Question #271368

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.

1
Expert's answer
2021-11-29T19:32:18-0500

Solution;

Given;

L=2HL=2H

R=16ΩR=16\Omega

C=0.02FC=0.02F

E=100sin33tE=100sin33t

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

Also from E;

V0=100VV_0=100V

w=33w=33

From Kirchoff's Loop Law;

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

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

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.

One possible solution of the above differential equation is;

Q(t)=Q0cos(wtϕ)Q(t)=Q_0cos(wt-\phi)

Q0=V0wR2+(XLXC)2Q_0=\frac{V_0}{w\sqrt{R^2+(X_L-X_C)^2}}

tanϕ=XLXCRtan\phi=\frac{X_L-X_C}{R}

XL=wLX_L=wL

XC=1wCX_C=\frac{1}{wC}

By direct substitution of values;

XL=33×2=66X_L=33×2=66

XC=133×0.02=1.515X_C=\frac{1}{33×0.02}=1.515

tanϕ=661.51516=4.03tan\phi=\frac{66-1.515}{16}=4.03

Q0=10033162+(661.515)2=0.0456Q_0=\frac{100}{33\sqrt{16^2+(66-1.515)^2}}=0.0456 ϕ=tan14.03=76.07\phi=tan^{-1}4.03=76.07

Hence , charge at any time t;

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

But ;

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

Hence;

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




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