EE11L Experiment 1

Experiment 1

Introductory Laboratory and Kirchoff’s Laws

Max Woo
10/27/14
Lab 1D, Professor Babaie

Objectives

The purpose of this experiment is to become familiar with circuit construction and the myDAQ measurement instrument. We will also be verifying equivalent resistance and Kirchoff’s Laws. Lastly, we will learn and apply the concepts behind the Wheatstone Bridge.

Theory

The Theory section should include statements and explanation of concepts, equations, and formulas which are necessary for analyzing the experimental situation.

For part 1, we will be verifying Ohm’s law, given in Eq. 1, which states the relationship between the voltage, current, and resistance of a circuit.

Ohm’s Law: $V = IR$ (Eq. 1)

For the rest of the parts, we will combine Ohm’s law with Kirchhoff’s voltage and current laws to analyze resistor networks. Kirchhoff’s voltage law states that, for any given current through a loop, the voltage changes between each component in the loop will sum to 0. Kirchhoff’s current law states that, for any given node in a circuit, if we denote the currents going into the node as positive and currents leave the node as negative, the sum of the currents will equal 0.

To aid us in parts 2 through 5, we can derive the expressions for the equivalent resistance of resistors in parallel and series. We start with 2 resistors in series connected to a voltage source, as shown in Fig. 2

Fig. 2: 2 resistors in series

Using Ohm’s law and Kirchhoff’s voltage law, we know

$V_0 - R_1 I_0 - R_2 I_0 = 0$ (Eq. 2)

Solving this for $I_0$ gives

$I_0 = \frac{V_0}{R_1 + R_2}$ (Eq. 3)

Using Ohm’s law once again, we can calculate the total resistance of both resistors:

$R_{total} = \frac{V_0}{I_0} = R_1 + R_2$ (Eq. 4)

We can do the same for a parallel circuit, depicted in Fig. 3:

Fig. 3: 2 resistors in parallel

We define current $I_1$ and $I_2$ as the current going through $R_1$ and $R_2$ respectively. Once again using Ohm’s law and Kirchhoff’s voltage law, we get

$V_0 - R_1 I_1 = 0$ (Eq. 5)
$V_0 - R_2 I_2 = 0$ (Eq. 6)

Solving for $I_1$ and $I_2$ gives

$I_1 = \frac{V_0}{R_1}$ (Eq. 7)
$I_2 = \frac{V_0}{R_2}$ (Eq. 8)

From Kirchhoff’s current law, we know $I_0 = I_1 + I_2$. Combining this with Eq. 7 and 8 gives

$I_0 = \frac{V_0}{R_1} + \frac{V_0}{R_2}$ (Eq. 9)

Using Ohm’s law one last time to calculate the total resistance:

$R_{total} = \frac{V_0}{I_0} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$ (Eq. 10)

For part 5, we need to calculate the output voltage of a Wheatstone bridge, shown in Fig. 4, in terms of the 4 resistors and the voltage source.

Fig. 4: A Wheatstone bridge circuit

To do so, we first label the voltage source as $V$, and then denote the voltage at the positive terminal of the voltage output as $v_1$ and the negative terminal as $v_2$. Using Kirchhoff’s current law, we know that the currents going into the R1-R2 and R3-R4 junctions total zero, so using Ohm’s law to get the currents in terms of voltages and resistances gives us:

$\frac{v_1-V}{R_1} + \frac{v_1}{R_3} = 0$
$\frac{v_2-V}{R_2} + \frac{v_2}{R_4} = 0$

Solving this gives us:

$v_1 = \frac{VR_3}{R_1+R_3}$ (Eq. 11)
$v_2 = \frac{VR_4}{R_2+R_4}$ (Eq. 12)

Part I

Procedure

For this part, we build the circuit shown in Fig. 5, using the myDAQ for the 5V power supply, and changing between 3 different resistors (100 Ohm, 470 Ohm, and 1000 Ohm). We measured the voltage, current, and resistance across each resistor we tested.

Fig 5: Circuit diagram for part 1

Data

Data Analysis

To calculate the theoretical resistances given by Ohm’s law, we used the voltage and current for each circuit with Eq. 1 (Reprinted below).

$V = IR$ (Eq. 1)

The calculated theoretical values, along with their associated experimental results and error values, are shown below:

Discussion

The actual resistance values of the resistors did deviate from their marked values by a bit, but still within the given variance of each resistor.
The results are enough to verify Ohm’s law because, in all 3 circuits, the error values for the experimental results were very small, and could have just been the result of resistances in the wire, multimeter inaccuracy, etc.
Although our power supply may have had internal resistance, causing the actual voltage output of the power supply to deviate from 5V and to change for each circuit, it still would not effect our experiment because we measured the actual voltage drop across the resistor for each circuit.

Part II

Procedure

First we build the circuit shown in Fig. 6, and measure the resistance across A and B, comparing the value to the calculated theoretical resistance.

