Networks I - Lab Section 6: Mondays 12.15-3.00 pm

Lab Assignment 2 - __due:
next Monday 25 September 2005__

__Laboratory 2 – KVL and KCL__

** Objectives:** The objective of this lab is to
practice using KVL and KCL for circuit analysis. Using KVL will show that the sum of the
voltages around a loop is zero. Using KCL will show that the sum of currents
entering and leaving a node sum to zero.

__Procedure:__

**Part 1:**

Build the circuit below using three different resistors. When building the circuit, select resistor values such that the power generated in each resistor in the circuit will be well below its ¼ Watt power dissipation rating. This will involve doing some calculations first.

Select three different resistances for R1, R2, and R3 and
also a DC voltage for Vin. Using KVL,
calculate the expected voltages across R1, R2, and R3. Then using KCL, calculate the expected
currents through R1, R2, and R3. Then
calculate the power dissipation in each resistor. ** Show all your calculations (even those
you will not use)**. If the power
is much less than ¼ Watt for each resistor, build the circuit using those
values. If the power is more than ¼
Watt, then try different values and recalculate the power. Once the circuit is built, measure the actual
values using the multi-meter and compare these values with the
calculations. Explain how the
measurement process itself will affect the resistances, voltages and currents
that are being measured. Comment on your
results.

**Part 2:**

Build the circuit in Mentorgraphics and display all the electrical parameters of the circuit. Compare these parameters with the values you measured empirically, and also the values calculated analytically. Explain differences in your results. Next, measure the actual values of the resistors using the Keithley multi-meter. Input these ‘measured’ resistor values into your Mentorgraphics circuit and recalculate the voltages and currents. How do these values now compare with the values measured, and all the values you calculated analytically?