A capacitor of capacitance 10^3 farad is connected in series with a resistor of 10ohms and induction of 0.05 henry. The current in the circuit at t=O and the change on the capacity is also 0. Determine the change at any time t after the circuit is closed given that the emf is 50sin 200t
To find the charge on the capacitor at any time t after the circuit is closed, we need to solve the differential equation that describes the behavior of the circuit. The equation is:
Q''(t) + (1/(RC))Q'(t) + (1/(LC))Q(t) = (1/L)E(t)
Where Q(t) is the charge on the capacitor at time t, R is the resistance, C is the capacitance, L is the inductance, E(t) is the emf, and '' represents the second derivative with respect to time.
Given that the capacitance is 10^3 F, the resistance is 10 ohms, the inductance is 0.05 H, and the emf is 50sin(200t), we can substitute these values into the differential equation:
Q''(t) + (1/(10 * 10^3))(Q'(t)) + (1/(0.05 * 10^3))(Q(t)) = (1/(0.05))(50sin(200t))
Simplifying the equation:
Q''(t) + 0.0001Q'(t) + 0.02Q(t) = 1000sin(200t)
To solve this differential equation, we first assume a particular solution in the form:
Qp(t) = A*sin(200t) + B*cos(200t)
Taking the derivatives and substituting into the differential equation, we get:
-40A*cos(200t) - 40B*sin(200t) + 0.0001(200A*cos(200t) - 200B*sin(200t)) + 0.02(A*sin(200t) + B*cos(200t)) = 1000sin(200t)
Equating the coefficients of sin(200t) and cos(200t), we get:
-40A + 0.0001*200A + 0.02B = 0
-40B - 0.0001*200B + 0.02A = 1000
Simplifying these equations, we have:
(200*0.0001 - 40)*A + 0.02B = 0
-0.0001*200*B + (0.02 - 40)*A = 1000
Solving these equations simultaneously, we can find the values of A and B. Once we have A and B, we can determine the charge on the capacitor at any time t using the particular solution Qp(t) = A*sin(200t) + B*cos(200t).
Note that to find the initial conditions (Q(0) = 0 and Q'(0) = 0), we need to find the general solution of the homogeneous equation by assuming Q(t) = e^(at). After substituting this into the homogeneous equation and solving for a, we can find the general solution and then determine the constants using the initial conditions.
However, since you mentioned that the current in the circuit at t=0 and the change in the capacitance are also 0, it implies that the circuit has been in a steady state before t=0. In this scenario, the equation simplifies to a simple harmonic oscillator equation, and the behavior of the circuit is given by:
Q(t) = Qmax * sin(ωt + φ)
Where Qmax is the maximum charge on the capacitor, ω is the angular frequency (ω = 200 rad/s), and φ is the phase angle.
To determine the change in charge at any time t after the circuit is closed, we need to know the values of Qmax and φ. These can be obtained by using the initial conditions Q(0) = 0 and Q'(0) = 0.
Given Q(0) = 0, we have:
Q(0) = Qmax * sin(φ) = 0
Since the sine function is zero when the argument is zero, this implies that φ = 0 or π.
Given Q'(0) = 0, we have:
Q'(0) = Qmax * ω * cos(φ) = 0
For this equation to be satisfied, cos(φ) must be zero. This occurs when φ = π/2 or 3π/2.
So, the possible values for the phase angle φ are 0, π/2, and 3π/2. Using these values, we can determine the values of Qmax at any time t using the equation Q(t) = Qmax * sin(ωt + φ).