Lab 4 Report Example

Part 1:

According to Kirchoff’s nodal rule, the current that flows inside the node is distributed among the paths. This means that the sum of all currents coming in and out of a node is always zero (Young, Freedman and Ford 855-860). In this particular case, if we consider the node on the left side of the schematic diagram (fig. 1), then: I1 = I2 – I8.2k (eq. 1).
Now let us consider the two loops generated by the parallel circuit. According to Kirchoff’s Voltage Loop (KVL), the sum of all the voltages in a certain loop is zero (Young, Freedman and Ford 855-860). Therefore, applying KVL to Loop 1, we would have the equation written below.
KVL: Loop 1 6V-I19.1kΩ-I38.2kΩ - I11.1kΩ=0 (eq. 2)
Note that the direction of the voltage decreases in the direction of the positive current flow. In this case, I1 and I3 are both flowing as shown by the arrows in the schematic diagram. For the purposes of solving the values for the current later, we must simplify (eq. 2) in terms of I1 and I2 using (eq. 1). Therefore:
6V-I110.2kΩ-I1- I21.1kΩ=0
6V-I118.4kΩ+ I28.2kΩ=0 (eq. 3)
Moreover, KVL must also apply to Loop 2. However, it is important to note that the voltage flows in the direction of the negative current flow (I2) but flows in the positive direction of another current (I3). Therfore:

KVL: Loop 2

9V+ I21.2kΩ-I38.2kΩ=0
9V+ I29.4kΩ-I18.2kΩ=0 (eq. 4)
Again, note that the equation is simplified in terms of I1 and I2. Now, we must have three linear equations in terms of I1, I2 and I3.
I1-I2-I3=06V-I118.4kΩ+ I28.2kΩ=09V+ I29.4kΩ-I18.2kΩ=0
We can now solve for the values of the currents I1, I2 and I3. After calculations (shown in the appendix section), the values of the currents are: I1 = -0.16 mA, I2 = -1.1 mA and I3 = 0.94 mA. Using these values, we can now compute the values for the voltage per resistor using Ohm’s Law (V = IR). The measured and simulated values are shown in the table below:
It is important to note the calculated values may not be exactly equal to the other two values. Errors may be a great factor in this particular laboratory experiment. Errors in making the actual circuit together with not accounting certain values (resistance of the wires, imperfection of other materials) can cause deviations from the theoretical values. Nevertheless, the calculated values are closer to the measured values (except for the value of I1) than the simulated values.

Part 2

In this part, consider just one loop (see fig. 2). However, there exists a node before the 4k Ω resistor. In this node, according to Kirchoff’s nodal rule, I2 = I1 + I4k (eq. 5) with I1 = 3 mA. Using KVL on the loop:

KVL:

12V-I21kΩ-I4k4kΩ-I25kΩ-8V= 0 (eq. 6)
12V-I21kΩ-I2-I14kΩ-I25kΩ-8V= 0 12V-I21kΩ-I2)4kΩ +(I14kΩ-I25kΩ-8V= 0
12V-I21kΩ-I2)4kΩ+3 mA4kΩ-I25kΩ-8V= 0
16V-I210kΩ= 0
I2= 1.6mA
The theoretical value for I2 is 1.6 mA. We can compute for the value of I4k using (eq. 5). Thus, I4k = -1.6mA. The values do not have any conflict with the simulated values because the calculated and simulated values are almost similar.

Reference:

Young, H., Freedman, R., and Ford, AL. Sears and Zemansky’s University Physics with Modern Physics. 13th Edition. San Francisco, CA: Pearson Education, Inc., publishing as Addison-Wesley, 2012. Print.
Appendix:
Figure 1: Schematic diagram of the first part of the experiment showing the direction of the loop and the values of resistors and cells.
Figure 2: Schematic diagram for the second part of the experiment showing the values for resistors, cells and the current I1.
Calculation in Part 1:
I1-I2-I3=06V-I118.4kΩ+ I28.2kΩ=09V+ I29.4kΩ-I18.2kΩ=0
Using the second linear equation:
6V-I118.4kΩ= -I28.2kΩ
I2= 6V-I118.4kΩ-8.2kΩ
(eq. 7)
Use (eq. 7) in the third linear equation:
9V+ 6V-I118.4kΩ-8.2kΩ9.4kΩ-I18.2kΩ=0
I1≈ -0.16 mA
Use this value to solve for I2 (eq. 7), then solve I3 using (eq. 1).
I2= 6V--0.16mA18.4kΩ-8.2kΩ ≈ -1.1mA
I3=I1-I2= -0.16mA+1.1 mA=0.94 mA