Chapter 20, Problem 46P

(a)

To determine

The voltage drop across the resistor. $\Delta V_R$ at $t=0$

Answer to Problem 46P

Solution: The voltage drop across the resistor. $\Delta V_R$ at $t=0$ is 0

Explanation of Solution

Given Info: Electromotive force $\epsilon$ is $6.0\text{V}$, Resistance R is $8.0\text{MΩ}$, Inductance $L$ is $25\text{mH}$, time t is $0$.

Formula to calculate the voltage drop across the resistor $\Delta V_R$

$\Delta V_R=RI$

• $\Delta V_R$ is the voltage drop across the resistor
• R is the resistance
• $I$ is the current

As the voltage across the inductor decrease, the current through the inductor increases.

$I=\frac{\epsilon \left(1-e^{\left(-t/\tau \right)}\right)}{R}$

• $\epsilon$ is the electromotive force
• t is the time elapsed
• $\tau$ is the time constant

Use $\epsilon \left(1-e^{\left(-t/\tau \right)}\right)/R$ for I in the above equation to find expression for $\Delta V_R$

$\begin{array}{c}\Delta V_{\text{R}}=R\left(\frac{\epsilon \left(1-e^{\left(\frac{-t}{\tau }\right)}\right)}{\text{}R}\right)\\ =\epsilon \left(1-e^{\left(\frac{-t}{\tau }\right)}\right)\end{array}$

Substitute $6.0\text{V}$ for $\epsilon$ and $0$ for t in the above equationto find $\Delta V_R$

$\begin{array}{c}\Delta V_{\text{R}}=6.0\text{V}\left(1-e^0\right)\\ =6.0\text{V}\left(1-1\right)\\ =0\end{array}$

Conclusion:

The voltage drop across the resistor $\Delta V_R$ at $t=0$ is 0

(b)

To determine

The voltage drop across the resistor. $\Delta V_R$ after one time constant has passed. $\left(t=\tau \right)$

Answer to Problem 46P

Solution: The voltage drop across the resistor after one time constant has passed. $\Delta V_R$ at $\left(t=\tau \right)$ is $3.8\text{V}$

Explanation of Solution

Given Info: Electromotive force $\epsilon$ is $6.0\text{V}$, Resistance R is $8.0\text{MΩ}$, Inductance $L$ is $25\text{mH}$, time t is $\tau$

Formula to calculate the voltage drop across the resistor $\Delta V_R$

$\Delta V_R=RI$

Use $\epsilon \left(1-e^{\left(-t/\tau \right)}\right)/R$ for I in the above equation to find expression for $\Delta V_R$

$\begin{array}{c}\Delta V_R=R\left(\frac{\epsilon \left(1-e^{\left(\frac{-t}{\tau }\right)}\right)}{\text{}R}\right)\\ =\epsilon \left(1-e^{\left(\frac{-t}{\tau }\right)}\right)\end{array}$

Substitute $6.0\text{V}$ for $\epsilon$ and $\tau$ for t in the above equationto find $\Delta V_R$

$\begin{array}{c}\Delta V_R=\epsilon \left(1-e^{\left(\frac{-\tau }{\tau }\right)}\right)\\ =\left(6.0\text{V}\right)\left(1-e^{-1}\right)\\ =\left(6.0\text{V}\right)\left(0.632\right)\\ =3.8\text{V}\end{array}$

Conclusion:

The voltage drop across the resistor after one time constant has passed. $\Delta V_{\text{R}}$ at $\left(t=\tau \right)$ is $3.8\text{V}$

(c)

To determine

The voltage drop across the inductor $\Delta V_L$ at $t=0$.

Answer to Problem 46P

Solution: The voltage drop across the inductor $\Delta V_L$ at $t=0$ is $6.0\text{V}$

Explanation of Solution

Given Info: Electromotive force $\epsilon$ is $6.0\text{V}$, Resistance R is $8.0\text{MΩ}$, Inductance $L$ is $25\text{mH}$, time t is $0$

Formula to calculate the voltage drop across the resistor $\Delta V_L$

$\Delta V_L=RI$

• $\Delta V_L$ is the voltage drop across the inductor

Use $\epsilon \left(e^{\left(-t/\tau \right)}\right)/R$ for I in the above equation to find expression for $\Delta V_L$

$\begin{array}{c}\Delta V_L=R\left(\frac{\epsilon \left(e^{\left(\frac{-t}{\tau }\right)}\right)}{\text{}R}\right)\\ =\epsilon \left(e^{\left(\frac{-t}{\tau }\right)}\right)\end{array}$

Substitute $6.0\text{V}$ for $\epsilon$, and $0$ for t in the above equationto find $\Delta V_L$

$\begin{array}{c}\Delta V_L=\epsilon \left(e^{\left(\frac{-t}{\tau }\right)}\right)\\ =\left(6.0\text{V}\right)\left(e^{-0}\right)\\ =\left(6.0\text{V}\right)\left(1\right)\\ =6.0\text{V}\end{array}$

Conclusion:

The voltage drop across the inductor $\Delta V_L$ at $t=0$ is $6.0\text{V}$

(d)

To determine

The voltage drop across the resistor. $\Delta V_{\text{L}}$ after one time constant has passed. $\left(t=\tau \right)$

Answer to Problem 46P

Solution: The voltage drop across the resistor after one time constant has passed. $\Delta V_L$ at $\left(t=\tau \right)$ is $2.2\text{V}$

Explanation of Solution

Given Info: Electromotive force $\epsilon$ is $6.0\text{V}$, Resistance R is $8.0\text{MΩ}$, Inductance $L$ is $25\text{mH}$, time t is $\tau$

Formula to calculate the voltage drop across the resistor $\Delta V_L$

$\Delta V_L=RI$

Use $\epsilon \left(e^{\left(-t/\tau \right)}\right)/R$ for I in the above equation to find expression for $\Delta V_L$

$\begin{array}{c}\Delta V_{\text{L}}=R\left(\frac{\epsilon \left(e^{\left(\frac{-t}{\tau }\right)}\right)}{\text{}R}\right)\\ =\epsilon \left(e^{\left(\frac{-t}{\tau }\right)}\right)\end{array}$

Substitute $6.0\text{V}$ for $\epsilon$ and $\tau$ for t in the above equationto find $\Delta V_L$

$\begin{array}{c}\Delta V_L=\epsilon \left(e^{\left(\frac{-\tau }{\tau }\right)}\right)\\ =\left(6.0\text{V}\right)\left(e^{-1}\right)\\ =\left(6.0\text{V}\right)\left(0.368\right)\\ =2.2\text{V}\end{array}$

Conclusion:

The voltage drop across the resistor after one time constant has passed. $\Delta V_L$ at $\left(t=\tau \right)$ is $2.2\text{V}$