Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
Question
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Chapter 27, Problem 27.80AP

(a)

To determine

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K,300K,and320K .

(a)

Expert Solution
Check Mark

Answer to Problem 27.80AP

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.015932 25.1067
0.405 0.019602 20.66116
0.41 0.024117 17.00046
0.415 0.029673 13.98578
0.42 0.036508 11.50433
0.425 0.044918 9.461686
0.43 0.055264 7.780834
0.435 0.067995 6.397529
0.44 0.083657 5.259572
0.445 0.102927 4.323453
0.45 0.126637 3.553464
0.455 0.155807 2.92028
0.46 0.191697 2.39962
0.465 0.235855 1.97155
0.47 0.290184 1.619662
0.475 0.357027 1.330432
0.48 0.439268 1.092727
0.485 0.540454 0.897394
0.495 0.818117 0.605048
0.5 1.006569 0.496737
0.505 1.238432 0.407774
0.51 1.523704 0.334711
0.515 1.874688 0.274712
0.52 2.306521 0.225448
0.525 2.837827 0.185001
0.53 3.491518 0.151796
0.535 4.295787 0.124541
0.54 5.285319 0.10217
0.545 6.502788 0.08381
0.55 8.000701 0.068744
0.555 9.843657 0.056381
0.56 12.11114 0.046238
0.565 14.90093 0.037917
0.57 18.33335 0.031091
0.575 22.55642 0.025492
0.58 27.75228 0.020899
0.585 34.145 0.017133
0.59 42.01028 0.014044
0.595 51.68732 0.011512
0.6 63.59346 0.009435

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.005274 75.84672
0.405 0.0064 63.28565
0.41 0.007766 52.79679
0.415 0.009423 44.03979
0.42 0.011435 36.72991
0.425 0.013876 30.62901
0.43 0.016838 25.53795
0.435 0.020432 21.29022
0.44 0.024793 17.74668
0.445 0.030086 14.79101
0.45 0.036508 12.32605
0.455 0.044301 10.27061
0.46 0.053758 8.556892
0.465 0.065233 7.128278
0.47 0.079158 5.937492
0.475 0.096055 4.945067
0.48 0.11656 4.118066
0.485 0.141441 3.428998
0.495 0.20827 2.376718
0.5 0.252728 1.978408
0.505 0.306677 1.646686
0.51 0.372141 1.370449
0.515 0.451579 1.140443
0.52 0.547974 0.948949
0.525 0.664947 0.789537
0.53 0.806888 0.656844
0.535 0.979129 0.546404
0.54 1.188137 0.454493
0.545 1.44176 0.37801
0.55 1.749522 0.314372
0.555 2.122981 0.261425
0.56 2.576159 0.217378
0.565 3.126073 0.180738
0.57 3.793374 0.150262
0.575 4.603119 0.124915
0.58 5.585715 0.103836
0.585 6.778058 0.086308
0.59 8.224923 0.071733
0.595 9.98064 0.059615
0.6 12.11114 0.049541

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.002004 199.5582
0.405 0.002403 168.5349
0.41 0.002881 142.3127
0.415 0.003454 120.1526
0.42 0.004141 101.4283
0.425 0.004964 85.60991
0.43 0.005952 72.24847
0.435 0.007135 60.96416
0.44 0.008554 51.43551
0.445 0.010256 43.39059
0.45 0.012295 36.59933
0.455 0.014741 30.86719
0.46 0.017672 26.02967
0.465 0.021187 21.9477
0.47 0.0254 18.50372
0.475 0.030452 15.59839
0.48 0.036508 13.14778
0.485 0.043769 11.08098
0.495 0.062909 7.868498
0.5 0.07542 6.629515
0.505 0.09042 5.585066
0.51 0.108402 4.704703
0.515 0.129961 3.962729
0.52 0.155807 3.337456
0.525 0.186794 2.810585
0.53 0.223943 2.366674
0.535 0.26848 1.992698
0.54 0.321875 1.67767
0.545 0.385889 1.412324
0.55 0.462633 1.188846
0.555 0.554641 1.000647
0.56 0.664947 0.842173
0.565 0.79719 0.70874
0.57 0.955733 0.596401
0.575 1.145807 0.50183
0.58 1.373682 0.422223
0.585 1.646877 0.355218
0.59 1.974404 0.298824
0.595 2.367069 0.251366
0.6 2.837827 0.211429

Explanation of Solution

Given information: Th first symbol i.e. Euler’s number is e , the second symbol i.e. magnitude of electron charge is e , Boltzmann’s constant is kB , the absolute temperature is T , the value of current across a semiconductor diode temperature 0K is 1.00nA .

