TF EIMLab1 Writeup
Kirchoff's Law (50 pnts)
Construct the circuit below
Enter the values of the DC voltage and Resisters that you used.
Use a voltmeter to measure the potential difference and resistances.
| Variable | Measured Value |
| [math]\displaystyle{ V_A }[/math] | 20 Volts |
| [math]\displaystyle{ R_1 }[/math] | 902 [math]\displaystyle{ \Omega }[/math] |
| [math]\displaystyle{ R_2 }[/math] | 10.2 [math]\displaystyle{ \Omega }[/math] |
| [math]\displaystyle{ R_3 }[/math] | 10.6 [math]\displaystyle{ \Omega }[/math] |
Enter the measured and predicted quantities in the table below
Given [math]\displaystyle{ V_A }[/math] and the values of all resistors, use Kirchoff's laws to predict
a.) Predict the value of [math]\displaystyle{ V_B }[/math]
[math]\displaystyle{ V_A - I_1R_1 = V_b }[/math]
I_1 =?
Kirchoff's Loop theorem (Voltage Law)
[math]\displaystyle{ V_A -I_1(R_1+R_{tot}) = 0 }[/math] where R[math]\displaystyle{ _{tot} }[/math] =resistance for [math]\displaystyle{ R_2 }[/math] and [math]\displaystyle{ R_3 }[/math] in series.
[math]\displaystyle{ R_{tot} = \frac{1}{\frac{1}{R_2} + \frac{1}{R_3}} = 5.2 \Omega }[/math]
[math]\displaystyle{ I_1 = \frac{V_A}{R_1+R_{tot}} = \frac{ 20 \mbox{Volts}}{902 \Omega + 5.2 \Omega} = 22 \mbox{mA} }[/math]
[math]\displaystyle{ V_b = V_A - I_1 R_1 = 20 - 0.022 \times 902 = 156 }[/math] mV
b.) Predict the values of the three currents.
[math]\displaystyle{ I_1 = I_2 + I_3 }[/math]
I[math]\displaystyle{ _2 R_2 - I_3 R_3 = 0 }[/math]
2 equations and 2 unkowns
[math]\displaystyle{ I_3 = \frac{I_1}{\frac{R_3}{R_2} +1} = \frac{22 mV}{\frac{10.6}{10.2}+1} = 10.8 }[/math] mA
[math]\displaystyle{ I_2 = I_1 - I_3 = 22 - 10.8 = 11.2 }[/math] mA
c.) compare your predictions and measurements by filling in the table below.
| Variable | Measured Value | Predicted Value | % Difference |
| [math]\displaystyle{ V_B }[/math] | 103.5 +/- 50 mV | 156 | 50% |
| [math]\displaystyle{ I_1 }[/math] | 20.4 mA | 22 | 7% |
| [math]\displaystyle{ I_2 }[/math] | 9.5 mA | 11.2 | 18% |
| [math]\displaystyle{ I_3 }[/math] | 9.0 mA | 10.8 | 20% |
Internal resistance (30 pnts)
Measure the internal resistance of your power source by graphing the potential difference on the x-axis and the current on the y-axis for several values of the resistance [math]\displaystyle{ R_L }[/math] shown in the circuit below. Begin with [math]\displaystyle{ R_L = 1k\Omega }[/math] and then decrease it by a factor of 5 for each subsequent measurement. You can use a volt meter to measure the current and potential difference.
| R_{Load} ([math]\displaystyle{ \Omega }[/math]) | V (Volts) | I (mA) |
| 10 | 1.42 [math]\displaystyle{ \pm }[/math] 0.06 | 118 [math]\displaystyle{ \pm }[/math] 3 |
| 93 | 1.55[math]\displaystyle{ \pm }[/math] 0.01 | 16.15 [math]\displaystyle{ \pm }[/math] 0.05 |
| 990 | 1.575 [math]\displaystyle{ \pm }[/math] 0.005 | 1.54 [math]\displaystyle{ \pm }[/math] 0.04 |
| 10110 | 1.579 [math]\displaystyle{ \pm }[/math] 0.005 | 0.146 [math]\displaystyle{ \pm }[/math] 0.005 |
| 15060 | 1.579 [math]\displaystyle{ \pm }[/math] 0.005 | 0.101 [math]\displaystyle{ \pm }[/math] 0.003 |
[math]\displaystyle{ V_A - V_B = IR_{Load} = V_{bb} - I R_{int} }[/math]
[math]\displaystyle{ (V_A - V_B) = - R_{int} I + V_{bb} }[/math]
[math]\displaystyle{ y = mx+b \Rightarrow R_{int} = }[/math] slope of V-vs- I plot
- Since the current is in mA the resistance must be in kOhms in order to get volts.
- [math]\displaystyle{ R_{int} = 1.33 \pm 0.04 \Omega }[/math]
Book suggests 1.5 flashlight battery could have an internal resistance of 0.5 Ohms.
Below I set R= 95 [math]\displaystyle{ \Omega }[/math] and changed V
| V (mv) | I (mA) |
| 126.1 | 0.79 |
| 300 | 2.64 |
| 500 | 4.62 |
| 1000 | 9.29 |
| 2000 | 18.78 |
| 3000 | 30.4 |
Now lets try to fix V and change R
Questions (20 pnts)
- What conservation law is involved in Kirchoff's Loop Theorem?
- What does the slope in the internal resistance plot above represent?