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4 years ago

15

A circuit consists of a 6 ohm resistor, a 0.2 farad capacitor, and an AC voltage source supplying V(t) = 120 cos(20 t) volts. Wr

ite the differential equation for the charge on the capacitor. Write a formula for the solution, assuming q(0) = 0.

1 answer:

Zinaida [17]4 years ago

7 0

Answer:

q = 24 cos (20t) (1- e (-t / 1.2))

Explanation:

To work this circuit we use the mesh equation (Kirchoff) that states that the voltage in a closed circuit is zero

Vg + Vr + Vc = 0

Where Vg is the generator voltage, Vr the resistance voltage and Vc the capacitor voltage

Vg = 120 cos 20t = V

Vr = i R

Vc = q / C

We see that in one term we have the current (i) and in another the load (q), but there is a relationship between the two

i = dq / dt

Let's replace in the initial equation

V + R dq / dt + q / C = 0

Reorder the terms

Rdq / dt + q / C - V = 0

dq / dt + q / rC - V / R = 0

dq / dt = V / r -q / RC

dq / (V / R -q / RC) = dt

we integrate

∫I dq / (V / R - 1 / RC q) = ∫ dt

We change variables

u = (V / R - 1 / RC q)

du = -1 / RC dq

∫ dq / (V / R - 1 / RC q) = -RC ∫ du / u

-RC ln u = -RC ln (V / R - 1 / RC q)

We evaluate between the limits of integration, the lower t = 0 q (0) = 0

-RC [ln (V / R - 1 / RC q) - ln (V / R)] = t

[ln (V / R - 1 / RC q) - ln (V / R)] = -t / RC

Ln (V / R - 1 / RC q) / (V / R)] = -t / RC

Ln (1 - 1 / VC q)) = (-t / RC)

Ln (VC- q) = ln (VC) (-t / RC)

VC-q = VC e (-t / RC)

q = VC - Vc e (-t / RC)

q = VC (1- e (-t / RC)

We substitute the values they give us

q = (120 cos (20t) 0.2) (1- e (-t / 6 0.2))

q = 24 cos (20t) (1- e (-t / 1.2))

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3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

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Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

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