# An Example of Solving a Circuit

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Connexions module: m0016 1 An Example of Solving a Circuit ∗ Don Johnson This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License † Abstract An example of solving a circuit (a) (b) Figure 1: The circuit shown is perhaps the simplest circuit that performs a signal processing function. The input is provided by the voltage source vin and the output is the voltage vout across the resistor labelled R2. 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 = vin (1) v-i: v1 = R1i1 (2) vout = R2iout (3) ∗ Version 2.8: Jun 8, 2005 2:58 pm -0500 † http://creativecommons.org/licenses/by/1.0 http://cnx.org/content/m0016/2.8/ Connexions module: m0016 2 KCL: (−i)− i1 = 0 (4) i1 − iout = 0 (5) KVL: −v + v1 + vout = 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 vout . The KVL equation can be rewritten as vin = v1 + vout. Substituting into it the resistor's v-i relations, we have vin = R1i1 + R2iout. 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 i1 = iout, which means that vin = R1iout +R2iout = (R1 +R2) iout. Solving for the current in the output resistor, we have iout = vinR1+R2 . 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. Vout = R2 R1 +R2 vin (7) Exercise 1 (Solution on p. 3.) Referring back to Figure 1, a circuit should serve some useful purpose. What kind of system do

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