In this problem we will investigate a fun idea called "duality."
Consider the series circuit in the diagram shown. We are given device parameters \(V=$V\)V, \(R_1=$R1\Omega\), and \(R_2=$R2\Omega\). All of the unknown voltages and currents are labeled in associated reference directions. Solve this circuit for the unknowns and enter them into the boxes given.
The value (in Volts) of \(v_1\) is:
The value (in Amperes) of \(i_1\) is:
The value (in Volts) of \(v_2\) is:
The value (in Amperes) of \(i_2\) is:
The value (in Volts) of \(v_V\) is:
The value (in Amperes) of \(i_V\) is:
The sum of the powers (in Watts) entering all of the elements is:

Now, let's turn our attentions to the parallel circuit. The numerical value of the strength of the current source is set to be the same as the numerical value of the strength of the voltage source: \(I=$I\)A. The resistances are set to be reciprocals of the resistances in the series circuit: \(R_3=$R3\Omega\) and \(R_4=$R4\Omega\).
Solve this circuit for the unknowns and enter them into the boxes given.
The value (in Volts) of \(v_3\) is:
The value (in Amperes) of \(i_3\) is:
The value (in Volts) of \(v_4\) is:
The value (in Amperes) of \(i_4\) is:
The value (in Volts) of \(v_I\) is:
The value of (in Amperes) \(i_I\) is:
The sum of the powers (in Watts) entering all of the elements is:

Look carefully at the numbers you have derived. Compare the series and parallel circuit. Do you see the pattern?