Question #273058

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

1
Expert's answer
2021-11-30T09:42:29-0500

Solution;

Given;

L=2H

R=16 ohms

C=0.02farads

E(t)=100sin33t

From which;

V0=100V_0=100

w=33tw=33t

From Kirchoff's Law;

LdIdt+IR+QC=V0sinwtL\frac{dI}{dt}+IR+\frac{Q}{C}=V_0sinwt

But since the capacitor is initially uncharged,

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

By substitution;

Ld2Qdt2+RdQdt+QC=V0sinwtL\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)=Q0cos(wtϕ)Q(t)=Q_0cos(wt-\phi)

Q0=V0wZQ_0=\frac{V_0}{wZ}

Z=R2+(XLXC)2Z=\sqrt{R^2+(X_L-X_C)^2}

Z=16+(33×2133×0.02)2Z=\sqrt{16+(33×2-\frac{1}{33×0.02})^2}

Z=66.44Z=66.44

Q0=10033×66.44=0.0456Q_0=\frac{100}{33×66.44}=0.0456

tanϕ=XLXCR=661.51516=4.03tan\phi=\frac{X_L-X_C}{R}=\frac{66-1.515}{16}=4.03

ϕ=tan1=76.07\phi=tan^{-1}=76.07

Therefore;

Q(t)=0.0456sin(33t76.07)Q(t)=0.0456sin(33t-76.07)

But ;

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

I(t)=1.505sin(33t76.07)AI(t)=1.505sin(33t-76.07)A


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