A voltage source of 10 V is connected to a series RC circuit where R = 2.0 ´ 106 W , and C = 3.0 µF. Find the amount of time required for th

Question

A voltage source of 10 V is connected to a series RC circuit where R = 2.0 ´ 106 W , and C = 3.0 µF. Find the amount of time required for the current in the circuit to decay to 5% of its original value. Hint: This is the same amount of time for the capacitor to reach 95% of its maximum charge.

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Yến Oanh 6 months 2021-08-22T05:28:49+00:00 1 Answers 0 views 0

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    2021-08-22T05:30:37+00:00

    Answer:

    0.5 s

    Explanation:

    The discharge current in a series RC circuit is given by

    i = \frac{V_{0} }{R} e^{-\frac{t}{RC} } where i₀ = V₀/R

    For the current i to decay to 5%i₀, i/i₀ = 0.05. Where R = resistance = 2.0 × 10⁶ Ω and C = capacitance = 3.0 μF = 3.0 10⁻⁶ F

    So,

    i = {i_{0}e^{-\frac{t}{RC} }

    {\frac{i}{i_{0} } = e^{-\frac{t}{RC}

    0.05 = e^{-\frac{t}{2 X 10^{6} X 3 X 10^{-6}}\\

    0.05 = e^{-\frac{t}{6}

    taking natural logarithm of both sides

    -t /6 = ln(0.05)

    t = -ln(0.05)/6 = 0.5 s

    So it takes the current 0.5 s to reach 5% of its original value

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