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The question is:

Replace the circuit with a Thevenin Equivalent Circuit and find the Thevenin Resistance (Rt) in ohms.

I've used Thevenin before and solved quite a few equivalent circuits as well.

Unfortunately, with this circuit, I'm getting very confused on how you would exactly go about combining these resistors. Are they in series or parallel? Or are two of them in series with the other one? That's where I get confused.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I'm not incredibly familiar with Thevenin's Theorem but would we not also need to know the voltage of Vin? \$\endgroup\$
    – rob j loranger
    Mar 15, 2013 at 23:54
  • \$\begingroup\$ @rob I do not see a node labeled Vin in the diagram. We can assume that the open terminals on the right are the vantage point from which we are "looking into" the circuit. We have everything we need to reduce the circuit to its Thevenin equivalent: a single voltage source in series with a resistance. \$\endgroup\$
    – Kaz
    Mar 16, 2013 at 1:01
  • \$\begingroup\$ that's not the original image, the original was here link \$\endgroup\$
    – rob j loranger
    Mar 16, 2013 at 2:06
  • \$\begingroup\$ @rob I see. In that case, Vin basically refers to the Thevenin voltage that we want to find. It's not a great choice of symbol, obviously, since it is actually a voltage exhibited by the circuit. I.e. an output voltage. Usually we label driving inputs as Vin. \$\endgroup\$
    – Kaz
    Mar 16, 2013 at 2:58

2 Answers 2

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There are two steps to finding the Thevenin equivalent circuit: finding the Thevenin voltage and the Thevenin resistance.

Thevenin voltage is the voltage across the two points you interested in (Vin). In this case it is easy to calculate as there is no current flowing in the 43 and 60 \$\Omega\$ resistors thus no voltage drop. Thus the voltage at Vin is the same as the voltage form the source, 72 V.

Thevenin resistance is calculated by 'turning off' all independent current and independent voltage sources and calculating the resistance between the two points. Turning off a voltage source sets the voltage across it to 0, which results in a short (0 \$\Omega\$) in parallel with the 275 \$\Omega\$ resistor. Any resistor combined in parallel with a short results in a short, leaving you with the 43 and 60 \$\Omega\$ resistors now in series, giving a Thevenin resistance of 103 ohms.

Putting the two together gives you a voltage source of 72 V in series with a 103 \$\Omega\$ resistor for you Thevenin equivalent circuit.

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There is another way to calculate the Thevenin Equivalent: \$R_{th}=\dfrac{V_{oc}}{I_{sc}}\$

In this case, \$V_{oc}=72V\$ because there is no current flow in the 60\$\Omega\$ and 43\$\Omega\$.

\$I_{sc}\$ is calcualted by shorting \$V_{in}\$ so the 60\$\Omega\$ and 43\$\Omega\$ resistors are in series, and solving the original circuit. This puts \$R_2\$ and \$R_3\$ in parallel with \$R_1\$.

\$I_{sc}=\dfrac{72V}{(43\Omega+60\Omega)}=699mA\$

Finally, \$R_{th}=\dfrac{V_{oc}}{I_{sc}}=\dfrac{72V}{699mA}=103\Omega\$

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