Lab Report DC Circuits

Lab Report: DC Circuits Introduction The objective of this laboratory experiment was to study Kirchhoff’s Rules and to be able to apply them to complex DC circuits. When we are discussing a simple DC circuit consisting of a wire, a power source, and a single resistor, it is relatively easy to predict the flow of current in that circuit. After all, we know that the current would flow from a positive end of a power source to a negative end of a power source, traveling along the wire with a resistor in-line with it, but not all the circuits are so straight-forward. More complex circuits have multiple junctions where multiple currents can meet and split. Complex circuits can also have multiple power sources and switches. In such cases it can be much more difficult to predict the direction of a current along one segment of a wire within a circuit. When we know the value of the resistance as well as the magnitude and the direction of the electric potential difference along a single resistor, we can calculate the current running through that resistor using the equation I = V/R, but sometimes the only thing we know is the resistance of each resistor in a circuit and the electric potential difference of a power source. Kirchhoff’s Rules are very useful in such cases, as they can help us determine and calculate the directions and magnitudes of currents running through each individual resistor. Kirchhoff’s Rules state the following: 1.) The sum of the currents entering any junction in a circuit must equal the sum of the currents leaving that junction, and 2.) The sum of the potential differences across all elements around any closed circuit loop must be zero. Considering that the current will flow through a resistor from the point of a higher potential to the point of a lower potential, we can easily set up the equations that would allow us to solve for the magnitudes and the directions of individual currents flowing through a resistor. As a rule of thumb only three Kirchhoff’s equations are required in order to solve the circuit. Initially we would have to predict the directions of the currents running through each segment, and then carefully follow it, keeping all the signs in check. It is very important to consider the signs when we are performing our calculations, and to be consistent with the choice of the direction we have chosen in the beginning. Depending on the assumed direction, a positive result for the current would mean the direction was correct, and a negative result would mean the current was opposite of the assumed direction. The apparatus used in this laboratory experiment consisted of a circuit board, a 100Ω resistor (R1), a 10Ω plug-in resistor (R2), a 50Ω plug-in resistor (R3), a 22Ω resistor (R4), a decade resistance box with variable resistance (Rdecade), a switch, a power supply, and a digital multimeter. All the components are set-up in a specific arrangement. We have examined two circuits in this laboratory experiment: the one with a closed switch (Diagram 1), and the one with an open switch (Diagram 2). Procedures and Calculations First we had measured the resistance of each of the resistors separately. We have used a digital multimeter, set for the most sensitive ohms range, to measure the resistance of R1, R2, R3, and R4. Then we have wired the circuit, as shown in Diagram 1, with a closed switch. The output voltage of a power supply was measured using a multimeter and was found to be 4.98V. Then we had measured an electrical potential drop across R1 by adjusting the resistance of a decade box and making sure that the potential drop across R1 is as close to 2.00V as possible. The resistance of a decade box was measured as well. Finally we had measured the potential drop across each resistor using a digital multimeter. The values of resistance and electric potential drop across each resistor were recorded in Table 1. Currents flowing through each resistor, including the decade box, were calculated using the equation I = V/R and recorded in Table 1 with the units of milliamperes. The same procedures were performed for the circuit depicted in Diagram 2, where the switch is open. The resistance value on the decade box was kept constant. The values of electric potential differences across each resistor, including the decade box were calculated and recorded in Table 2. Current, flowing through each resistor, was calculated and recorded in Table 2, as well. To demonstrate the validity of the Kirchhoff’s Rules, we used the junction and the loop rules for the specific junctions and loops pictured in both Diagram 1 and Diagram 2, respectively. For Diagram 1, where the switch was closed, the loop rule (potential difference) for the first loop was calculated as follows R1 – R2 – R3 = 2.00 – 0.24 – 1.76 = 0.00 V. The percent error difference was calculated to be 0%. The second loop was calculated as follows (R4 + R3 + Rdecade) – Power Supply = (0.25 + 1.76 + 2.98) – 4.98 = 0.01 V. The percent error difference was calculated to be 1%. The third loop was calculated as follows Power Supply – Rdecade – R1= 4.98 – 2.98 – 2.00 = 0.00 V. The percent error difference was calculated to be 0%. The junction rule involves the currents calculated for each resistor. The first junction was calculated as follows I3 = I2 + I4  35.3 mA = 22.9 mA + 11.4 mA  35.3 mA ≈ 34.3 mA. The percent error difference was calculated to be 2.9%. The second junction was calculated as follows ID = I1 + I3  55.8 mA = 20.5 mA + 35.3 mA  55.8 mA = 55.8 mA. The percent error difference was calculated to be 0%. For Diagram 2, where the switch was open, the loop rule (potential difference) for the first loop was calculated as follows R2 + R1 – R3 = 0.15 + 1.41 – 1.56 = 0.00 V. The percent error difference was calculated to be 0%. The second loop was calculated as follows (R4 + R3 + Rdecade) – Power Supply = (0.99 + 1.56 + 2.42) – 4.98 = -0.01 V. The percent error difference was calculated to be 1%. The third loop was calculated as follows Power Supply – (Rdecade + R1 + R2 + R4) = 4.98 – (2.42 + 1.41 + 0.15 + 0.99) = 0.01 V. The percent error difference was calculated to be 1%. The first junction was calculated as follows I4 = I2 + I3  45.0 mA = 14.3 mA + 31.3 mA  45.0 mA ≈ 45.6 mA. The percent error difference was calculated to be 1.3%. The second junction was calculated as follows ID = I1 + I3  45.3 mA = 14.5 mA + 31.3 mA  45.3 mA ≈ 45.8 mA. The percent error difference was calculated to be 1.1%. From the above calculations we can see that the sums of potential differences around several closed loops was calculated to be either zero or nearly zero, thus conforming to the second Kirchhoff’s rule. The sums of all outgoing currents was either equal or nearly equal to the sum of all entering currents at a specific junction, thus conforming to the first Kirchhoff’s rule. Conclusion The results obtained in this laboratory experiment were in agreement with Kirchhoff’s Rules. If we were to take the percent error difference threshold at 5%, we can conclude that all out results were statistically valid. The highest percent error difference was calculated to be 2.9%, which is well below the established threshold. The lowest percent error difference was calculated to be 0%, which means that in some cases our results were completely in agreement with Kirchhoff’s Rules and there was no deviation. Altogether, such results mean that Kirchhoff’s Rules is a good tool in analyzing the more complex circuits, where often the direction and magnitude of the current is hard to establish when a limited amount of information is present. Table 1 Switch R (ohms) V (volts) I (milliamps) R1 97.5 2.00 20.5 R2 10.5 0.24 22.9 R3 49.9 1.76 35.3 R4 22.0 0.25 11.4 Rdecade 53.4 2.98 55.8 R (ohms) V (volts) I (milliamps) R1 97.5 1.41 14.5 R2 10.5 0.15 14.3 R3 49.9 1.56 31.3 R4 22.0 0.99 45.0 Rdecade 53.4 2.42 45.3 Closed Table 2 Switch Open Answers to the questions: 1.) The measurements agreed with the Kirchhoff’s Rules. Slight experimental differences could be attributed to the internal resistances of the measuring device, as well as the power supply. 2.) The measured and calculated values were very closely in agreement with each other, as the highest percent error difference was calculated to be 2.9%, which is very low, considering a threshold of 5%. Again, discrepancies can be attributed to any unaccounted internal resistances of the conducting materials, measuring device, and the power supply. 3.) The results would be affected, as the effective voltage would drop, causing the currents to drop sufficiently. Thus if the internal resistance would not be accounted for, the currents calculated using the value of the power output would be sufficiently higher than they really would be. 4.) The power dissipated in a resistor is calculated using the equation P = IV, where I is in amperes and V is in volts. Tables below provide the calculated results: Switch Closed P (watts) Switch Open P (watts) R1 0.041 R1 0.020 R2 0.005 R2 0.002 R3 0.062 R3 0.049 R4 0.003 R4 0.045 Rdecade 0.166 Rdecade 0.110 5.) In both cases Rdecade dissipates the most power. This is due to the fact that Rdecade has a relatively high resistance, but what is more important is that it has the highest amount of current going through it, as it is positioned right next to the power source, and current is not being split anywhere before the decade box. From equation P = IV we see that power is the product of resistance and current, thus the higher both of these values, the more power is being dissipated across that specific resistor.