Find the Thévenin equivalent with respect to the 7k ohm resistor.

SolutionRemove the 7k ohm, since it is not part of the circuit we wish to simplify. Keep the terminals open since we are finding the Thevenin.

Find Vth, the voltage across the terminals (in this case it is the voltage over the 3k ohm). Combine the 1k and 2k in parallel.

1k || 2k = (1k*2k) / (1k+2k) = 2M/3k = 2/3k = 667 ohms

Now use a voltage divider to compute Vth across the 3k ohm.

Vth = [3k/(667+3k)] * 5V = 4.1V

Find the Thevenin Resistance by deactivating all sources and computing the total resistance across the terminals. The voltage sources is shorted, as shown:

Now let's redraw the circuit, bringing the 1k and 2k into a vertical position (but still keeping them connected the same way electrically).

They are all in parallel, so:

Rth = 1k || 2k || 3k = 1 / (1/1k + 1/2k + 1/3k) = 545 ohms

Note, as a check, the equivalent resistance for parallel resistors is always smaller than the smallest resistor in the combination. For example, 545 is smaller than 1k.

The final Thevenin equivalent is then:

Find the Norton Equivalent with respect to the 20uF capacitor

SolutionRemove the capacitor, since it is not part of the circuit we wish to simplify.

In order to find the Norton Short-circuit current, short the terminals where the capacitor used to be, since we are finding the Norton

Now the 2k ohm resistor is shorted-out, so we can eliminate it. Then we'll find the current through the short. Here are two different ways to solve for the current:

Mesh-Current Solution for Isc

Mesh-current analysis to find the current through the short.

Notice that all the mesh currents were drawn counter-clockwise. I3 is the current we are particularly interested in. Here are the mesh current equations:

KVL for i1:

12k*i1 -1k*i4 -12 = 0

KVL for i2:

-15 + 4k*i2 - 4k*i3 = 0

KVL for i3:

-4k*i2 + 10k*i3 -6k*i4 + 12 = 0

KVL for i4: We have trouble writing the voltage over the current source, so we either must add another variable, or simply write:

i4 = 20mA

Now solve the system of equations.

Solving the second equation for i2, we get:

i2 = (15 + 4k*i3) / 4k

Now rewrite equation 3, plugging in our formulas for i2 and i4:

-4k*(15 + 4k*i3)/4k + 10k*i3 - 6k*20mA + 12 = 0

Solve for i3:

-15 - 4k*i3 + 10k*i3 -120 + 12 = 0

i3 = 123/6k = 20.5mA

Superposition Solution for Isc

Superposition analysis to find the current through the short means we solve several simpler circuits (one for each source. Since there are three sources, we have three simpler circuits to solve.

In each of these circuits, all sources but one have been deactivated (voltage sources are shorted, current sources are opened). Our total answer is the sum of the answers we get from each of the circuits:

Isc = Isc15 + Isc12 + Isc20

First, let's combine the 8k, 3k, and 1k in series:

R = 8k + 3k + 1k = 12k

and also drop the 5k, since no current can flow through it.

The 12k is shorted out by the horizontal wire, so it can be removed.

Now we can see that the current through the 6k ohm resistor will be the current Isc. Thus,

Isc15 = 15V/6k = 2.5mA

That's not the final answer though! We have to find the other two parts of Isc.

The 5k ohm is again left hanging, so we can remove it. The 8k, 3k, and 1k can be combined in series (12k). The 4k ohm is shorted out, so it can be removed.

Isc12 = -12V/6k = -2mA

We can remove the 4k and 6k since they are both shorted out. We can also combine the 8k and 3k in series (11k).

Isc20 = 20mA

Now we add the three sub-answers together to get our final answer:

Isc = Isc15 + Isc12 + Isc20

= 2.5mA - 2mA + 20mA

= 20.5 mA

Now find the Norton resistance (same as Thevenin resistance). First, open the terminals where the capacitor used to be:

Now deactivate all sources (short voltage sources, open current sources):

The 4k ohm is shorted out, so it can be removed, and the 5k cannot have any current through it, so it can also be removed.

