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I'm struggling with analyzing this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

\$R_1 = 5\Omega\$

\$R_2 = 5\Omega\$

\$R_3 = 2.5\Omega\$

\$R_4 = 5\Omega\$

Voltage source = 5 V

The task is to determine the Thevenin equivalent (Vth and Rth) between a and b respectively (black boxes). I'm not sure how to approach this question. Could anyone help me to get started?

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  • \$\begingroup\$ The voltage source you put there is a sine wave source. You're saying it's 5VDC? \$\endgroup\$
    – tokamak
    Mar 10 '16 at 20:02
  • \$\begingroup\$ Oops, yes that's what I meant. I'm not used to drawing schematics online. \$\endgroup\$
    – Steve
    Mar 10 '16 at 20:03
  • \$\begingroup\$ Also this is clearly a homework problem, which is ok, but you're required to explain what you've tried yourself to solve the problem first before anyone can help you. \$\endgroup\$
    – tokamak
    Mar 10 '16 at 20:04
  • \$\begingroup\$ Okay. I've tried determining Req by removing the voltage source from its current position and moved it around so the voltage now goes through R1. This means R3 and R4 are now in series and are parallel with R2, which allows me to determine Req. By doing this, I get Req to be 3 Ohms. This is equivalent to Rth as far as I understand. I can now calculate Vth: 5 V*3 Ohms/(3 Ohms + 5 Ohms) = 1.875 V = Vth \$\endgroup\$
    – Steve
    Mar 10 '16 at 20:15
  • \$\begingroup\$ @Steve No, you place a short where voltage source is, then you calc the equivalent resistance. And you calculate the voltage A-B as it is now, then you have Veq, Req. \$\endgroup\$ Mar 10 '16 at 20:40
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The best way to solve these examples is usually by using source transformations.

  1. Convert V1 into a current source V1/R2 --> I1 = 5V/5ohm = 1A. R2 is now parallel to the current source.
  2. Combine R1 and R2 (they are now parallel) -> R = 2.5 ohm
  3. Convert the current source back to a voltage source, V = I1 * R = 2.5V. R is now in series with the voltage source.
  4. Combine R with R3 --> R' = 5 ohm
  5. Convert this back to a current source I = 2.5V / 5ohm = 0.5A. R' is now parallel to the current source and parallel to R4.
  6. Combine R' and R4: Rth = 2.5ohm
  7. Convert the current source back to voltage source Vth=0.5A*Rth = 1.25V

This seems difficult, but with a little practice this is the quickest way.

The result is Vth=1.25V and Rth=2.5, which agrees with the result from the simulation.

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You can rearrange the resistors like this to help visualize how to simplify:

schematic

simulate this circuit – Schematic created using CircuitLab

Tyler's answer explains how to find the thevenin resistance. To then find the thevenin voltage, it's the open-circuit voltage between your terminals a and b. (source: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html#c3), which turns out to be EDIT: 1.25V (yes, as Mario points out, the circuit simulator is correct).

Voltage divider: top resistance is R2, bottom resistance (across a and b) is R13 (R4 does not come into play in the voltage divider because we're looking for the voltage across and b terminals. Think about it from the bottom-up, from the zero volt terminal up). So:

1 / R13 = 1/R1 + 1/R3

R13 = 1/(1/5 + 1/2.5) = 5/3 Ω

Vth = 5V * (R13 / (R13+R2))

Vth = 5V * (5/3Ω / (5/3Ω + 5Ω)) = 1.25V

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Whenever we are asked to find Thevenins euivalent circuit...WE have to first of all remove the load and find the voltage seen from load and the resistance seen from load In this case we assume that load was across A and B which is already removed. NOw have a look at this image which shows how to find Vthenter image description here

Vth= 1.25V To find Rth ..REplace the source (May be current or Voltage) by their internal resistance. For voltage source, internal resistance is zero that is we short circuit it. And then from left to right
(1) (R2 parallel R1)=2.5 (2) 2.5 series with R3(2.5) = 5 (3) 5 paralle R4(5) = 2.5 ohm therefore Rth= 2.5

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