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The equation is:

\$I_{in} = \frac{V_{th} - V_{in}} { R_{th}}\$

This equation is from a one-port network.

enter image description here

Variables:

I = input current

Vth = thevenin voltage

Vin = input voltage

Rth = thevenin resistance

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  • \$\begingroup\$ Where is the schematic? \$\endgroup\$
    – dirac16
    Jun 15 '17 at 23:50
  • \$\begingroup\$ @dirac question edited. \$\endgroup\$
    – 21rw
    Jun 15 '17 at 23:55
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    \$\begingroup\$ You added a single line, without a schematic. What did you miss about "where is the schematic"? \$\endgroup\$
    – Jashaszun
    Jun 16 '17 at 0:10
  • \$\begingroup\$ @Jashazun I'm sorry. Reedited the question. \$\endgroup\$
    – 21rw
    Jun 16 '17 at 0:14
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Here's the equivalent circuit in the situation you're asking about

schematic

simulate this circuit – Schematic created using CircuitLab

"X" is whatever device is outside the device being studied. It could be a source, a resistor, or anything else.

Regardless of what is connected to the port, if the input voltage is \$V_{in}\$, then you can see the voltage across the thevenin equivalent resistor is \$V_{in}-V_{th}\$, so the input current can be obtained from Ohm's law as

$$\frac{V_{in}-V_{th}}{R_{th}}.$$

Which, incidentally, means there is a sign error in the formula you presented.

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