It is given that the expression for the current-voltage characteristic curve for a semiconductor diode as a function of temperature T is,

I=I0(eeΔVkBT1) (1)

Here,

I is the current across a semiconductor diode temperature TK .

I0 is the current across a semiconductor diode temperature 0K .

e is the first symbol i.e.Euler’s number.

e is the second symbol i.e. magnitude of electron charge.

kB is the Boltzmann’s constant.

ΔV is the voltage across the diode.

T is the absolute temperature.

Formula to calculate the resistance across the diode is,

R=ΔVI (2)

Here,

R is the resistance across the diode.

The value of magnitude of electron charge is 1.602×1019C .

The value of Boltzmann’s constant is 1.38×1023J/K .

The value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V .

From equation (1), formula to calculate the current across a semiconductor diode temperature 280K is,

I1=I0(eeΔV1kBT1) (3)

Here,

I1 is the current across a semiconductor diode temperature 280K .

ΔV1 is the initial voltage across the diode for temperature 280K .

Substitute 1.602×1019C for e , 1.38×1023J/K for kB , 0.400V for ΔV1 , 280K for T in equation (3) to find I1 ,

I1=(1.00nA×1A109nA)(e(1.602×1019C)(0.400V)(1.38×1023J/K)(280K)1)=0.15932A

Thus, the current across a semiconductor diode temperature 280K is 0.15932A .

From equation (2), formula to calculate the resistance across the diode is,

R1=ΔV1I1 (4)

Here,

R1 is the resistance across the diode.

Substitute 0.15932A for I1 , 0.400V for ΔV1 in equation (4) to find R1 ,

R1=0.400V0.15932A=25.1067Ω

Thus, the resistance across the diode is 25.1067Ω .

As the value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V , the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.015932 25.1067
0.405 0.019602 20.66116
0.41 0.024117 17.00046
0.415 0.029673 13.98578
0.42 0.036508 11.50433
0.425 0.044918 9.461686
0.43 0.055264 7.780834
0.435 0.067995 6.397529
0.44 0.083657 5.259572
0.445 0.102927 4.323453
0.45 0.126637 3.553464
0.455 0.155807 2.92028
0.46 0.191697 2.39962
0.465 0.235855 1.97155
0.47 0.290184 1.619662
0.475 0.357027 1.330432
0.48 0.439268 1.092727
0.485 0.540454 0.897394
0.495 0.818117 0.605048
0.5 1.006569 0.496737
0.505 1.238432 0.407774
0.51 1.523704 0.334711
0.515 1.874688 0.274712
0.52 2.306521 0.225448
0.525 2.837827 0.185001
0.53 3.491518 0.151796
0.535 4.295787 0.124541
0.54 5.285319 0.10217
0.545 6.502788 0.08381
0.55 8.000701 0.068744
0.555 9.843657 0.056381
0.56 12.11114 0.046238
0.565 14.90093 0.037917
0.57 18.33335 0.031091
0.575 22.55642 0.025492
0.58 27.75228 0.020899
0.585 34.145 0.017133
0.59 42.01028 0.014044
0.595 51.68732 0.011512
0.6 63.59346 0.009435

From equation (1), formula to calculate the current across a semiconductor diode temperature 300K is,

I1=I0(eeΔV1kBT1) (5)

Here,

I1 is the current across a semiconductor diode temperature 300K .

ΔV1 is the voltage across the diode for temperature 300K .

Substitute 1.602×1019C for e , 1.38×1023J/K for kB , 0.400V for ΔV1 , 300K for T in equation (5) to find I1 ,

I1=(1.00nA×1A109nA)(e(1.602×1019C)(0.400V)(1.38×1023J/K)(300K)1)=0.005274A

Thus, the current across a semiconductor diode temperature 300K is 0.005274A .

