Lab Report
Abstract:
In this lab, we used our existing knowledge of electrical circuits and laws, to analyze and practice voltage division and current division theories. For the first part, we set up the voltage divider circuit as from the pre-lab shown in Figure 1, and measured all the voltages, and currents for every resistor.
Figure 1: Voltage division circuit
We then analyzed the circuit from the point of view of Kirchhoff’s voltage law (KVL), and calculated the error between the measured and predicted values. According to KVL, the sum of voltage drops across each of the resistors must equal the supply voltage of 15 V; and according to voltage division theorem, the voltage drop across any resistor is given by:
Vn=Vs×(RnR1+R2++Rn)
Where Vnis the voltage across the nth resistor; Vsis the supply voltage; and R1, R2,,and Rn are the resistance values of n resistors in a circuit.
In the second part of the lab, we set up the current divider circuit from pre-lab as shown in Figure 2, and again measured all the voltages and currents.
Figure 2: Current division circuit
We then removed the load resistor R3 and measured the voltage and current for R2. We then replaced R3 with a 100 kΩ resistor and then by a 1MΩ resistor, and again measured the voltage and current for R2 for each case. The purpose of this exercise was to study the effect of load on the circuit. For calculation, the current division theorem according to the following equation was used:
I2=Is×(R3R2+R3)
Where Inis the current through the 1st resistor; Isis the supply current and equals VsRe; and R1, R2, R3 are the resistance values of resistors in the circuit, Reis the equivalent resistance.
Data Table:
Voltage Division:
Current Division:
Sample Calculations – Predicted voltage based on voltage division:
Vn=Vs×(RnR1+R2++Rn)
V1=15×1010+20+30=2.5V
Predicted current based on current division:
I2=Is×(R3R2+R3)
I2=0.68×3050= 0.408 mA
Percent error (%)=|Imeasured-Ipredicted|Imeasured×100
Percent error %= |0.409-0.408|0.409 ×100=0.2%
Questions:
1) According to KVL, V1 + V2 + V3 = V4
2) The larger the load resistance, lesser the current through it, therefore more is the current through resistor R2. Also, the supply current decreases with increase in load resistance.
Error/Conclusion: The calculations show that the predicted and measured values of currents and voltages are almost error free. However, as explained in Question 1, there is a 0.04 V difference in the total voltages. This can be attributed to the voltage drops across the connecting wires. Though they are assumed to be ideal, they possess a non-zero resistance. When a current of 0.25 mA passes through the wires (series), there will definitely be non-zero voltage drops across them. Also, load analysis shows that more the resistance of the load, lesser is the current drawn from the supply.