We can combine the 8k, 3k, and 1k in series.

However, the 12k combination is shorted out by the wire, so it can be eliminated.

Now let's redraw the circuit:

The resistors are in parallel, so the total resistance seen by the capacitor is

Rth = 6k || 2k = 6k*2k/(6k+2k) = 12M/8k = 1.5k ohms

Thus the final Norton is shown below:

Pspice Simulation

Find the Thévenin equivalent with respect to the capacitor in the circuit shown. Then replace the capacitor with a resistor chosen for maximum power transfer. What is the value of the resistor? What is the power absorbed by this resistor?

SolutionThe Thevenin equivalent has two parts, Vth and Rth. We will do the easier one first -- Rth. To find the Thevenin resistance, deactivate all sources (short voltages and open currents). Since we are finding the Thevenin with respect to the capacitor, we also take the cap out of the circuit and consider the resistance seen from the terminals where the cap was.

The 3K and 1K are in series, but then that combination is shorted out by the wire that replaced the voltage source. Another way to think about it is that we have 1K+3K =4K, and then 4K || 0 = 0. That is, a 0 ohm resistor in parallel with anything else is still 0. Thus we have:

Redrawing it slightly (but maintaining the same connections):

We now see the 4K is in parallel to the 5K, so

Rth = 4K || 5K = 4K*5K/(4K+5K) = 2.22K ohms

Now we must find Vth. For this, we must find the open circuit voltage at the terminals:

Note that there is 2V across the 4K and 5K in series. It does not matter that the 2V is also across the 3K and 1K in series. We will use a voltage divider for the 4K and 5K in series with a know total voltage of 2V:

Vth = 5K/(4K+5K)*2V = 1.11V

So the final Thevenin equivalent is:

The Norton could be found directly by computing Rth in the same way and finding Isc by shoring the terminals and computing the current. Now that we have the Thevenin, we can also find the Norton simply by Ohm's law: Isc = Vth/Rth = 1.11V/2.22Kohm = 0.5mA. The Norton is then:

Find the Thévenin equivalent with respect to the capacitor in the circuit shown. Then replace the capacitor with a resistor chosen for maximum power transfer. What is the value of the resistor? What is the power absorbed by this resistor?

SolutionSince we are finding the Thevenin with respect to the capacitor, we also take the cap out of the circuit and consider the resistance seen from the terminals where the cap was.

The Thevenin equivalent has two parts, Vth and Rth. We will do the easier one first -- Rth. To find the Thevenin resistance, deactivate all sources (short voltages and open currents).

From the point of view of the capacitor terminals, the 1K and 2K are shorted out. The 7K also is not included because no current can flow through it. If current was fed into the top terminal, it would flow through the 4K and 5K and then come back through the other terminal. Thus

Rth = 4K || 5K = (4K*5K)/(4K+5K) = 2.2K ohms

Next, we'll find Vth using node-voltage analysis, with one node (the bottom wire is the reference node).

Writing KCL at the node V1 (current leaving):

Solve for V by multiplying through by 20K:

5V - 15 + 4V - 120 = 0

9V = 135

V = 135/9 = 15V

So the final Thevenin Equivalent is:

Find the Thévenin equivalent circuit with respect to the capacitor. You must use superposition to find Vth

SolutionSince we are finding the Thevenin with respect to the capacitor, we also take the cap out of the circuit and consider the resistance seen from the terminals where the cap was.

The Thevenin equivalent has two parts, Vth and Rth. We will do the easier one first -- Rth. To find the Thevenin resistance, deactivate all sources (make their value 0).

Voltage sources of value 0 can be modeled as shorts. Current sources of value 0 can be modeled as opens.

If current was fed into the top terminal, it would split left and right. The current going left would flow through the 4K (skipping the 1K and 2K because they are shorted out), then split again

through the 8K and 9K. The current going right would flow through the 5K (skipping the 7k because of the open).

