Ask Question
26 December, 15:37

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

+3
Answers (1)
  1. 26 December, 16:04
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers