.

It has two parts, Vth and Rth. We'll find them each below. First, let's remove the capacitor, since we are finding the equivalent with respect to the capacitor (and thus it is not included in the circuit we are reducing). Since we are finding the Thévenin, we leave a gap (an open) because we will be finding the open-circuit voltage for Vth.

The terminals shown in the circuit below are the connections from the removed capacitor to the rest of the circuit (sliding the 600K resistor to the left a bit, but keeping its electrical connections the same).

V

The current source of 35uA will flow down through the 300K, then split between two branches: (a) the 400K and (b) the 1M and 600K in series. These two branches (a) and (b) are in parallel because they are connected electrically at the head (where the 400K, and 1M are conected) and the tail (where the 400K and 600K are connected). We can use a current divider to find how much of the 35uA goes down the (b) branch:

Now we can use the 7uA in branch (b) to find the voltage across the 600K (which is also the open-circuit voltage across the terminals of the capacitor). Using Ohm's law, we get:

V

Note that the voltage has polarity with the "+" at the bottom of the 600K and the "-" at the top of the 600K, because the current must flow in the "+" terminal for the passive sign convention.

In this reduced circuit, the 300K is not connected on the left side, so we can safely ignore it. The 40V source now forms two independent voltage divider circuits:

- Above it: the series combination of the 200K and then the combined parallel 500K and 700K
- Below it: the series combination of the 400K, 1M, and 600K

V

We computed the voltage in each subcircuit with the "+" at the bottom of the 600K and the "-" at the top of the 600K, so we can add them directly now.

V

V

The 500K is in parallel with the 700K and that combination is in series with the 200K. However, that entire combination is shorted out by the wire where the 40V source used to be. So with respect to the capacitor, if current would flow from the capacitor into the top terminal, it would completely bypass those three resistors.

Current flowing from the capacitor into the top terminal would thus split down through the 400K and the 600K. The fraction of current through the 400K would then be forced to also go through the 1M, so the 400K and 1M are in series, and then that combination is in parallel with the 600K.

Rth = 600K || (400K + 1M)

Rth = 600K || 1.4M

Rth = (600K * 1.4M) / (600K + 1.4M)

Rth = 420K

Back to Questions

© 2012, Steven H. VanderLeest