Rth = 5K || (4K + 8K || 9K) = 3.1K ohms

 

Next, we'll find Vth using superposition. There are three sources, so we'll have three subproblems to solve. The total Vth can be found as the sum of the individual effects:

Subproblem 1

Vth due to the 3V can be found with this circuit:

Use a voltage divider to find Vth:

Vth = 5K/(4K + 5K + 8K||9K) * 3V = 1.13V

Subproblem 2

Vth due to the 10V can be found with this circuit:

We can find the voltage across the 8K (on our way to finding the voltage across the terminals) by combining the 4K and 5K in series and that in parallel with the 8K:

Req = (5K+4K)||8K = 4.24K

V1 across Req can be found by a voltage divider:

V1 = Req/(Req+9K)*10V = 3.20V

Then the voltage over the 5K (which is also the voltage over the terminals) is another voltage divider on the 3.2V:

Vth = 5K/(4K+5K)*3.2 = 1.78V

Subproblem 3

Vth due to the 6mA can be found with this circuit:

Now combine the 8K and 9K in parallel and that in series with the 4K:

Req = 4K + 8K||9K = 8.24K

Use a current divider to find the current through the 5K branch:

I = Req / (Req + 5K)*6mA = 3.73mA

Use Ohm's law to find the voltage over the 5K:

Vth = 3.73mA * 5K = 18.67V

Now add up the subanswers to get the total Vth:

Vth = 1.13 + 1.78 + 18.67 = 21.58V

The Thevenin equivalent circuit is then:

Find the Norton Equivalent with respect to the 3 Kohm resistor in the middle of the circuit, i.e., the 3 Kohm resistor itself should not be part of the equivalent that you compute.

SolutionSince we are finding the Norton with respect to the 3 Kohm, we take the 3 Kohm out of the circuit and consider the resistance seen from the terminals where the 3K was.

The circuit to the left of the 3K is already a Norton equivalent, where the Norton current is -10 mA (because it is facing down). The resistance is infinite. That is, when you open the current source to deactivate it, the 1K and 2K are left disconnected.

The circuit to the right of the 3K is already a Thevenin, where the voltage is 6V and the equivalent resistance is 9 Kohms. Converting to a Norton, we get Norton current of 0.667 mA and a resistance of 9 Kohms.

Now combine the two Nortons. The total curent will be -10mA + 0.667mA = -9.33 mA. The total resistance is infinite in parallel with 9K, which is simply 9K.

Find the Thévenin equivalent with respect to the 1nF capacitor. You must use super-position to find Vth, the Thévenin voltage.

SolutionA Thévenin equivalent is a circuit, like the one shown here.

.

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).

Find Vth The question requires that we use superposition to find Vth. There are two sources in the circuit, so we will have a reduced circuit for each source (with all other sources deactivated). The total Vth will be the sum (super-imposing) of the two subcircuit answers: Vtotal = V35uA + V40V

V35uA = Vth due to 35 microamp source We deactivate the 40V source by shorting it. The resulting circuit is:

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: V35uA = 7uA * 600K = 4.2V 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.

V40V Vth due to 40V source We deactivate the 35uA source by opening it. The resulting circuit is:

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

These are independent, just like mountain climbers climbing up to the 40 thousand foot peak of Mt. Himalaya on the north face and another group of climbrs on the south face. The fraction of the total height for one group has no effect on the other group. So we will use a voltage divider just for the combination of 400K, 1M, and 600K, which goes across the entire voltage (height of the mountain) of 40V. The voltage across just the 600K (which is also the open circuit voltage across the capacitor) is:

Notice that the 12V has polarity with the "+" at the bottom of the 600K and the "-" at the top of the 600K, because the voltage is higher at the "+" side of the voltage

source and lower at the "-" side of the voltage source (where the "-" of the voltage source is at the top of the 600K).

Vtotal by Superposition Using the answers to the subcircuits above, we now have: Vtotal = V35uA + V40V 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. Vtotal = 4.2V + 12V Vtotal = 16.2V

Find Rth To find Rth, we deactivate all the sources, so open the current source and short the voltage source. The resulting circuit is:

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

Draw Circuit The final equivalent circuit is then: