Find the charge on the capacitor in an LRC series at t = 0.02 sec when L = 0.05 h, R = 2
Ohm, C = 0.01 farad. The initial conditions are q(0) = 5 C, i(0) = 0 Ampere. Determine the firt
time at which the charge on the capacitor is equal to zero. Also, find the current at any time t
To find the charge on the capacitor at t = 0.02 sec, we can use the following formula for the charge on a capacitor in an LRC series circuit:
q(t) = q(0) * e^(-Rt/2L) * cos(sqrt(1/(LC) - (R/(2L))^2) * t - φ)
where:
q(t) is the charge on the capacitor at time t.
q(0) is the initial charge on the capacitor.
e is the base of the natural logarithm (approximately 2.71828).
R is the resistance in Ohms.
L is the inductance in Henrys.
C is the capacitance in Farads.
φ is the phase angle, which can be determined from the given initial conditions.
Let's calculate the charge on the capacitor at t = 0.02 sec:
q(0.02) = 5 * e^(-(2/0.05)*0.02/2*0.05) * cos(sqrt(1/(0.05*0.01) - ((2/(2*0.05)) ^2)*0.02 - φ)
However, we also need to find the phase angle (φ) to compute the charge on the capacitor accurately.
Given:
R = 2 Ohm
L = 0.05 H
C = 0.01 F
q(0) = 5 C
i(0) = 0 A
To find φ, we need to calculate the initial current (i(0)) using the formula:
i(0) = (1/L) * ∫(0 to ∞) (V(t)/R - q(t)/RC) dt
Since i(0) = 0 A, the integral of the expression must be zero. This implies:
∫(0 to ∞) (V(t)/R - q(0)/RC) dt = 0
As a result, we can find φ by substituting q(0) = 5 C into the equation and solving for φ.
To find the charge on the capacitor in an LRC series at t = 0.02 sec, we can use the equations that govern the behavior of an LRC circuit. The general equation for the charge on the capacitor, q(t), is given by:
q(t) = q(0) * e^(-t / RC) * cos(ωt + φ)
Where:
- q(0) is the initial charge on the capacitor
- R is the resistance in the circuit
- C is the capacitance
- t is the time in seconds
- e is the base of the natural logarithm
- ω is the angular frequency, given by ω = 1 / sqrt(LC)
- φ is the phase angle, given by φ = tan^(-1)(ωL/R)
Given the values in the problem:
- q(0) = 5 C
- t = 0.02 sec
- L = 0.05 H
- R = 2 Ω
- C = 0.01 F
First, calculate the angular frequency using the values of L and C:
ω = 1 / sqrt(LC) = 1 / sqrt(0.05 * 0.01) = 1 / sqrt(0.0005) = 1 / 0.02236 ≈ 44.74 rad/s
Next, calculate the phase angle using the values of ω, L, and R:
φ = tan^(-1)(ωL/R) = tan^(-1)(44.74 * 0.05 / 2) = tan^(-1)(0.8935) ≈ 40.60°
Now, substitute the values into the equation for q(t) to find the charge on the capacitor at t = 0.02 sec:
q(t) = q(0) * e^(-t / RC) * cos(ωt + φ)
= 5 * e^(-0.02 / (2 * 0.01)) * cos(44.74 * 0.02 + 40.60)
≈ 2.82 C
Therefore, the charge on the capacitor at t = 0.02 sec is approximately 2.82 C.
To find the first time at which the charge on the capacitor is equal to zero, we need to find the time when the expression q(t) = 0:
0 = q(0) * e^(-t / RC) * cos(ωt + φ)
Since cos(ωt + φ) can only be zero when ωt + φ = (2n + 1) * π / 2 (n is an integer), we can set:
ωt + φ = (2n + 1) * π / 2
Substituting the values of ω and φ, we have:
44.74t + 40.60 = (2n + 1) * π / 2
Simplifying, we find:
t = ((2n + 1) * π / 2 - 40.60) / 44.74
To find the smallest positive value of t, we can substitute n = 0:
t = ((2 * 0 + 1) * π / 2 - 40.60) / 44.74
= (π / 2 - 40.60) / 44.74
≈ -0.021 sec
Since time cannot be negative, there is no first time at which the charge on the capacitor is equal to zero.
Lastly, to find the current at any time t, we can differentiate the equation for q(t) with respect to time:
i(t) = dq(t) / dt
Differentiating q(t) = q(0) * e^(-t / RC) * cos(ωt + φ) with respect to t:
i(t) = (-q(0) / RC) * e^(-t / RC) * cos(ωt + φ) - (q(0) / RC^2) * e^(-t / RC) * sin(ωt + φ)
Substituting the given values into the equation, we can calculate the current at any time t.