In the circuit shown, the switch is opened at t = 0, after having been closed for a long time. Find the current I as a function of time for t > 0.

Solution:

- Concepts:

RC circuits - Reasoning:

While the switch is closed the charge on the capacitor is Q = CV.

When the switch is opened the capacitor discharges through the resistor. - Details of the calculation:

While the switch is closed the voltage across the capacitor is 4 V.

When the switch is opened the capacitor discharges through a 250 Ω resistor.

V = V_{0}exp(-t/τ), τ = RC.

V = (4 V)exp(-t/(2.5*10^{-5 }s)).

I = V/R = (1.6*10^{-2 }A) exp(-t/(2.5*10^{-5 }s)).

Consider the RC circuit below with a switch S that can be opened and closed.

The capacitor is uncharged when, at t = 0, the switch is
closed. Current begins to flow and the capacitor will charge.

(a) Write the equation relating the charge on the
capacitor Q(t) and the current in the circuit I(t) to V_{0} and time.

(b) Solve for the voltages V_{C}(t) across the
capacitor and V_{R}(t) across the resistor.

(c) Sketch V_{C}(t) and V_{R}(t).

** **Solution:

- Concepts:

RC circuits - Reasoning:

We are asked to analyze the transient behavior of an RC circuit. - Details of the calculation:

(a) V_{0}- RdQ/dt - Q/C = 0, dQ/dt = I(t) = V_{0}/R - Q/(RC).

(b) Q(t) = CV_{0}(1 - exp(-t/(RC)), V_{C}(t) = V_{0}(1 - exp(-t/(RC)),

I(t) = (V_{0}/R)exp(-t/(RC)), V_{R}(t) = V_{0}exp(-t/(RC).

(c)

(a) Suppose a capacitor is charged by a voltage source, and then switched to a resistor for discharging. Would a larger capacitance value result in a slower discharge, or a faster discharge? How about a larger resistance value?

(b) Now consider an inductor, "charged" by a current source and then switched to a resistor for discharging. Would a larger inductance value result in a slower discharge, or a faster discharge? How about a larger resistance value?

Solution:

- Concepts:

Transients in circuits - Reasoning:

The transient behavior of a circuit is characterized by is time constant τ. - Details of the calculation:

(a) τ = RC.

Larger capacitance --> slower discharge

Larger resistance --> slower discharge

(a) τ = L/R.

Larger inductance --> slower discharge

Larger resistance --> faster discharge

(a) In the RC circuit shown, V abruptly changes value from V_{1} to
V_{2} at t = 0 after having been at V_{1} for a long time. Find the expression for the capacitor voltage as
a function of time.

(b) In the RC circuit shown, V abruptly changes value from V_{1} = 6 V to
V_{2} = 15 V at t = 0 after having been at V_{1} for a long
time. Find the expression for the capacitor voltage as
a function of time.

Solution:

- Concepts:

Transients in circuits - Reasoning:

We are asked to analyze the transient behavior of an RC circuit. - Details of the calculation:

(a) For t < 0 we have V_{C}(t) = V_{1}. Q(t) = CV_{1}.

For t > 0 we have

dQ/dt = I(t) = V_{2}/R - Q/(RC).

Q(t) = CV_{2}- C(V_{2}- V_{1})exp(-t/RC).

V_{C}(t) = V_{2}- (V_{2}- V_{1})exp(-t/RC).

(b) There are different approaches to analyzing this circuit.

(i) Using Kirchhoff's rules:

Use SI units, measure resistance in Ω, voltages in V and C in Farad.

For t > 0 we have

15 - 5*10^{3}I_{1}- 10^{4}I = 0.

Q/C - (20*10^{3}/3)I_{2}- 10^{4}I = 0.

I_{1}+ I_{2}= I. I_{2}= -dQ/dt.

Eliminate I_{1}= I - I_{2}:

15 - 15*10^{3}I_{ }+ 5*10^{3}I_{2}= 0.

Eliminate I:

I = 1*10^{-3}+ I_{2}/3.

Q/C - (10^{4})I_{2}- 10I = 0.

dQ/dt = 10^{-3}- Q/(10^{4}C).

Assume Q = A - Bexp(-t/x).

dQ/dt = -Q/x + A/x

x = 10^{4}C = 2.5*10^{-3}s, A = 10C.

Q(t) = 10C - Bexp(-t/2.5*10^{-3}s).

V_{C}(t) = 10 V - (B/C)exp(-t/2.5*10^{-3}s).

Initial condition: V(0) = (2/3)6 V = 4 V.

B/C = 6 V.

V_{C}(t) = 10 V - 6 V exp(-t/2.5*10^{-3}s).

(ii) Using Thevenin equivalent circuits:

ε_{eff}= VR_{2}/(R_{1}+ R_{2}) = 2V/3.

Z_{eff}= R_{3}+ R_{1}R_{2}/(R_{1}+ R_{2}) = 10 kΩ.

The equivalent circuit looks like this,

V_{C}(t) = 10 V - 6 V exp(-t/(RC)) = 10 V - 6 V exp(-t/2.5*10^{-3}s) .