A \( 0.04 \mathrm{~F} \) capacitor's current is \( i(t)=-0.6 \pi \sin (\pi t) \mathrm{A} \). Assuming \( i(t)=0=v(t) \) for \( t \leq 0 \), the voltage for \( t>0 \) is: \[ v(t)=\square[\cos (\pi t)-1] \mathrm{V} \]
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The current \(i(t)\) through a capacitor is related to the voltage \(v(t)\) across the capacitor and its capacitance \(C\) by the equation: \[ i(t) = C \frac{dv(t)}{dt} \] Given that the capacitance \(C = 0.04 \) F and the current \(i(t) = -0.6 \pi \sin(\pi t) \) Show more…
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