Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 31, Problem 16P

The switch in Figure P31.15 is open for t < 0 and is then thrown closed at time t = 0. Find (a) the current in the inductor and (b) the current in the switch as functions of time thereafter.

(a)

Expert Solution
Check Mark
To determine
The current in the inductor.

Answer to Problem 16P

The current in inductor in the terms of time is ε5R(1e2.5RLt) .

Explanation of Solution

Given info: The value of resistance R is 4.00Ω , inductance of the circuit is 1.00H and emf of the battery is 10.0V .

Explanation:

Formula to calculate current in a loop as per Kirchhoff law is,

i=0IsIIL=0I=IsIL (1)

Here,

Is is current flowing through switch s .

I is current flowing through resistance R .

Il is the current flowing through the inductance L .

Write the expression for net voltage in loop 1,

εRIsRI=0 (2)

Write the expression to calculate net voltage in loop 2,

εRIs2RILLdILdt=0 (3)

Here,

R is resistance of circuit.

L is inductance of circuit.

dILdt rate of change of current in inductance.

Substitute IsIL for I in equation (II).

εRIsR(IsIL)=0ε2RIsRIL=0Is=ε2R+IL2 (4)

Substitute ε2R+IL2 for Is in equation (3).

εR(ε2R+IL2)2RILLdILdt=0ε(ε2+RIL2)2RILLdILdt=0ε2RIL22RILLdILdt=0ε25RIL2LdILdt=0

Arrange the terms of above equation to simplify for integration.

LdILdt=ε22.5RILL2.5RdILdt=ε5RILL2.5RdILdt=(ε5R+IL)dIL(ε5R+IL)=2.5RLdt

On integrate,

dILε5R+IL=(2.5RL)dt (5)

Assume ε5R+IL=T .

Differentiate above equation,

dIL=dT

Substitute dT for dIL and T for ε5R+IL in equation (5).

dILT=(2.5RL)dtdILT=2.5RLdtlnT=2.5RLt+c

Substitute ε5R+IL for T in above equation,

ln(ε5R+IL)=2.5RLt+c (6)

Apply boundary condition,

Substitute 0 for t and 0 for IL in above equation.

ln(ε5R+0)=2.5RL×0+cc=ln(ε5R)

Substitute ln(ε5R) for c in equation (VI)

ln(ε5R+IL)=2.5RLt+ln(ε5R)ln(ε5R+IL)ln(ε5R)=2.5RLtln(ε5R+IL)(ε5R)=2.5RLtln(1ILε5R)=2.5RLt

Further solve the above expression.

1ILε5R=e2.5RLtIL=ε5R(1e2.5RLt)

Thus, the current in inductor in the terms of time is IL=ε5R(1e2.5RLt) .

Conclusion:

Therefore, the current in inductor in the terms of time is IL=ε5R(1e2.5RLt) .

(b)

Expert Solution
Check Mark
To determine
The current in the switch as a function of time.

Answer to Problem 16P

The current in the switch as the function of time is 3ε5Rε10Re2.5RLt .

Explanation of Solution

Formula to calculate current in inductor, from equation (VII)

IL=ε5R(1e2.5RLt)

Formula to calculate current in switch, from equation, from equation (IV)

Is=ε2R+IL2

Substitute ε5R(1e2.5RLt) for IL in above equation to calculate Is .

Is=ε2R+12(ε5R)(1e2.5RLt)=ε2R+ε10R(1e10.0t)=ε2R+ε10R+ε10Re10.0t=3ε5Rε10Re2.5RLt

Thus, current in switch is 3ε5Rε10Re2.5RLt .

Conclusion:

Therefore, the current in switch is 3ε5Rε10Re2.5RLt .

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Chapter 31 Solutions

Physics for Scientists and Engineers

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