PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
10th Edition
ISBN: 9781337888462
Author: SERWAY
Publisher: CENGAGE L
Question
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Chapter 26, Problem 46AP

(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 46AP

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 46AP

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

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , additional homework tip  1

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

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , additional homework tip  2

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

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , 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 SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , 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 SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , 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 SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , additional homework tip  6

Conclusion:

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

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , additional homework tip  7

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

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , additional homework tip  8

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

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 26, Problem 46AP , additional homework tip  9

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Chapter 26 Solutions

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN

Ch. 26 - The quantity of charge q (in coulombs) that has...Ch. 26 - A Van de Graaff generator (see Problem 24)...Ch. 26 - An electric current in a conductor varies with...Ch. 26 - Prob. 10PCh. 26 - An electric heater carries a current of 13.5 A...Ch. 26 - You are working at a company that manufactures...Ch. 26 - Prob. 13PCh. 26 - Prob. 14PCh. 26 - Prob. 15PCh. 26 - Prob. 16PCh. 26 - Prob. 17PCh. 26 - Prob. 18PCh. 26 - An aluminum wire with a diameter of 0.100 mm has a...Ch. 26 - Plethysmographs are devices used for measuring...Ch. 26 - At what temperature will aluminum have a...Ch. 26 - You are working in a laboratory that studies the...Ch. 26 - Assume that global lightning on the Earth...Ch. 26 - The Van de Graaff generator, diagrammed in Figure...Ch. 26 - A 100-W lightbulb connected to a 120-V source...Ch. 26 - The potential difference across a resting neuron...Ch. 26 - The cost of energy delivered to residences by...Ch. 26 - Residential building codes typically require the...Ch. 26 - Assuming the cost of energy from the electric...Ch. 26 - An 11.0-W energy-efficient fluorescent lightbulb...Ch. 26 - A 500-W heating coil designed to operate from 110...Ch. 26 - Why is the following situation impossible? A...Ch. 26 - Make an order-of-magnitude estimate of the cost of...Ch. 26 - Lightbulb A is marked 25 W 120 V, and lightbulb B...Ch. 26 - One wire in a high-voltage transmission line...Ch. 26 - You are working with an oceanographer who is...Ch. 26 - A charge Q is placed on a capacitor of capacitance...Ch. 26 - An experiment is conducted to measure the...Ch. 26 - Prob. 39APCh. 26 - Prob. 40APCh. 26 - Review. An office worker uses an immersion heater...Ch. 26 - The strain in a wire can be monitored and computed...Ch. 26 - A close analogy exists between the flow of energy...Ch. 26 - The dielectric material between the plates of a...Ch. 26 - Review. A parallel-plate capacitor consists of...Ch. 26 - Prob. 46APCh. 26 - Why is the following situation impossible? An...Ch. 26 - Prob. 48CPCh. 26 - A spherical shell with inner radius ra and outer...Ch. 26 - Material with uniform resistivity is formed into...
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