FIg. 6: Circuit diagram for part 2

Then, we take a resistor with a resistance of over $1 M\Omega$, and measure the resistance first by connecting the multimeter as usual, and then measure the resistance when we grip the multimeter leads and resistor leads together.

Data

Table 2: The measured resistance of individual resistor and the total resistance of the network (measured from A to B as depicted in Fig 6)

Table 3: The resistance of the large resistor when measured normally and when measured by gripping the multimeter leads to the resistor leads

Data Analysis

To start, we label the 470$\Omega$, 1000$\Omega$, and 680$\Omega$ resistors as $R_1$,$R_2$, and $R_3$, respectively. We can use the equivalent resistance of series and parallel resistors derived earlier to find the total resistance of this circuit. From Eq. 10, we know that the total resistance of $R_1$ and $R_2$, denoted by $R_{1+2}$, is

$R_{1+2} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$ (Eq. 13)

We can also see that $R_{1+2}$ and $R_3$ are in series, so using Eq. 4, we know the total resistance of the whole circuit is:

$R_{total} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} + R_3$ (Eq. 14)

Using this, we were able to calculate the theoretical values for the circuit, shown below alongside the experimental and error values:

Table 4: The theoretical, experimental, and error values for the total resistance of the circuit shown in Fig 6, measured from A to B.

For calculating the skin resistance, we know that the skin acts a resistor in parallel to the large resistor. We denote the large resistor as $R_{large}$ and the skin as $R_{skin}$. Using Eq. 10, we can once again find the total resistance to be:

$R_{total} = \frac{1}{\frac{1}{R_{large}} + \frac{1}{R_{skin}}}$ (Eq. 15)

Using the measured values for $R_{total}$ and $R_{large}$, we calculate $R_{skin}$ to be $2.390 M\Omega$.

Discussion

Using the skin resistance of $2.390 M\Omega$ calculated earlier, we can use Ohm’s law to find how much current is needed to produce 50V. We calculate this value to be $I = 0.021 mA$.

Part III

Procedure

Construct a voltage divider following the circuit diagram in Fig. 7, with R1 = 22$\Omega$ and R2 = 100$\Omega$, and measure the voltage across each resistor.

Fig. 7: Voltage divider circuit

Then construct a current divider as shown in Fig. 7, with R1 = 22$\Omega$, R2 = 100$\Omega$, and R3 = 470$\Omega$, and measure the current across each resistor.

Fig. 8: Current divider circuit

Then, build a circuit that involves a photoresistor, so that in normal room light, the voltage output is about 50% of the 5V input, and in darkness the voltage output is about 0. To do this, we build a simple voltage divider as shown in Fig. 9:

Fig. 9: Voltage divider with photoresistor

The resistor $R$ in Fig. 9 represents a resistor with the same resistance as the photoresistor during normal light. We let the output voltage be the voltage across the normal resistor. This way, during normal room light the resistor and photoresistor have the same resistance, so the voltage across the resistor will be half the total voltage. However, in darkness the photoresistor will have a much higher resistance, so the voltage across the photoresistor will be much larger, reducing the voltage across the resistor.

Data

Table 5: Measure voltages across the voltage divider

Table 6: Measured currents across the current divider

Table 7: Measured resistances and voltages of the photoresistor circuit in light and darkness. Note that in darkness, the resistance of the photoresistor exceeded the maximum measurable resistance of the myDAQ, hence the value of “> 200 M” in the table.

Data Analysis

The voltage divider in Fig. 7 looks exactly like the resistors-in-series circuit we analyzed in the Theory section. Thus, the current is given by Eq. 3, reprinted below:

$I_0 = \frac{V_0}{R_1 + R_2}$ (Eq. 3)

We know from Ohm’s law that the voltage along each resistor is the current multiplied by the resistance, so using the values measured in the lab we can calculate the theoretical voltages across each resistor, shown below along with the measured values and error values:

Table 8: Theoretical and measured voltages for the voltage divider

For the current divider, we first start by finding the current of the whole circuit. To do so, we first calculate the total resistance by noticing the similarity between the current divider and the circuit in Fig. 6. Thus, we can infer that the total resistance is

$R_{total} = \frac{1}{\frac{1}{R_2} + \frac{1}{R_3}} + R_1$ (Eq. 16)

We can then find the total current $I_{total} = \frac{V_{total}}{R_{total}}$, which also happens to be the current through R1. To calculate the current through R2 and R3, we first find the voltage drop across R1 using Ohm’s law:

$V_1 = I_1 R_1 = I_{total} R_1$ (Eq. 15)

We know from Kirchhoff’s voltage law that the voltage across R2 and R3 is the total voltage minus the voltage across R1

$V_2 = V_3 = V_{total} - V_1$ (Eq. 17)

Finally, we can calculate the current through R2 and R3 by dividing the voltage by the resistance of each. Thus, following this process using the values given in the lab, we calculate the theoretical values for the currents, shown alongside the measured values and the errors:

Table 9: The theoretical and experimental values for the currents through each resistor

Discussion

The measured voltage and currents through the voltage divider and current divider fell within the error bounds of the theoretical values for each, verifying the validity of each circuit.
The resistance of the photoresistor changed drastically between light and darkness, from a resistance of about 4$K\Omega$ in room light to an unmeasurable resistance in darkness, meaning a resistance of over 200$M\Omega$.

