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
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Chapter 28, Problem 28.78AP

The circuit shown in Figure P28.78 is set up in the laboratory to measure an unknown capacitance C in series with a resistance R = 10.0 MΩ powered by a battery whose emf is 6.19 V. The data given in the table are the measured voltages across the capacitor as a function of lime, where t = 0 represents the instant at which the switch is thrown to position b. (a) Construct a graph of In (ε/Δv) versus I and perform a linear least-squares fit to the data, (b) From the slope of your graph, obtain a value for the time constant of the circuit and a value for the capacitance.

Δv(V) t(s) In (ε/Δv)
6.19 0
5.56 4.87
4.93 11.1
4.34 19.4
3.72 30.8
3.09 46.6
2.47 67.3
1.83 102.2

Chapter 28, Problem 28.78AP, The circuit shown in Figure P28.78 is set up in the laboratory to measure an unknown capacitance C

(a)

Expert Solution
Check Mark
To determine

To draw: The graph of ln(ε/Δv) versus time t and perform a linear least squares fit to the data.

Answer to Problem 28.78AP

The graph of ln(ε/Δv) versus time t , where ln(ε/Δv) on y axis and time t on x axis is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 28, Problem 28.78AP , additional homework tip  1

And the equation of best fit is ln(ε/Δv)=(0.01178)t+0.08798 .

Explanation of Solution

Introduction:

The linear least square is a technique of fitting a statistical model to a set of variables. This shows the data points linearly in terms of unknown parameters.

Given: The capacitance is C and the resistance R is 10.0MΩ , the emf ε of a battery is 6.19V , the time t at which the switch is thrown to position b is 0 and the data given in the table is,

Δv(V) t(s)
6.19 0
5.55 4.87
4.93 11.1
4.34 19.4
3.72 30.8
3.09 46.6
2.47 67.3
1.83 102.2

Write the equation of line which gives a linear relationship.

y=mx+c

Here,

y is the ordinate.

x is the abscissa.

m is the slope of the line.

c is the intercept on y axis.

Take ln(ε/Δv) on y axis and time on x axis due to which y=ln(ε/Δv) and x=t

Substitute ln(ε/Δv) for y and t for x in above equation.

ln(ε/Δv)=mt+c (1)

Calculate the values of ln(ε/Δv) for each values of Δv .

Δv(V) x=t(s) y=ln(ε/Δv)
6.19 0 0
5.55 4.87 0.109
4.93 11.1 0.227
4.34 19.4 0.355
3.72 30.8 0.509
3.09 46.6 0.694
2.47 67.3 0.918
1.83 102.2 1.218

Write the formula of slope.

m=N(xiyi)(xi)(yi)N(xi2)(xi)2

Here,

N is the number of points.

The number of data points N is 8 .

Calculate (xiyi) , (xi) , (yi) and (xi2) from the table.

x=t(s) y=ln(ε/Δv) x2 xy
0 0 0 0
4.87 0.109 23.7169 0.53083
11.1 0.227 123.21 2.5197
19.4 0.355 376.36 6.887
30.8 0.509 948.64 15.6772
46.6 0.694 2171.56 32.3404
67.3 0.918 4529.29 61.7814
102.2 1.218 10444.84 124.4796
xi=282.27 yi=4.03 x2=18617.617 xiyi=244.216

Substitute 8 for N , 244.216 for (xiyi) , 282.27 for (xi) , 4.03 for (yi) and 18617.617 for (xi2) in above equation to find m .

m=(8)(244.216)(282.27)(4.03)(8)(18617.617)(282.27)2=0.01178

Thus, the slope of line is 0.01178 .

Write the formula of intercept.

c=(xi2)(yi)(xi)(xiyi)N(xi2)(xi)2

Substitute 18617.617 for (xi2) , 4.03 for (yi) , 282.27 for (xi) , 244.216 for (xiyi) and 8 for N in above equation to find c .

c=(18617.617)(4.03)(282.27)(244.216)(8)(18617.617)(282.27)2=0.08798

Thus, the value of intercept on y axis is 0.08798 .

Substitute 0.01178 for m and 0.08798 for c in equation (1).

ln(ε/Δv)=(0.01178)t+0.08798

Thus, the equation of best fit is ln(ε/Δv)=(0.01178)t+0.08798 .

The graph of ln(ε/Δv) versus time t , where ln(ε/Δv) on y axis and time t on x axis is,

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 28, Problem 28.78AP , additional homework tip  2

Figure 1

(b)

Expert Solution
Check Mark
To determine

The value of time constant of the circuit and the value of the capacitance.

Answer to Problem 28.78AP

The value of time constant of the circuit is 84.8s and the value of the capacitance is 8.48μF .

Explanation of Solution

Given: The capacitance is C and the resistance R is 10.0MΩ , the emf ε of a battery is 6.19V and the time t at which the switch is thrown to position b is 0 .

Write the equation of time constant.

τ=RC (2)

Here,

τ is the time constant.

R is the resistance.

C is the capacitance.

The time constant from the slope is,

τ=1m

Substitute 0.01178 for m in above equation to find τ .

τ=10.01178=84.8s

Thus, the time constant is 84.8s .

Substitute 84.8s for τ and 10.0MΩ for R in equation (2) to find C .

(84.8s)={10.0MΩ(106Ω1MΩ)}CC=8.48×106F(1μF106F)C=8.48μF

The capacitance is 8.48μF .

Conclusion:

Therefore, the value of time constant of the circuit is 84.8s and the value of the capacitance is 8.48μF .

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Chapter 28 Solutions

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

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