1

Solution;

Given;

L=2H

R=16 ohms

C=0.02farads

E(t)=100sin33t

From which;

$V_0=100$

$w=33t$

From Kirchoff's Law;

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

But since the capacitor is initially uncharged,

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

By substitution;

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

Which is a second order differential equation whose solution is;

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

$Q_0=\frac{V_0}{wZ}$

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$Z=\sqrt{16+(33×2-\frac{1}{33×0.02})^2}$

$Z=66.44$

$Q_0=\frac{100}{33×66.44}=0.0456$

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

$\phi=tan^{-1}=76.07$

Therefore;

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

But ;

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

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