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

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Answer 1

Textbook Question

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

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 $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ , 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

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 $84.8 s$ and the value of the capacitance is $8.48 μF$ .

Explanation of Solution

Given: The capacitance is $C$ and the resistance $R$ is $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ 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 $τ$ .

$τ=10.01178=84.8 s$

Thus, the time constant is $84.8 s$ .

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

$(84.8 s)={10.0 MΩ(106 Ω1 MΩ)}CC=8.48×10−6 F(1 μF10−6 F)C=8.48 μF$

The capacitance is $8.48 μF$ .

Conclusion:

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

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Answer 2

Textbook Question

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

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 $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ , 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

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 $84.8 s$ and the value of the capacitance is $8.48 μF$ .

Explanation of Solution

Given: The capacitance is $C$ and the resistance $R$ is $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ 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 $τ$ .

$τ=10.01178=84.8 s$

Thus, the time constant is $84.8 s$ .

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

$(84.8 s)={10.0 MΩ(106 Ω1 MΩ)}CC=8.48×10−6 F(1 μF10−6 F)C=8.48 μF$

The capacitance is $8.48 μF$ .

Conclusion:

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

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Answer 3

Textbook Question

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

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 $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ , 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

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 $84.8 s$ and the value of the capacitance is $8.48 μF$ .

Explanation of Solution

Given: The capacitance is $C$ and the resistance $R$ is $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ 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 $τ$ .

$τ=10.01178=84.8 s$

Thus, the time constant is $84.8 s$ .

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

$(84.8 s)={10.0 MΩ(106 Ω1 MΩ)}CC=8.48×10−6 F(1 μF10−6 F)C=8.48 μF$

The capacitance is $8.48 μF$ .

Conclusion:

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

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Answer 4

Textbook Question

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

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 $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ , 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

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 $84.8 s$ and the value of the capacitance is $8.48 μF$ .

Explanation of Solution

Given: The capacitance is $C$ and the resistance $R$ is $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ 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 $τ$ .

$τ=10.01178=84.8 s$

Thus, the time constant is $84.8 s$ .

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

$(84.8 s)={10.0 MΩ(106 Ω1 MΩ)}CC=8.48×10−6 F(1 μF10−6 F)C=8.48 μF$

The capacitance is $8.48 μF$ .

Conclusion:

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

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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 $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ , 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

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 $84.8 s$ and the value of the capacitance is $8.48 μF$ .

Explanation of Solution

Given: The capacitance is $C$ and the resistance $R$ is $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ 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 $τ$ .

$τ=10.01178=84.8 s$

Thus, the time constant is $84.8 s$ .

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

$(84.8 s)={10.0 MΩ(106 Ω1 MΩ)}CC=8.48×10−6 F(1 μF10−6 F)C=8.48 μF$

The capacitance is $8.48 μF$ .

Conclusion:

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

Answer 8

(a)

Expert Solution

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 $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ , 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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

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

Answer 13

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 $ln(ε/Δv)=(0.01178)t+0.08798$ .

Answer 14

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 $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ , 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

Answer 15

(b)

Expert Solution

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 $84.8 s$ and the value of the capacitance is $8.48 μF$ .

Explanation of Solution

Given: The capacitance is $C$ and the resistance $R$ is $10.0 MΩ$ , the emf $ε$ of a battery is $6.19 V$ 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 $τ$ .

$τ=10.01178=84.8 s$

Thus, the time constant is $84.8 s$ .

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

$(84.8 s)={10.0 MΩ(106 Ω1 MΩ)}CC=8.48×10−6 F(1 μF10−6 F)C=8.48 μF$

The capacitance is $8.48 μF$ .

Conclusion:

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

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

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

Answer 20

Answer to Problem 28.78AP

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

Answer 21

Explanation of Solution

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