College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 20, Problem 46P

(a)

To determine

The voltage drop across the resistor. ΔVR at t=0

(a)

Expert Solution
Check Mark

Answer to Problem 46P

Solution: The voltage drop across the resistor. ΔVR at t=0 is 0

Explanation of Solution

Given Info: Electromotive force ε is 6.0V , Resistance R is 8.0 , Inductance L is 25mH , time t is 0 .

Formula to calculate the voltage drop across the resistor ΔVR

ΔVR=RI

  • ΔVR 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=ε(1e(t/τ))R

  • ε is the electromotive force
  • t is the time elapsed
  • τ is the time constant

Use ε(1e(t/τ))/R for I in the above equation to find expression for ΔVR

ΔVR=R(ε(1e(tτ))R)=ε(1e(tτ))

Substitute 6.0V for ε and 0 for t in the above equationto find ΔVR

ΔVR=6.0V(1e0)=6.0V(11)=0

Conclusion:

The voltage drop across the resistor ΔVR at t=0 is 0

(b)

To determine

The voltage drop across the resistor. ΔVR after one time constant has passed. (t=τ)

(b)

Expert Solution
Check Mark

Answer to Problem 46P

Solution: The voltage drop across the resistor after one time constant has passed. ΔVR at (t=τ) is 3.8V

Explanation of Solution

Given Info: Electromotive force ε is 6.0V , Resistance R is 8.0 , Inductance L is 25mH , time t is τ

Formula to calculate the voltage drop across the resistor ΔVR

ΔVR=RI

Use ε(1e(t/τ))/R for I in the above equation to find expression for ΔVR

ΔVR=R(ε(1e(tτ))R)=ε(1e(tτ))

Substitute 6.0V for ε and τ for t in the above equationto find ΔVR

ΔVR=ε(1e(ττ))=(6.0V)(1e1)=(6.0V)(0.632)=3.8V

Conclusion:

The voltage drop across the resistor after one time constant has passed. ΔVR at (t=τ) is 3.8V

(c)

To determine

The voltage drop across the inductor ΔVL at t=0 .

(c)

Expert Solution
Check Mark

Answer to Problem 46P

Solution: The voltage drop across the inductor ΔVL at t=0 is 6.0V

Explanation of Solution

Given Info: Electromotive force ε is 6.0V , Resistance R is 8.0 , Inductance L is 25mH , time t is 0

Formula to calculate the voltage drop across the resistor ΔVL

ΔVL=RI

  • ΔVL is the voltage drop across the inductor

Use ε(e(t/τ))/R for I in the above equation to find expression for ΔVL

ΔVL=R(ε(e(tτ))R)=ε(e(tτ))

Substitute 6.0V for ε , and 0 for t in the above equationto find ΔVL

ΔVL=ε(e(tτ))=(6.0V)(e0)=(6.0V)(1)=6.0V

Conclusion:

The voltage drop across the inductor ΔVL at t=0 is 6.0V

(d)

To determine

The voltage drop across the resistor. ΔVL after one time constant has passed. (t=τ)

(d)

Expert Solution
Check Mark

Answer to Problem 46P

Solution: The voltage drop across the resistor after one time constant has passed. ΔVL at (t=τ) is 2.2V

Explanation of Solution

Given Info: Electromotive force ε is 6.0V , Resistance R is 8.0 , Inductance L is 25mH , time t is τ

Formula to calculate the voltage drop across the resistor ΔVL

ΔVL=RI

Use ε(e(t/τ))/R for I in the above equation to find expression for ΔVL

ΔVL=R(ε(e(tτ))R)=ε(e(tτ))

Substitute 6.0V for ε and τ for t in the above equationto find ΔVL

ΔVL=ε(e(ττ))=(6.0V)(e1)=(6.0V)(0.368)=2.2V

Conclusion:

The voltage drop across the resistor after one time constant has passed. ΔVL at (t=τ) is 2.2V

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