Question
Determine the current through the AC source as a function of time in the long run. The capacitor has a capacitance $C=\frac{1}{2 \omega^2 L}$.(GRAPH CAN'T COPY)
Step 1
Step 1: Identify steady-state: for sinusoidal long-run, current i(t)=I_m cos(ωt+φ). Show more…
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When capacitor of capacitance $C$ is charged by a source of voltage $V,$ the power expended at time $t$ is $$ P(t)=\frac{V^{2}}{R}\left(e^{-t / R C}-e^{-2 t / R C}\right) $$ where $R$ is the resistance in the circuit. The total energy stored in the capacitor is $$ W=\int_{0}^{\infty} P(t) d t $$ Show that $W=\frac{1}{2} C V^{2}$
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When a capacitor of capacitance $C$ is charged by a source of voltage $V,$ the power expended at time $t$ is $$P(t)=\frac{V^{2}}{R}\left(e^{-t / R C}-e^{-2 t / R C}\right)$$ where $R$ is the resistance in the circuit. The total energy stored in the capacitor is $$W=\int_{0}^{\infty} P(t) d t$$ Show that $W=\frac{1}{2} C V^{2}$
Consider the RLC circuit with $R=16 \Omega, L=8 \mathrm{H}$ $C=\frac{1}{40} \mathrm{F},$ and $E(t)=17 \cos 2 t \mathrm{V} .$ Determine the current in the circuit for $t > 0,$ given that at $t=0,$ the capacitor is uncharged and there is no current flowing.
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