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.
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Answer:
Step-by-step explanation:
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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
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