From equation (2), formula to calculate the resistance across the diode is,

R1=ΔV1I1 (6)

Here,

R1 is the resistance across the diode.

Substitute 0.005274A for I1 , 0.400V for ΔV1 in equation (6) to find R1 ,

R1=0.400V0.005274A=75.84672Ω

Thus, the resistance across the diode is 75.84672Ω .

As the value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V , the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.005274 75.84672
0.405 0.0064 63.28565
0.41 0.007766 52.79679
0.415 0.009423 44.03979
0.42 0.011435 36.72991
0.425 0.013876 30.62901
0.43 0.016838 25.53795
0.435 0.020432 21.29022
0.44 0.024793 17.74668
0.445 0.030086 14.79101
0.45 0.036508 12.32605
0.455 0.044301 10.27061
0.46 0.053758 8.556892
0.465 0.065233 7.128278
0.47 0.079158 5.937492
0.475 0.096055 4.945067
0.48 0.11656 4.118066
0.485 0.141441 3.428998
0.495 0.20827 2.376718
0.5 0.252728 1.978408
0.505 0.306677 1.646686
0.51 0.372141 1.370449
0.515 0.451579 1.140443
0.52 0.547974 0.948949
0.525 0.664947 0.789537
0.53 0.806888 0.656844
0.535 0.979129 0.546404
0.54 1.188137 0.454493
0.545 1.44176 0.37801
0.55 1.749522 0.314372
0.555 2.122981 0.261425
0.56 2.576159 0.217378
0.565 3.126073 0.180738
0.57 3.793374 0.150262
0.575 4.603119 0.124915
0.58 5.585715 0.103836
0.585 6.778058 0.086308
0.59 8.224923 0.071733
0.595 9.98064 0.059615
0.6 12.11114 0.049541

From equation (1), formula to calculate the current across a semiconductor diode temperature 320K is,

I1=I0(eeΔV1kBT1) (7)

Here,

I1 is the current across a semiconductor diode temperature 320K .

ΔV1 is the voltage across the diode for temperature 320K .

Substitute 1.602×1019C for e , 1.38×1023J/K for kB , 0.400V for ΔV , 320K for T in equation (7) to find I1 ,

I1=(1.00nA×1A109nA)(e(1.602×1019C)(0.400V)(1.38×1023J/K)(320K)1)=0.002004A

Thus, the current across a semiconductor diode temperature 320K is 0.002004A .

From equation (2), formula to calculate the resistance across the diode is,

R1=ΔV1I1 (8)

Here,

R1 is the resistance across the diode.

Substitute 0.002004A for I1 , 0.400V for ΔV1 in equation (8) to find R1 ,

R1=0.400V0.002004A=199.5582Ω

Thus, the resistance across the diode is 199.5582Ω .

As the value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V , the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.002004 199.5582
0.405 0.002403 168.5349
0.41 0.002881 142.3127
0.415 0.003454 120.1526
0.42 0.004141 101.4283
0.425 0.004964 85.60991
0.43 0.005952 72.24847
0.435 0.007135 60.96416
0.44 0.008554 51.43551
0.445 0.010256 43.39059
0.45 0.012295 36.59933
0.455 0.014741 30.86719
0.46 0.017672 26.02967
0.465 0.021187 21.9477
0.47 0.0254 18.50372
0.475 0.030452 15.59839
0.48 0.036508 13.14778
0.485 0.043769 11.08098
0.495 0.062909 7.868498
0.5 0.07542 6.629515
0.505 0.09042 5.585066
0.51 0.108402 4.704703
0.515 0.129961 3.962729
0.52 0.155807 3.337456
0.525 0.186794 2.810585
0.53 0.223943 2.366674
0.535 0.26848 1.992698
0.54 0.321875 1.67767
0.545 0.385889 1.412324
0.55 0.462633 1.188846
0.555 0.554641 1.000647
0.56 0.664947 0.842173
0.565 0.79719 0.70874
0.57 0.955733 0.596401
0.575 1.145807 0.50183
0.58 1.373682 0.422223
0.585 1.646877 0.355218
0.59 1.974404 0.298824
0.595 2.367069 0.251366
0.6 2.837827 0.211429

Conclusion:

