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  • 5 years ago

The current applied to a 10μF capacitor is i=0.01 sin100t. If the capacitor was initially discharged, what is the voltage across the capacitor at time t= pi/200sec?

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  1. anonymous
    • 5 years ago
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    The current of a capacitor is related to the voltage as follows: \[ i(t) = C\frac{dv}{dt} \] Thus, we can get an equation for voltage: \[ \begin{align*} v(t) &= C\int i(t) dt \\ v(t) &= C\int0.01\sin(100t) + K\\ v(t) &= \frac{-0.01C}{100}\cos(100t) + K \end{align*} \] Solve for the constant of integration with the initial condition, and then plug in \(t = \frac{\pi}{200}\).

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