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 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.
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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).
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.
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:
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
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