Therefore, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.015932 25.1067
0.405 0.019602 20.66116
0.41 0.024117 17.00046
0.415 0.029673 13.98578
0.42 0.036508 11.50433
0.425 0.044918 9.461686
0.43 0.055264 7.780834
0.435 0.067995 6.397529
0.44 0.083657 5.259572
0.445 0.102927 4.323453
0.45 0.126637 3.553464
0.455 0.155807 2.92028
0.46 0.191697 2.39962
0.465 0.235855 1.97155
0.47 0.290184 1.619662
0.475 0.357027 1.330432
0.48 0.439268 1.092727
0.485 0.540454 0.897394
0.495 0.818117 0.605048
0.5 1.006569 0.496737
0.505 1.238432 0.407774
0.51 1.523704 0.334711
0.515 1.874688 0.274712
0.52 2.306521 0.225448
0.525 2.837827 0.185001
0.53 3.491518 0.151796
0.535 4.295787 0.124541
0.54 5.285319 0.10217
0.545 6.502788 0.08381
0.55 8.000701 0.068744
0.555 9.843657 0.056381
0.56 12.11114 0.046238
0.565 14.90093 0.037917
0.57 18.33335 0.031091
0.575 22.55642 0.025492
0.58 27.75228 0.020899
0.585 34.145 0.017133
0.59 42.01028 0.014044
0.595 51.68732 0.011512
0.6 63.59346 0.009435

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.005274 75.84672
0.405 0.0064 63.28565
0.41 0.007766 52.79679
0.415 0.009423 44.03979
0.42 0.011435 36.72991
0.425 0.013876 30.62901
0.43 0.016838 25.53795
0.435 0.020432 21.29022
0.44 0.024793 17.74668
0.445 0.030086 14.79101
0.45 0.036508 12.32605
0.455 0.044301 10.27061
0.46 0.053758 8.556892
0.465 0.065233 7.128278
0.47 0.079158 5.937492
0.475 0.096055 4.945067
0.48 0.11656 4.118066
0.485 0.141441 3.428998
0.495 0.20827 2.376718
0.5 0.252728 1.978408
0.505 0.306677 1.646686
0.51 0.372141 1.370449
0.515 0.451579 1.140443
0.52 0.547974 0.948949
0.525 0.664947 0.789537
0.53 0.806888 0.656844
0.535 0.979129 0.546404
0.54 1.188137 0.454493
0.545 1.44176 0.37801
0.55 1.749522 0.314372
0.555 2.122981 0.261425
0.56 2.576159 0.217378
0.565 3.126073 0.180738
0.57 3.793374 0.150262
0.575 4.603119 0.124915
0.58 5.585715 0.103836
0.585 6.778058 0.086308
0.59 8.224923 0.071733
0.595 9.98064 0.059615
0.6 12.11114 0.049541

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.002004 199.5582
0.405 0.002403 168.5349
0.41 0.002881 142.3127
0.415 0.003454 120.1526
0.42 0.004141 101.4283
0.425 0.004964 85.60991
0.43 0.005952 72.24847
0.435 0.007135 60.96416
0.44 0.008554 51.43551
0.445 0.010256 43.39059
0.45 0.012295 36.59933
0.455 0.014741 30.86719
0.46 0.017672 26.02967
0.465 0.021187 21.9477
0.47 0.0254 18.50372
0.475 0.030452 15.59839
0.48 0.036508 13.14778
0.485 0.043769 11.08098
0.495 0.062909 7.868498
0.5 0.07542 6.629515
0.505 0.09042 5.585066
0.51 0.108402 4.704703
0.515 0.129961 3.962729
0.52 0.155807 3.337456
0.525 0.186794 2.810585
0.53 0.223943 2.366674
0.535 0.26848 1.992698
0.54 0.321875 1.67767
0.545 0.385889 1.412324
0.55 0.462633 1.188846
0.555 0.554641 1.000647
0.56 0.664947 0.842173
0.565 0.79719 0.70874
0.57 0.955733 0.596401
0.575 1.145807 0.50183
0.58 1.373682 0.422223
0.585 1.646877 0.355218
0.59 1.974404 0.298824
0.595 2.367069 0.251366
0.6 2.837827 0.211429

(b)

To determine

To draw: The graph for R versus ΔV for T=280K,300K,and320K .

