Question

a simple series circuit has an inductor of 1 Henry a capacitor of 10^-6 farad's and a resistor of 1000 ohms.the initial charge on the capacitor is zero.if a 12 volt battery is connected to the circuit and the circuit is closed at t=0, find the charge on the capacitor 11 second later and the steady state charge.​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Thus the limiting charge is lim Q t - ∞ = 9 / 4 x 10^-6 coulombs

Step-by-step explanation:

We are given that:

• Inductance = 1 H
• Resistance = 1000 Ohms
• Initial charge = 0
• Voltage of battery = 12 V

Solution:

We have

C2 = 3 /4 x 10^-6

C1 = - 3 x 10^-6

Then Q = - 3 x 10^-6 e ^-10^3 t + 3 /4 x 10^-6 e ^-4 x 10^3 t + 9 / 4 x 10^-6

At t = 0.001 s The charge on capacitor Q = 1.16 x 10^-6 Coulombs

At t = 0.01 sec Charge on capacitor is Q = 2.249 x 10^-6 coulombs

Limiting charge:

lim Q t - ∞ = 9 / 4 x 10^-6 coulombs

Thus the limiting charge is lim Q t - ∞ = 9 / 4 x 10^-6 coulombs