an LC circuit with a 3 uF capacitor and a 3 H inductor has a current I(t)=10*sin(2t) supplied to it . after 2 s, how much charge is stored on a plate of the capacitor?
To find out how much charge is stored on a plate of the capacitor, we need to calculate the charge Q(t) at a specific time t.
The charge on a capacitor can be found using the equation: Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.
In an LC circuit, the voltage across the capacitor can be found using the equation: V(t) = -L * (dI(t)/dt), where V(t) is the voltage across the capacitor, L is the inductance, and dI(t)/dt is the derivative of the current with respect to time.
Given that the current I(t) = 10*sin(2t), we can find its derivative as follows:
dI(t)/dt = 10*cos(2t)*(d(2t)/dt) = 20*cos(2t)
Now, we can substitute this derivative into the equation for capacitance to find the voltage across the capacitor at a particular time t:
V(t) = -L * (dI(t)/dt) = -3H * (20*cos(2t))
Next, we can substitute this voltage into the equation for charge to find the charge on the capacitor:
Q(t) = C * V(t) = (3uF) * (-3H * (20*cos(2t)))
Finally, to find the charge after 2 seconds (t = 2), we can substitute t = 2 into the expression for Q(t):
Q(2) = (3uF) * (-3H * (20*cos(4)))
Calculating the expression above will give us the amount of charge stored on a plate of the capacitor after 2 seconds.