Calculate voltage and current in time dependent circuit

Question

I'm working with this time dependent circuit, and i'm having troubles with some of the assignments. The following values are given R1 = 1kΩ, R2 = 2kΩ, R3 = 1kΩ, C1 = 22uF and C2 = 47uF. Is delivers 20mA, and when t=0 the switch goes from connected to disconnected. The circuit is stationary for t < 0 (when sure what that means?)

• First of i shall calculate the current through R2, when T<0. Is it right when i assume i should use current division? Calculate the total Ω of the circuit, and divide by R2, times the 20mA?

• Next, i shall calculate the voltage across C1. That one i'm uncertain about. I have found a formula V = q/C, but i can't seem to find q, the charge, or neither how to calculate what?

• Lastly i shall find an expression for the current i(t) for t>0. Here i'm a little lost, so hoping for some guidance

Might be rather easy, but for a newbie, who have been introduced to all at once, its hard to find head and tail,

Hoping for help!

Answer 1

Typical homework problem. I think of it as two steps:

1. Setup of Cap. Voltage via introduced energy
2. Discharge of Cap. Through Connected Impedance

The current division between R3 and R1 + R2 in series gives you the setup current (I label it I_0) (since R3 < R1+R2, the most current should go into R3)

I0 * R2 Gives you the voltage in the node that is connecting R2, C1, C2 (I name the node n1 and the voltage V_0)

Total capacitance of two parallel caps is the sum C1 + C2 = C'

the time constant is then R2*C' = tau

Classic RC-exponential discharge curve, R2 is the current path:

V(t) = V_0 * e ^ -(t/tau)

and V = R*I gives V(t)/R2 = i(t)

easy as pie.

All these tools are easily searchable in any electronics schoolbook or the internet.

Answer 2

Most probably the phrase about the circuit being stationary for \$t<0\$ means the circuit has reached steady state. Since it is powered only by a DC current source, that means that there is no time variation of any quantity in the circuit until the switch opens.

So this allows us to determine the initial conditions of the circuit just before the switch opens: no time variations means the current in C1 and C2 is zero, hence they may be neglected in the circuit. Therefore you have the current \$I_s\$ flowing through a 2-branches current divider. First branch is \$R_3\$, second branch is the series \$R_1,R_2\$.

Applying current divider formula to obtain \$i(0-)\$ you get:

$$ i(0^-) = I_s \cdot \dfrac{R_3}{R_1+R_2+R_3} = 20mA \times \dfrac{1k\Omega}{1k\Omega + 2k\Omega + 1k\Omega} = 5mA $$

Let's tackle the voltage across the caps at \$t = 0^-\$, let's call it \$v(0^-)\$. Since the caps are in parallel to \$R_2\$, the voltage is the same, hence:

$$ v(0^-) = i(0^-) \cdot R_2 = 5mA \times 2k\Omega = 10V $$

As for the expression for \$i(t)\$ after the switch opens: notice that with the switch open you are left with two charged caps connected in parallel with \$R_2\$, hence the current in the resistor will be a classic exponential discharge curve. The time constant of the exponential is \$ \tau = R_2 \cdot C_{tot}\$ where \$ C_{tot} = C_1 + C_2 \$ is the total capacitance seen by the resistor. Therefore:

$$ i(t) = i(0^-) \cdot e^{-\frac{t}{\tau}} $$

Answer 3

1. There are many ways. I suggest following: Transform the current source with parallel resistor (Norton source) into a voltage source with series resistor (Thevenin source). Now you have a voltage source with three resistors in series which makes it very easy to calculate the current.
2. For the stationary case you have to treat the Cs as open circuits (i.e. as insulators). You can use either the voltage divider formular or the result of 1. and apply Ohm's law to R2 to get the voltage across R2 (which is also across the Cs).
3. Just a hint: \$V_c = q/C\$ is not enough. You need also \$\frac{dq}{dt} = I\$.
   i.e. \$I_c = C\frac{dV_c}{dt}\$.
   You'll get a (very simple) ordinary differential equation you must solve.