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]V_A [/math]	20 Volts
[math]R_1[/math]	902 [math]\Omega[/math]
[math]R_2[/math]	10.2 [math]\Omega[/math]
[math]R_3[/math]	10.6 [math]\Omega[/math]

Enter the measured and predicted quantities in the table below

Given [math]V_A[/math] and the values of all resistors, use Kirchoff's laws to predict

a.) Predict the value of [math]V_B[/math]

[math]V_A - I_1R_1 = V_b [/math]

I_1 =?

Kirchoff's Loop theorem (Voltage Law)

[math]V_A -I_1(R_1+R_{tot}) = 0[/math] where R[math]_{tot}[/math] =resistance for [math]R_2[/math] and [math]R_3[/math] in series.

[math]R_{tot} = \frac{1}{\frac{1}{R_2} + \frac{1}{R_3}} = 5.2 \Omega[/math]

[math]I_1 = \frac{V_A}{R_1+R_{tot}} = \frac{ 20 \mbox{Volts}}{902 \Omega + 5.2 \Omega} = 22 \mbox{mA}[/math]

[math]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]I_1 = I_2 + I_3[/math]

I[math]_2 R_2 - I_3 R_3 = 0[/math]

2 equations and 2 unkowns

[math]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]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]V_B[/math]	103.5 +/- 50 mV	156	50%
[math]I_1[/math]	20.4 mA	22	7%
[math]I_2[/math]	9.5 mA	11.2	18%
[math]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]R_L[/math] shown in the circuit below. Begin with [math]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]\Omega[/math])	V (Volts)	I (mA)
10	1.42 [math]\pm[/math] 0.06	118 [math]\pm[/math] 3
93	1.55[math] \pm[/math] 0.01	16.15 [math]\pm[/math] 0.05
990	1.575 [math]\pm[/math] 0.005	1.54 [math]\pm[/math] 0.04
10110	1.579 [math]\pm[/math] 0.005	0.146 [math]\pm[/math] 0.005
15060	1.579 [math]\pm[/math] 0.005	0.101 [math]\pm[/math] 0.003

[math]V_A - V_B = IR_{Load} = V_{bb} - I R_{int}[/math]

[math](V_A - V_B) = - R_{int} I + V_{bb}[/math]

[math]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]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]\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)

1. What conservation law is involved in Kirchoff's Loop Theorem?
2. What does the slope in the internal resistance plot above represent?

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