You measured the resistance values and using an ohmmeter directly, and you also determined the values of and from Ohm's law in the two circuit configurations. In addition, the resistance was given from the resistor color-code. Which of these give the most accurate resistance value? Briefly explain your answer. Examine the power use determined for the individual resistors , and the combined resistances for both series and parallel configuration. Based on your observations, write a general equation that describes the "equivalent power" for each configuration. For the series circuit, based on your results, write an equation relating the voltage from the power source to the voltage across each component resistor. For the parallel circuit, based on your results, write an equation relating the current flow from the power source to the currents flow in each component resistor.

Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter27: Current And Resistance
Section: Chapter Questions
Problem 27.5CQ: How does the resistance for copper and for silicon change with temperature? Why are the behaviors of...
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  1. You measured the resistance values and using an ohmmeter directly, and you also determined the values of and from Ohm's law in the two circuit configurations. In addition, the resistance was given from the resistor color-code. Which of these give the most accurate resistance value? Briefly explain your answer.
  2. Examine the power use determined for the individual resistors , and the combined resistances for both series and parallel configuration. Based on your observations, write a general equation that describes the "equivalent power" for each configuration.
  3. For the series circuit, based on your results, write an equation relating the voltage from the power source to the voltage across each component resistor.
  4. For the parallel circuit, based on your results, write an equation relating the current flow from the power source to the currents flow in each component resistor.                                                                               
Purpose:
Color Code
Ohmmeter
unit
R₁
R₂
Rs
unit
R₁
R₂
Rr
Table 1: Resistance Determinations
R₁
x 10%
1.048
10.0210
V
volt
6.79
3.03
10.0
V
SR₁
x 10%
volt
9.40
9.40
1.000
10.1000
Table 2: Resistors in Series
I
Amp
0.592 x 10-3
0.592 x 10-3
0.592 x 10-3
Table 3: Resistors in Parallel
I
Amp
0.840 x 10-3
1.730 x 10-3
2.620 x 10-3
R₂
x 10³
5.270
10.105
R
52
1.147 x 10*
5.118 x 10³
1.689 x 10+
R
52
1.119 x 10*
5.434 x 10³
3.591 x 10*
SR₂
x 10³
4.700
+0.470
P
4.097 x 10-3
1.794 x 10-3
5.920 x 10-³
P
W
7.896 x 10-³
1.626 x 10-²
2.465 x 10-²
1. 8R₁(Color Code) = Brown, black, orange, silver 10,000(10%)
5R₁(Ohmmeter) = silver - (10 %) = 10,000 x
10
100±1000
2. Calculate R:
V
6.79volt
R₁T 0.592 x 10-³ Amp
R₂
V
V
R₁ ===
3.03 volt
0.592 x 10 Amp
10. Ovolt
T0.592 x 10 Amp
- 1.147 x 10
Calculate P:
3. Calculate R:
R₁ = P = VxI= (6.79volt) (0.592 x 10-³ Amp) = 4.097 x 10-³W
R₂ = P =V x1 = (3.03 volt) (0.592 x 10-³Amp) = 1.794 x 10-³W
Rs = P = VxI= (10.0 volt)(0.592 x 10-³ Amp) = 5.920 x 10-³W
V
9.40volt
R₁-0.840 x 10-³ Amp
9.40volt
R₂1.730 x 10 Amp
V
5.118 x 10³
V
9.41 volt
Rp T 2.620 x 10-³ Amp
= 1.689 x 10*
= 1.119x10*0
= 5.434 x 10³0
= 3.591 x 10*
Calculate P:
R₁ = P = VxI=(9.40volt)(0.840 x 10-3 Amp) = 7.896 x 10-³W
R₂ =P = VxI=(9.40volt)(1.730 x 10-3 Amp) = 1.626 x 10-²W
R₂ =P =VxI= (9.41volt) (2.620 x 10-³ Amp) = 2.465 x 10-²W
Transcribed Image Text:Purpose: Color Code Ohmmeter unit R₁ R₂ Rs unit R₁ R₂ Rr Table 1: Resistance Determinations R₁ x 10% 1.048 10.0210 V volt 6.79 3.03 10.0 V SR₁ x 10% volt 9.40 9.40 1.000 10.1000 Table 2: Resistors in Series I Amp 0.592 x 10-3 0.592 x 10-3 0.592 x 10-3 Table 3: Resistors in Parallel I Amp 0.840 x 10-3 1.730 x 10-3 2.620 x 10-3 R₂ x 10³ 5.270 10.105 R 52 1.147 x 10* 5.118 x 10³ 1.689 x 10+ R 52 1.119 x 10* 5.434 x 10³ 3.591 x 10* SR₂ x 10³ 4.700 +0.470 P 4.097 x 10-3 1.794 x 10-3 5.920 x 10-³ P W 7.896 x 10-³ 1.626 x 10-² 2.465 x 10-² 1. 8R₁(Color Code) = Brown, black, orange, silver 10,000(10%) 5R₁(Ohmmeter) = silver - (10 %) = 10,000 x 10 100±1000 2. Calculate R: V 6.79volt R₁T 0.592 x 10-³ Amp R₂ V V R₁ === 3.03 volt 0.592 x 10 Amp 10. Ovolt T0.592 x 10 Amp - 1.147 x 10 Calculate P: 3. Calculate R: R₁ = P = VxI= (6.79volt) (0.592 x 10-³ Amp) = 4.097 x 10-³W R₂ = P =V x1 = (3.03 volt) (0.592 x 10-³Amp) = 1.794 x 10-³W Rs = P = VxI= (10.0 volt)(0.592 x 10-³ Amp) = 5.920 x 10-³W V 9.40volt R₁-0.840 x 10-³ Amp 9.40volt R₂1.730 x 10 Amp V 5.118 x 10³ V 9.41 volt Rp T 2.620 x 10-³ Amp = 1.689 x 10* = 1.119x10*0 = 5.434 x 10³0 = 3.591 x 10* Calculate P: R₁ = P = VxI=(9.40volt)(0.840 x 10-3 Amp) = 7.896 x 10-³W R₂ =P = VxI=(9.40volt)(1.730 x 10-3 Amp) = 1.626 x 10-²W R₂ =P =VxI= (9.41volt) (2.620 x 10-³ Amp) = 2.465 x 10-²W
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