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20 December, 02:50

A capacitor that is initially uncharged is connected in series with a resistor and an emf source with E = 120 V and negligible internal resistance. Just after the circuit is completed, the current through the resistor is 6.7*10-5 A. The time constant for the circuit is 4.4 s.'What are the resistance of the resistor and the capacitance of the capacitor?

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Answers (2)
  1. 20 December, 04:44
    0
    R = 1.79*10^6 Ω

    C = 2.46*10^-6 F

    Explanation:

    Given

    emf of the source, ε = 120 V

    Current passing through the resistor, I = 6.7*10^-5 A

    Time constant for the circuit, τ = 4.4 s

    From the information above, we can say that RC = 4.4 s

    Also, on applying the loop rule, we get

    ε - IR = 0

    ε = IR

    R = ε / I

    R = 120 / 6.7*10^-5

    R = 1.79*10^6 Ω

    Using the first equation, we can thus solve for C

    RC = 4.4 s

    C = 4.4 / R

    C = 4.4 / 1.79*10^6

    C = 2.46*10^-6 F

    C = 2.46 μF

    Therefore, the resistance and capacitance of the capacitor is respectively, 1.79 MΩ and 2.46 μF
  2. 20 December, 05:48
    0
    Resistance = 1.791 x 10^ (6) Ω

    Capacitance = 2.46 x 10^ (-6) F

    Explanation:

    We are given;

    EMF; E = 120V

    Current; I = 6.7 * 10^ (-5) A

    Time constant; τ = 4.4 s

    Now, just after the circuit is completed, the capacitor acts like a wire and thus we use the loop rule;

    So,

    E - IR = 0

    Let's make the resistance R the subject.

    R = E/I = 120 / (6.7 * 10^ (-5)) = 1.791 x 10^ (6) Ω

    Now formula for time constant is given as;

    τ = RC

    Where C is capacitance.

    Thus, C = τ/R = 4.4 / (1.791 x 10^ (6))

    C = 2.46 x 10^ (-6) F
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