Physics 6B Lab3

Physics 6B Lab3 - 20Vx(2.6kohms/4.3kohms)=12.09V Ohms law...

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Physics 6B Lab3 1.1) Power supply (V) Voltmeter Reading (V) Current Flow (mA) 2 1.99 0.91 4 4.05 1.86 6 5.98 2.75 8 8.03 3.69 10 10.04 4.62 12 11.95 5.5 14 14.03 6.46 16 15.97 7.36 18 18.01 8.31 20 19.34 8.93 The plot is linear. The entire plot is linear. 1.2)The slope is 0.46 mA/V 1.3) The reciprocal is 2.17k ohms 1.4) 10.2K 10.2x1000=10200 ohms 2mA 0.002A 10200(0.002)=20.4V 1.5) 9V=IR =I(10200) I=8.82x10^-4A I=0.002A 2.1) Power supply (V) Voltage Across Resistors (V) Current Flow (mA) 2 1.98 0.45 4 3.99 0.91 6 5.97 1.37 8 8.01 1.84 10 9.18 2.29 12 12.02 2.67 14 14 3.21 16 15.99 3.67
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18 18 4.14 20 19.38 4.48 The slope is 0.24 V/A. The Recipricol of the slope is 4.17 k ohms 2.2) The slope was much lower in this reading. And the ohms reading was much higher. Therefore, Resistors lower V/A but increase ohms. 2.3) Source: 20V 2 resistance: 1.7k, 2.6k V=IR R= 1.7k+2.6k=4.3k ohms 20V = I(4300) Total current: I=0.00465 A 20Vx(1.7kohms/4.3kohms)=7.9V
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Unformatted text preview: 20Vx(2.6kohms/4.3kohms)=12.09V Ohms law holds because the two individual voltages add up to the total circuit current. 3.1 Voltage (V) Current (mA) 9.98 4.59 V=IR 9.98=(0.00459) R R=2174 ohms 3.2) new current = 9.1 mA (0.0091 A) 9.98V=(0.0091A)R R=1096.7 ohms The current increased. 3.3) As resistance decreases, current flow increases. 3.4)Current again increased. This time, to 13.73 mA 3.5) Again, as resistance decreases, and source voltage stays the same, then current flow will increase. 3.6) 1/2.2k+1/3.1k+1/6.7k=R R=1.08kohms 3.7) 18Vx2.2kohms/(2.2kohms+1.7kohms+2.6kohms)=6.09V 18Vx1.7kohms/(2.2kohms+1.7kohms+2.6kohms)=4.71V 18Vx6.7kohms/(2.2kohms+1.7kohms+2.6kohms)=18.6V I(2200ohms)=18V I=0.0082A I(1700ohms)=18V I=0.0106A I(2600ohms)=18V I=0.0069A...
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Physics 6B Lab3 - 20Vx(2.6kohms/4.3kohms)=12.09V Ohms law...

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