When a switch is closed, an uncharged 151 nF capacitor is connected in a series circuit with a 10.0 x 106 ohm resistor and a 9.00 V battery.

Question

When a switch is closed, an uncharged 151 nF capacitor is connected in a series circuit with a 10.0 x 106 ohm resistor and a 9.00 V battery. What is the charge on the capacitor 1.3 s after the switch is closed

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RuslanHeatt 6 months 2021-07-27T04:13:18+00:00 1 Answers 2 views 0

Answers ( )

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    2021-07-27T04:14:41+00:00

    Answer:

    Explanation:

    Given that,

    Capacitance of capacitor

    C = 151 nF

    Resistance of a resistor

    R = 10×10^6 ohms

    Battery EMF

    V = 9V

    Charge on capacitor after t = 1.3sec?

    This is an RC series circuit

    Current in the circuit is given as

    i= dq/dt

    dq = i•dt

    q = ∫i•dt

    The voltage in the capacitor is give as

    q = CV

    Vc = q/C, since q = ∫i•dt

    Vc = 1/C ∫i•dt

    Therefore applying KVL

    Vr + Vc = V.

    Vr is voltage across resistor = iR

    iR + 1/C ∫i•dt = V

    Note that, i=dq/dt and q = ∫i•dt

    R dq/dt + q/C = V

    Solving this differential equation

    Divide through by R

    dq/dt + q/RC = V/R

    Since, R = 10×10^6 C = 151×10^-9F

    Then, dq/dt + q/RC = V/R

    dq/dt + q/10×10^6 × 151×10^-9= 9/10×10^6

    dq/dt + 0.662q = 9×10^-7

    Using integrating factor method

    IF = e(0.662t)

    So,

    q•e(0.662t) = 9×10^-7∫(e(0.662t)dt

    q•e(0.662t) = 9 × 10^-7•e(0.662t) / 0.662 + K

    q•e(0.662t)=1.36×10^-7•e(0.662t) + K

    Divide through by e(0.662t)

    q = 1.36×10^-7 + K•e(-0.662t)

    At the beginning, there was no charge on the capacitor

    q(0) = 0

    0 = 1.36 × 10^-7 + k

    Then, k = -1.36 ×10^-7

    q =1.36×10^-7—1.36×10^-7•e(-0.662t)

    Now, at t = 1.3s

    q=1.36×10^-7— 1.36 ×10^-7•e(-0.662 × 1.3)

    q=1.36×10^-7-1.36×10^-7•e(-0.8606)

    q =1.36×10^-7—5.75 × 10^-8

    q = 7.85 × 10^-8

    Then, q = 78.5 nC

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