circuits test 1 review

8/3/2019 Circuits Test 1 Review

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o Thevenin’s Equivalent:

An independent voltage source in series with a resistor, which

replaces an interconnection of sources and resistors.

EX:

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

Find Vth, the voltage across the terminals (in this case it is the voltage over the 3kohm). 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 totalresistance 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 stillkeeping 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 smallerthan the smallest resistor in the combination. For example, 545 is smaller than 1k.

The final Thevenin equivalent is then:



o Norton’s Theorem

In order to find the Norton Short-circuit current, short the terminals where thecapacitor 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 analysis

http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/thev2.mesh.ans.htm

http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/thev2.mesh.ans.htm



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

Notice that all the mesh currents were drawn counter-clockwise. I3 is the current weare 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 eithermust 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



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:

o Source Transformations

P = iV

V = iRP = i2R

KCL: Sum of currents = 0 … OR … current in = current out

KVL: Change in Voltage = 0

(1)Count and label essential nodes and branches

(2) Label currents (i1, i2 … etc) with subscripts and arrows indicating

direction of current

(3)KVL & KCL enough for solution.

Batteries tend to send current out the positive side, unless they’re

being overpowered by another battery/ source.

Voltage Divider Theorem:

The voltage division rule (voltage divider) is a simple rule which can be used in

solving circuits to simplify the solution. Applying the voltage division rule can alsosolve simple circuits thoroughly. The statement of the rule is simple:

Voltage Division Rule: The voltage is divided between two series resistors in directproportion to their resistance.

It is easy to prove this. In the following circuit



Voltage Divider

the Ohm’s law implies that

(I)

(II)

Applying KVL

.

Therefore

.

Hence

.

Substituting in I and II

,

.

Consequently



,

.

which shows that the voltage is divided between two series resistors in direct

proportion to their resistance. The rule can be easily extended to circuits with morethan two resistors. For example,

Voltage Division among four resistors

,

,

,

.

The voltage division rule can be used solve simple circuits or to simplify solvingcomplicated circuits.

Problem 1-16: Voltage Divider

Find (or ) and (or ) using voltage division rule.a)

http://circuits.solved-problems.com/494/voltage-divider/

http://circuits.solved-problems.com/494/voltage-divider/



b)

c)



d)

Solution



a)

Voltage divider:

Ohm’s law:

b)

Voltage divider:

Ohm’s law:Please note that is leaving from the positive terminal of . Therefore, applyingthe Ohm’s law results in .



c)

Voltage divider:

Ohm’s law:

d)

The tricky part in this problem is the polarity of . In the defined formula forvoltage divider, the current is leaving the voltage source from the positive terminaland entering to resistors from positive terminals. In this problem, the current isentering to the the resistor from the negative terminal. Therefore, the voltage foris the negative of the voltage obtained from the voltage divider formula. The reasonis that another voltage can be defined with the inverse polarity and its value can befound using the voltage division rule. is the negative of the defined voltage



because it represents the voltage across the same nodes with inverse polarity.

Voltage divider:

Ohm’s law ( is entering from the negative terminal of ):

.

One of the common mistakes in using the voltage division rule is to use the formulafor resistors which are in parallel with other elements. For example, the voltagedivision rule cannot be used in the following circuit directly.

It will be incorrect if one tries to find using voltage divider by neglecting the otherresistor as

So, . However, if solving other parts of a circuits confirms that thecurrent of the other element/branch is zero, the voltage division rule can be still



applied. For example, suppose that the following network is a piece of a largercircuit.

Let’s assume that the analysis of the circuit shows that . In this case,

regardless of where A and B are connected.

Current Divider Theorem: Δ -> Y ; Y -> Δ Conversion:

Ammeters & Voltmeters:

Node Voltage Method (KCL)

Mesh Current Method (KVL)

circuits test 1 review

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