Part IV

Procedure

Build the circuit in Fig. 10 using resistor values R1 = 100$\Omega$, R2 = 470$\Omega$, R3 = 1000$\Omega$, R4 = 680$\Omega$, R5 = 2200$\Omega$, and R6 = 100$\Omega$. Measure the voltage and current through each resistor.

Fig. 10: A resistor network for Part 4

Data

Table 10: Measured voltages and currents from circuit in Fig. 10

Data Analysis

To start analyzing the circuit, we first identify 3 current loops, as shown below:

Fig. 11: circuit diagram for Part 4 with current loops

Using Kirchhoff’s voltage law, we find the equations for each of these loops to be

$5 - I_1 R_1 - (I_1 - I_3) R_2 - (I_1 - I_2) R_4 = 0$
$- (I_2 - I_1) R_4 - (I_2 - I_3) R_5 - I_2 R_6 = 0$
$- I_3 R_3 - (I_3 - I_2) R_5 - (I_3 - I_1) R_2 = 0$

Rearranging the equations to separate $I_1$, $I_2$, and $I_3$ gives:

$I_1 (R_1 + R_2 + R_4) - I_2 R_4 - I_3 R_2 = 5$
$I_1 R_4 - I_2 (R_4 + R_5 + R_6) + I_3 R_5 = 0$
$I_1 R_2 + I_2 R_5 - I_3 (R_3 + R_5 + R_2) = 0$

Solving this using a matrix solver gives us $I_1 = 7.9 mA$, $I_2 = 4.6 mA$, and $I_3 = 3.8 mA$. We can now look at the circuit, and we can determine what combination of currents gives the total current through each resistor:

Table 11: Shows how to calculate the theoretical current through each resistor using $I_1$, $I_2$, and $I_3$

And after determining the current through each resistor, we can find the theoretical voltage using Ohm’s law. All the results, theoretical and experimental, are shown below:

Table 11: The theoretical and measured values for current and voltage across each resistor

Discussion

The results of our experiment fell within the error bounds of our theoretical results, showing that our circuit obeyed Kirchoff’s laws.

Part V

Procedure

The goal is to bridge a Wheatstone bridge, as shown in Fig. 12, with a thermistor as R1 so that the output voltage is ~0V at room temperature and ~0.4V at body temperature.

Fig. 12: A Wheatstone bridge circuit

From the Theory section, we have the equations for the voltages at both terminals of the voltage output, restated below:

$v_1 = \frac{VR_3}{R_1+R_3}$ (Eq. 11)
$v_2 = \frac{VR_4}{R_2+R_4}$ (Eq. 12)

We can rearrange these equations to get:

$v_1 = \frac{1}{1+\frac{R_1}{R_3}}$ (Eq. 18)
$v_2 = \frac{1}{1+\frac{R_2}{R_4}}$ (Eq. 19)

These equations show that $v_1$ and $v_2$ are solely dependent on the ratio between R1-R3 and R2-R4, respectively. Thus, to make the voltage output zero, where $v_1 = v_2$, the ratio between R1 and R2 is equal to the ratio between R3 and R4. In addition, changing the temperature, and thus changing R1 and the R1-R3 ratio, will in turn change $v_1$ and cause a voltage difference between $v_1$ and $v_2$.
We measure the thermistor to have ~30$K\Omega$ in room temperature, so using a 20$K\Omega$ resistor for R3 dictates that our ratio for R2-R4 must be three-to-two. Thus, we choose R2 = 300$K\Omega$ and R4 = 200$K\Omega$, and measure the voltage with the thermistor at room temperature and body temperature.

Data

Table 13: The measured resistances and voltage output at room and body temperatures

Discussion

Using a Wheatstone bridge and a thermistor, we were successfully able to make a circuit that have ~0V output at room temperature and ~0.4V output at body temperature.
In comparison to the light sensing circuit made earlier, the temperature sensing circuit was much less susceptible to voltage changes, changing only 0.4V when the temperature increased from room to body temperature. This makes the temperature sensing circuit less prone to voltage spikes and large fluctuations, characteristics that are not desirable.

Conclusion

In all the of the parts of this experiment, our experimental data fell within the error bounds of our theoretical data. This verified the validity of both Ohm’s law, and Kirchhoff’s voltage and current laws. In addition, we were able to achieve our objectives for our light sensitive and temperature sensitive circuits.