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24 March, 10:08

Consider a series RC circuit at the left where C = 6 µ F, R = 2 MΩ, and ε = 20 V. You close the switch at t = 0. Find (a) the time constant for the circuit, (b) the half-life of the circuit, (c) the current at t = 0, (d) the voltage across the capacitor at t = 0, and (e) the voltage across the resistor after a very long time.

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  1. 24 March, 11:48
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    (a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V = 0

    Explanation:

    Solution

    Given that:

    C = 6 µ which is = 6 * 10^ ⁻6

    R = 2 MΩ, which is = 2 * 10^ 6

    ε = 20 V

    (a) When it is at the time constant we have the following:

    λ = CR

    = 6 * 10^ ⁻6 * 2 * 10^ 6

    λ = 12 seconds

    (b) We solve for the half life of the circuit which is given below:

    d₀ = d₀ [ 1 - e ^ ⁺t/CR

    d = decay mode]

    d₀/2 = d₀ 1 - e ^ ⁺t/12

    2^⁻1 = e ^ ⁺t/12

    Thus

    t/12 ln 2

    t = 12 * ln 2

    t = 12 * 0.693

    t = 8.31 seconds

    (c) We find the current at t = 0

    So,

    I = d₀/dt

    I = d₀/dt e ^ ⁺t/CR

    = CE/CR e ^ ⁺t/CR

    E/R e ^ ⁺t/CR

    Thus,

    at t = 0

    I E/R = 20 / 2 * 10^ 6

    = 10µ A

    (d) We find the voltage across the capacitor at t = 0 which is shown below:

    V = IR

    = 10 * 10^ ⁻6 * 2 * 10^ 6

    V = 20 V

    (e) We solve for he voltage across the resistor.

    At t = 0

    I = 0

    V = 0
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