A RC series circuit of R=15Ω and C=10μF is connected to 20 volt DC supply for very long time. Then capacitor is disconnected from circuit and connected to inductor of 10mH. Find amplitude of current.
Question
A RC series circuit of R=15Ω and C=10μF is connected to 20 volt DC supply for very long time. Then capacitor is disconnected from circuit and connected to inductor of 10mH. Find amplitude of current.
Solution
The problem involves two stages. First, the charging of the capacitor in an RC circuit, and then the discharging of the capacitor in an LC circuit.
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Charging of the Capacitor: When the capacitor is connected to the DC supply for a very long time, it gets fully charged. The voltage across the capacitor (Vc) will be equal to the supply voltage. So, Vc = 20V.
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Discharging of the Capacitor through an Inductor: When the fully charged capacitor is connected to the inductor, it starts discharging. The energy stored in the capacitor is transferred to the inductor.
The amplitude of the current (I) in an LC circuit can be calculated using the formula:
I = Vc / √(L/R² + 1/C)
Where: Vc = Voltage across the capacitor = 20V L = Inductance = 10mH = 10 x 10^-3 H R = Resistance = 15Ω C = Capacitance = 10μF = 10 x 10^-6 F
Substituting these values into the formula, we get:
I = 20 / √((10 x 10^-3)/(15^2) + 1/(10 x 10^-6)) I = 20 / √((10 x 10^-3)/225 + 10^5) I = 20 / √(4.44 x 10^-5 + 10^5) I = 20 / √10^5 I = 20 / 316.23 I = 0.0632 A
So, the amplitude of the current is approximately 0.0632 A.
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