Skip to content Skip to navigation

Connexions

You are here: Home » Content » An Example of Solving a Circuit

Navigation

Content Actions
  • Download module PDF
  • Add to ...
    Add the module to:
    • My Favorites
    • A lens
    • An external social bookmarking service
    • My Favorites (What is 'My Favorites'?)
      'My Favorites' is a special kind of lens which you can use to bookmark modules and collections directly in Connexions. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need a Connexions account to use 'My Favorites'.
    • A lens (What is a lens?)

      Definition of a lens

      Lenses

      A lens is a custom view of Connexions content. You can think of it as a fancy kind of list that will let you see Connexions through the eyes of organizations and people you trust.

      What is in a lens?

      Lens makers point to Connexions materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

      Who can create a lens?

      Any individual Connexions member, a community, or a respected organization.

    • External bookmarks
  • E-mail the author

Recently Viewed

This feature requires Javascript to be enabled.

An Example of Solving a Circuit

Module by: Don Johnson

Summary: An example of solving a circuit

Figure 1: The circuit shown is perhaps the simplest circuit that performs a signal processing function. The input is provided by the voltage source v in v in and the output is the voltage v out v out across the resistor labelled R 2 R 2 .
Subfigure 1.1Subfigure 1.2
, Subfigure 1.1 (circuit4.png), Subfigure 1.2 (circuit4a.png)

For the example circuit above, we have three v-i relations, two KCL equations, and one KVL equation for solving for the circuit's six voltages and currents.

v= v in v v in (1)
v-i:
v 1 = R 1 i 1 v 1 R 1 i 1 (2)
v out = R 2 i out v out R 2 i out (3)
KCL:
-i- i 1 =0 i i 1 0 (4)
i 1 - i out =0 i 1 i out 0 (5)
KVL:
-v+ v 1 + v out =0 v v 1 v out 0 (6)

We have exactly the right number of equations! Eventually, we will discover shortcuts for solving circuit problems; for now, we want to eliminate all the variables but v out v out . The KVL equation can be rewritten as v in = v 1 + v out v in v 1 v out . Substituting into it the resistor's v-i relations, we have v in = R 1 i 1 + R 2 i out v in R 1 i 1 R 2 i out . Yes, we temporarily eliminate the quantity we seek. Though not obvious, it is the simplest way to solve the equations. One of the KCL equations says i 1 = i out i 1 i out , which means that v in = R 1 i out + R 2 i out = R 1 + R 2 i out v in R 1 i out R 2 i out R 1 R 2 i out . Solving for the current in the output resistor, we have i out = v in R 1 + R 2 i out v in R 1 R 2 . We have now solved the circuit: We have expressed one voltage or current in terms of sources and circuit-element values. To find any other circuit quantities, we can back substitute this answer into our original equations or ones we developed along the way. Using the v-i relation for the output resistor, we obtain the quantity we seek.

V out = R 2 R 1 + R 2 v in V out R 2 R 1 R 2 v in (7)

Exercise 1

Referring back to Figure 1, a circuit should serve some useful purpose. What kind of system does our circuit realize and, in terms of element values, what are the system's parameter(s)?

Solution 1

The circuit serves as an amplifier having a gain of R 2 R 1 + R 2 R 2 R 1 R 2 .

Comments, questions, feedback, criticisms?

Send feedback