Answer to Question #272872 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;

$E(t)=100sin33t$

From which;

$V_0=100V$

$w=33$

Then current at time t is;

$I(t)=I_msin(wt-\phi)$

Where,

$I_m=\frac{V_0}{Z}$

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

$R=16\Omega$

$X_L=wL=33×2=66$

$X_c=\frac{1}{wC}=\frac{1}{33×0.02}=1.515$

By substitution;

$Z=\sqrt{16^2+(66-1.515)^2}=66.44$

Therefore;

$I_m=\frac{100}{66.44}=1.505A$

Also;

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

$\phi=tan^{-1}(4.03)=76.1$

And;

$I(t)=1.505sin(33t-76.1)$

Since the capacitor is intially uncharged;

$I(t)=+\frac{dQ}{dt}$

Therefore;

$Q=\int I(t)dt=\int1.505sin(33t-76.1)dt=1.505[\int sin(33t-76.1)$

$Q=1.505[-\frac{cos(33t-76.1}{33}]$

$Q=0.0456cos(33t-76.1)$