(b)

Expert Solution
Check Mark

Answer to Problem 27.80AP

The graph for R versus ΔV for T=280K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  1

The graph for R versus ΔV for T=300K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  2

The graph for R versus ΔV for T=320K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  3

Explanation of Solution

Given information: The first symbol i.e. Euler’s number is e , the second symbol i.e. magnitude of electron charge is e , Boltzmann’s constant is kB , the absolute temperature is T , the value of current across a semiconductor diode temperature 0K is 1.00nA .

The different values of the R and ΔV is given in spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K .

Thus, the graph for R versus ΔV for T=280K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  4

The different values of the R and ΔV is given in spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K .

Thus, the graph for R versus ΔV for T=300K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  5

The different values of the R and ΔV is given in spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K .

The graph for R versus ΔV for T=320K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  6

Conclusion:

Therefore, the graph for R versus ΔV for T=280K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  7

Therefore, the graph for R versus ΔV for T=300K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  8

Therefore, the graph for R versus ΔV for T=320K is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 27, Problem 27.80AP , additional homework tip  9

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The current–voltage characteristic curve for a semiconductor diode as a function of temperature T is given by                                  I = I0(eeΔV/kBT - 1)Here the first symbol e represents Euler’s number, the base of natural logarithms. The second e is the magnitude of the electron charge, the kB stands for Boltzmann’s constant, and T is the absolute temperature. (a) Set up a spreadsheet to calculate I and R = ΔV/I for ΔV = 0.400 V to 0.600 Vin increments of 0.005 V. Assume I0 = 1.00 nA. (b) Plot R versus ΔV for T = 280 K, 300 K, and 320 K.
In the classic model for electrical conduction, electron mobility u in a metal wire is defined as where v, is the drift velocity of the electrons and E is the electric field in the metal. This parameter is a measure of how freely electrons can travel through a metal when an electric field is applied, such as when the ends of a metal wire are connected to a potential difference. (a) Show that where a is the conductivity of the metal, n is the number density of the charge carriers, and q is the charge on each carrier. (Submit a file with a maximum size of 1 MB.) Choose File no file selected This answer has not been graded yet. (b) B = AVH is often written as AVH - R IB where R, is the Hall coefficient of the metal. Show that the following is true. (Submit a file with a maximum size of 1 MB.) Hyo -n (Choose File no file selected This answer has not been graded yet (c) The table below shows some measured values of Hall coefficients for some elemental materials. What is an average number…
The current I is measured through a sample of an ohmic material as a voltage V is applied. (a) What is the current when the voltage is doubled to 2V (assume the change in temperature of the material is negligible)? (b) What is the voltage applied is the current measured is 0.2I (assume the change in temperature of the material is negligible)? What will happen to the current if the material if the voltage remains constant, but the temperature of the material increases significantly?

