Solution

Part (a)

Before 100ms

To find out what happens at time 0, keep your eye on the capacitor voltage before time 0. Don't bother looking at the current through the resistor before time 0 because it will change. You need to find the current by backing it out from the capacitor's voltage (which is the only thing guaranteed not to change through an instantaneous event. Assume that everything was in steady-state before time 0, so the capacitor looks like an open. Thus the current through it is zero, but the voltage is definitely not zero! It is:

Vc(0-) = 2K / (2K+6K) * 2V = 0.5V

Vc(0+) = Vc(0-) = 0.5V

i(0+) = Vc(0+) / (8K + 2K) = 0.5V / 10K = 50uA

Now find final value of i:

i(infinite) = 0

Now find time constant. Note that the 2V source is not present in the circuit, so we have an open on the right, meaning the 2K resistor is not part of the Thevenin seen by the capacitor.

Tau1 = Rth * C = (8K + 2K) * (5uF) = 50ms

Put it all together:

i(t) = Final + (Init - Final) e^(-t/Tau1)

i(t) = 0 + (50uA - 0) e^(-t/50ms)

Note that we weren't asked to find Vc(t), but we'll need it for the next part, so here it is:

Vc(t) = 0.5e^(-t/50ms)

The time constant is the same, but we use the intial and final values for Vc.

After 100ms

To find current after time 100ms, again keep your eye on the capacitor voltage. We cannot use the steady-state trick here because the circuit has definitely not reached steady-state at 100ms. Since the time constant is 50ms, this is only 2 time constants after the switching event, which isn't enough time for everything to settle out. (Need around 5 or 6 time constants to get back to steady-state). So we'll need to use the capacitor voltage formula from the last time period to compute the value just before the new switching event at 100ms.

Vc(100ms-) = 0.5e^(-100ms/50ms) = 0.068V

So now we know the voltage on the cap just after the 100ms switching event. The capacitor voltage is the only value guaranteed to stay the same through an instantaneous switching event.

Vc(100ms+) = Vc(100ms-) = 0.068V

Here's the tricky part now. At time 100ms+ (momentarily after 100ms), we have the capacitor looking like a voltage source of 0.068V on the left and the voltage source on the right of 2V. So to find the current i(t), we'll need to do some type of more involved analysis, such as superposition, node-voltage analysis, or mesh-current. Let's try node-voltage, with a single node V at top-center and the bottom as the Reference node.

(V-2)/6K + V/2K + (V-0.068)/8K = 0

V [ 1/6K + 1/2K + 1/8K] = 2/6K + 0.068/8K

V = 342u / 792u = 0.432V

Knowing the voltage at top-center at 100ms, we can find the current by Ohm's Law.

i(100ms+) = (0.068V - 0.432V)/8K = -45.5uA

So that's the initial current. Now find the final current. After a long time (time infinity), the current fades back to 0.

The time constant is different after 100ms because the circuit is different. Now the 2V source is back. To find the Thevenin resistance seen by the capacitor, deactivate all sources, which in this case means short the voltage source.

Tau2 = Rth * C = [8K + (2K || 6K) ] * (5uF) = [9.5K] * (5uF) = 47.5ms

Now put it all together to get:

i(t) = Final + (Init - Final) e^(-t/Tau1)

i(t) = 0 + (45.5uA - 0) e^[-(t-100ms)/47.5ms]

Note that is important to do the time-shift by 100ms in the exponential so that 100ms appears to be time zero.

Part (b)

Part (c)

Plugging in to the equations above for 50ms (using first time period equation) and 150ms (using second time period equation), we get:

i(50ms) = 18.4Ua

i(150ms) = -15.9uA

Back to Transient Analysis Examples

© Calvin College, 2009
This page was written and is maintained by Steve VanderLeest.
It was last modified on 19-Apr-2009 .