Chapter 20, Problem 45P

(a)

To determine

The time constant $\tau$

Answer to Problem 45P

Solution: The time constant $\tau$ is $0.144\text{s}$

Explanation of Solution

Given Info: Voltage $\epsilon$ is $9.00\text{V}$, $t$ is $0.100\text{s}$, maximum value of $I\left(t\right)$ is $2.00\text{A}$

Since current value at $t=0.100\text{s}$ is half of its maximum current.

$I\left(t\right)=\frac{1}{2}{I}_{\max }$

• $I\left(t\right)$ is the current,
• $I_{\max }$ is the maximum current

Formula to calculate the time constant $\tau$ of the circuit,

$I\left(t\right)=I_{\max }\left(1-e^{\left(\frac{-t}{\tau }\right)}\right)$

• $\tau$ is the time constant,

Use $\frac{1}{2}{I}_{\max }$ for $I\left(t\right)$ in the above equation to find expression for $\tau$

$\begin{array}{l}\frac{1}{2}{I}_{\max }=I_{\max }\left(1-e^{\left(\frac{-t}{\tau }\right)}\right)\\ \frac{1}{2}=\left(1-e^{\left(\frac{-t}{\tau }\right)}\right)\\ -e^{\left(\frac{-t}{\tau }\right)}=\frac{1}{2}-1\\ e^{\left(\frac{-t}{\tau }\right)}=\frac{1}{2}\end{array}$

Take natural log on either sides to find $\tau$

$\begin{array}{l}-\frac{t}{\tau }=\ln \left(\frac{1}{2}\right)\\ \tau =\frac{-t}{\ln \left(\frac{1}{2}\right)}\end{array}$

Formula to calculate the time constant $\tau$ of the circuit

On rearrange the above equation

$\tau =\frac{-t}{\ln \left(\frac{1}{2}\right)}$

Substitute $0.100\text{s}$ for $t$ in the above equation to find $\tau$

$\begin{array}{c}\tau =\frac{-0.100\text{s}}{\ln \left(\frac{1}{2}\right)}\\ =0.144\text{s}\end{array}$

Conclusion:

The time constant $\tau$ is $0.144\text{s}$

(b)

To determine

The emf across the inductor $\Delta V_L$ at $t=0.100\text{s}$

Answer to Problem 45P

Solution: The emf across the inductor $\Delta V_L$ at $t=0.100\text{s}$ is $4.50\text{V}$

Explanation of Solution

Given Info: Voltage $\epsilon$ is $9.00\text{V}$, half the maximum current at t $\left(I_{\text{half}}\left(t\right)\right)$ is $1.00\text{A}$ , Current $I\left(t\right)$ is $2.00\text{A}$

Formula to calculate the resistance R

$I\left(t\right)=\frac{\epsilon }{R}$

• R is the Resistance
• $\epsilon$ is the voltage

Rearrange in terms of R.

$R=\frac{\epsilon }{I\left(t\right)}$

Substitute $2.00\text{A}$ for $I\left(t\right)$ and $9.00\text{V}$ for $\epsilon$ in the above equation to find R

$\begin{array}{c}R=\frac{9.00\text{V}}{2.00\text{A}}\\ =4.50\text{Ω}\end{array}$

Thus the Resistance R is $4.50\text{Ω}$

Conclusion:

Formula to calculate the emf across the inductor $\Delta V_L$ at $t=0.100\text{s}$ is

Apply Kirchhoff’s loop rule

$\epsilon -\left(I_{\text{half}}\left(t\right)\right)R-\Delta V_L=0$

Rearrange in terms of $\Delta V_L$.

$\Delta V_L=\epsilon -\left(I_{\text{half}}\left(t\right)\right)R$

Substitute $9.00\text{V}$ for $\epsilon$, $4.50\text{Ω}$ for R, and $1.00\text{A}$ for $\left(I_{\text{half}}\left(t\right)\right)$ in the above equation to find $\Delta V_{\text{L}}$

$\begin{array}{c}\Delta V_L=9.00\text{V}-\left(4.50\text{Ω}\right)\left(1.00\text{A}\right)\\ =4.50\text{V}\end{array}$

Therefore, the emf across the inductor $\Delta V_L$ at $t=0.100\text{s}$ is $4.50\text{V}$

(c)

To determine

The emf across the inductor $\Delta V_L$ at $t=0$

Answer to Problem 45P

Solution: The emf across the inductor $\Delta V_L$ at $t=0$ is $9.00\text{V}$

Explanation of Solution

Given Info: Voltage $\epsilon$ is $9.00\text{V}$, Resistance R is $4.50\text{Ω}$, When time t is 0, then Current I is also 0

Formula to calculate the emf across the inductor $\Delta V_L$ at $t=0$ is

Apply Kirchhoff’s loop rule

$\epsilon -IR-\Delta V_L=0$

On Rearrange

$\Delta V_L=\epsilon -IR$

Substitute $9.00\text{V}$ for $\epsilon$, $4.50\text{Ω}$ for R and $0$ for I in the above equation to find $\Delta V_L$

$\begin{array}{c}\Delta V_L=9.00\text{V}-\left(4.50\text{Ω}\right)\left(0\right)\\ =9.00\text{V}\end{array}$

Conclusion:

The emf across the inductor $\Delta V_L$ at $t=0$ is $9.00\text{V}$