Chapter 27 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 27 - Three wires are made of copper having circular...Ch. 27 - A metal wire of resistance R is cut into three...Ch. 27 - A metal wire has a resistance of 10.0 at a...Ch. 27 - The current-versus-voltage behavior of a certain...Ch. 27 - Two conductors made of die same material are...Ch. 27 - Two conducting wires A and B of the same length...Ch. 27 - Two lightbulbs both operate on 120 V. One has a...Ch. 27 - Wire B has twice the length and twice the radius...Ch. 27 - If you were 10 design an electric healer using...Ch. 27 - Prob. 27.2CQCh. 27 - When the potential difference across a certain...Ch. 27 - Over the lime interval after a difference in...Ch. 27 - How does the resistance for copper and for silicon...Ch. 27 - Use the atomic theory of matter to explain why the...Ch. 27 - If charges flow very slowly through a metal, why...Ch. 27 - Newspaper articles often contain statements such...Ch. 27 - Prob. 27.1PCh. 27 - A small sphere that carries a charge q is whirled...Ch. 27 - Prob. 27.3PCh. 27 - In the Bohr model of the hydrogen atom (which will...Ch. 27 - A proton beam in an accelerator carries a current...Ch. 27 - Prob. 27.6PCh. 27 - Prob. 27.7PCh. 27 - Figure P26.6 represents a section of a conductor...Ch. 27 - The quantity of charge q (in coulombs) that has...Ch. 27 - A Van de Graaff generator (see Problem 24)...Ch. 27 - The electron beam emerging from a certain...Ch. 27 - An electric current in a conductor varies with...Ch. 27 - A teapot with a surface area of 700 cm2 is to be...Ch. 27 - A lightbulb has a resistance of 240 when...Ch. 27 - Prob. 27.15PCh. 27 - A 0.900-V potential difference is maintained...Ch. 27 - An electric heater carries a current of 13.5 A...Ch. 27 - Prob. 27.18PCh. 27 - Prob. 27.19PCh. 27 - Prob. 27.20PCh. 27 - A portion of Nichrome wire of radius 2.50 mm is to...Ch. 27 - If the current carried by a conductor is doubled,...Ch. 27 - Prob. 27.23PCh. 27 - Prob. 27.24PCh. 27 - If the magnitude of the drill velocity of free...Ch. 27 - Prob. 27.26PCh. 27 - Prob. 27.27PCh. 27 - While taking photographs in Death Valley on a day...Ch. 27 - Prob. 27.29PCh. 27 - Plethysmographs are devices used for measuring...Ch. 27 - Prob. 27.31PCh. 27 - An engineer needs a resistor with a zero overall...Ch. 27 - An aluminum wire with a diameter of 0.100 mm has a...Ch. 27 - Review. Ail aluminum rod has a resistance of 1.23 ...Ch. 27 - At what temperature will aluminum have a...Ch. 27 - Assume that global lightning on the Earth...Ch. 27 - In a hydroelectric installation, a turbine...Ch. 27 - A Van de Graaff generator (see Fig. 25.23) is...Ch. 27 - A certain waffle iron is rated at 1.00 kW when...Ch. 27 - The potential difference across a resting neuron...Ch. 27 - Suppose your portable DVD player draws a current...Ch. 27 - Review. A well-insulated electric water healer...Ch. 27 - A 100-W lightbulb connected to a 120-V source...Ch. 27 - The cost of energy delivered to residences by...Ch. 27 - Prob. 27.45PCh. 27 - Residential building codes typically require the...Ch. 27 - Assuming the cost of energy from the electric...Ch. 27 - An 11.0-W energy-efficient fluorescent lightbulb...Ch. 27 - A coil of Nichrome wire is 25.0 m long. The wire...Ch. 27 - Review. A rechargeable battery of mass 15.0 g...Ch. 27 - A 500-W heating coil designed to operate from 110...Ch. 27 - Why is the following situation impossible? A...Ch. 27 - A certain toaster has a heating element made of...Ch. 27 - Make an order-of-magnitude estimate of the cost of...Ch. 27 - Review. The healing element of an electric coffee...Ch. 27 - A 120-V motor has mechanical power output of 2.50...Ch. 27 - Prob. 27.57APCh. 27 - Prob. 27.58APCh. 27 - Prob. 27.59APCh. 27 - Lightbulb A is marked 25 W 120 V, and lightbulb B...Ch. 27 - One wire in a high-voltage transmission line...Ch. 27 - An experiment is conducted to measure the...Ch. 27 - A charge Q is placed on a capacitor of capacitance...Ch. 27 - Review. An office worker uses an immersion heater...Ch. 27 - Prob. 27.65APCh. 27 - An all-electric car (not a hybrid) is designed to...Ch. 27 - Prob. 27.67APCh. 27 - Prob. 27.68APCh. 27 - An electric utility company supplies a customers...Ch. 27 - The strain in a wire can be monitored and computed...Ch. 27 - An oceanographer is studying how the ion...Ch. 27 - Why is the following situation impossible? An...Ch. 27 - Prob. 27.73APCh. 27 - A close analogy exists between the flow of energy...Ch. 27 - Review. When a straight wire is warmed, its...Ch. 27 - Prob. 27.76APCh. 27 - Review. A parallel-plate capacitor consists of...Ch. 27 - The dielectric material between the plates of a...Ch. 27 - Prob. 27.79APCh. 27 - Prob. 27.80APCh. 27 - The potential difference across the filament of a...Ch. 27 - Prob. 27.82CPCh. 27 - A spherical shell with inner radius ra and outer...Ch. 27 - Material with uniform resistivity is formed into...Ch. 27 - A material of resistivity is formed into the...
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