Question

A circuit is modeled as an RC circuit that has its output taken across the capacitor. Assuming a ramp input is supplied to this circuit. The differential equation representing this circuit is given by: $v_c(t) + RC \frac{dv_c(t)}{dt} = V_t$ $V_t = \text{ramped input voltage}$ $V = \text{voltage rise per second}$ $v_c(t) = \text{output voltage (taken across the capacitor)}$ At t=0, $v_c(t) = 0$. a. Solve the above single order differential equation for $v_c(t)$ using Laplace Transforms. (Hint: Use Partial Fractions) b. Sketch a graph of $V_t$ and $v_c(t)$. What is the physical meaning of this relationship?

          A circuit is modeled as an RC circuit that has its output taken across the capacitor. Assuming a ramp input is supplied to this circuit. The differential equation representing this circuit is given by:

$v_c(t) + RC \frac{dv_c(t)}{dt} = V_t$

$V_t = \text{ramped input voltage}$

$V = \text{voltage rise per second}$

$v_c(t) = \text{output voltage (taken across the capacitor)}$

At t=0, $v_c(t) = 0$.

a. Solve the above single order differential equation for $v_c(t)$ using Laplace Transforms. (Hint: Use Partial Fractions)

b. Sketch a graph of $V_t$ and $v_c(t)$. What is the physical meaning of this relationship?

A circuit is modeled as an RC circuit that has its output taken across the capacitor. Assuming a ramp input is supplied to this circuit. The differential equation representing this circuit is given by:

vc(t) + RC (dvc(t))/(dt) = Vt

Vt = ramped input voltage

V = voltage rise per second

vc(t) = output voltage (taken across the capacitor)

At t=0, vc(t) = 0.

a. Solve the above single order differential equation for vc(t) using Laplace Transforms. (Hint: Use Partial Fractions)

b. Sketch a graph of Vt and vc(t). What is the physical meaning of this relationship?

Added by Courtney S.

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