Question #272872

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
2021-11-29T13:59:42-0500

Solution;

Given;

E(t)=100sin33tE(t)=100sin33t

From which;

V0=100VV_0=100V

w=33w=33

Then current at time t is;

I(t)=Imsin(wtϕ)I(t)=I_msin(wt-\phi)

Where,

Im=V0ZI_m=\frac{V_0}{Z}

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

R=16ΩR=16\Omega

XL=wL=33×2=66X_L=wL=33×2=66

Xc=1wC=133×0.02=1.515X_c=\frac{1}{wC}=\frac{1}{33×0.02}=1.515

By substitution;

Z=162+(661.515)2=66.44Z=\sqrt{16^2+(66-1.515)^2}=66.44

Therefore;

Im=10066.44=1.505AI_m=\frac{100}{66.44}=1.505A

Also;

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

ϕ=tan1(4.03)=76.1\phi=tan^{-1}(4.03)=76.1

And;

I(t)=1.505sin(33t76.1)I(t)=1.505sin(33t-76.1)

Since the capacitor is intially uncharged;

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

Therefore;

Q=I(t)dt=1.505sin(33t76.1)dt=1.505[sin(33t76.1)Q=\int I(t)dt=\int1.505sin(33t-76.1)dt=1.505[\int sin(33t-76.1)

Q=1.505[cos(33t76.133]Q=1.505[-\frac{cos(33t-76.1}{33}]

Q=0.0456cos(33t76.1)Q=0.0456cos(33t-76.1)





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