General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Question
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Chapter 20, Problem 58E

(a)

To determine

The equation of the current for t>0.

(a)

Expert Solution
Check Mark

Answer to Problem 58E

The equation of current for t>0 is satisfied by i.

Explanation of Solution

Write the expression for voltage drop across resistance at position 2.

    VR=iR

Here, VR is the voltage drop across R, R is the resistance and i is the current.

Write the expression for voltage drop across inductor at position 2.

    VL=Ldidt

Here, VL is the voltage drop across L, L is the inductance and didt is the current change.

Apply Kirchoff’s volatge law at point 2.

    VL=VR        (1)

Conclusion:

Substitute (Ldidt) for VL and iR for VR in the equation (1).

    (Ldidt)=iR

Rearrange the above equation.

    iRLdidt=0

Thus, the equation for current is given above.

(b)

To determine

The value of current for t>0.

(b)

Expert Solution
Check Mark

Answer to Problem 58E

The value of the current is (εReRt/L).

Explanation of Solution

Write the equation of current for position 2.

    iRLdidt=0        (2)

Write the expression for the initial current at position 2.

    i0=εR

Here, i0 is the initial current and ε is the battery voltage.

Conclusion:

Rearrange equation (2).

    dii=RLdt

Integrate both side of the above equation to find the current i.

    0idii=RL0tdt[ln(i)]i0i=RL[t]0tln(ii0)=RtL

Solve the above equation.

    i=i0eRt/L

Substitute (εR) for i0 in the above equation.

    i=εReRt/L

Thus, the value of the current is (εReRt/L).

(c)

To determine

The value of current for 0s.

(c)

Expert Solution
Check Mark

Answer to Problem 58E

The current for 0s is (εR).

Explanation of Solution

Write the expression for the current in RL circuit for the position 2.

    i=εReRt/L        (3)

Conclusion:

Substitute 0 for t in equation (3).

    i=εReR(0)/L

Solve the above equation.

    i=εR

Thus, the current for 0s is (εR).

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Chapter 20 Solutions

General Physics, 2nd Edition

Ch. 20 - Prob. 11RQCh. 20 - Prob. 12RQCh. 20 - Prob. 13RQCh. 20 - Prob. 1ECh. 20 - Prob. 2ECh. 20 - Prob. 3ECh. 20 - Prob. 4ECh. 20 - Prob. 5ECh. 20 - Prob. 6ECh. 20 - Prob. 7ECh. 20 - Prob. 8ECh. 20 - Prob. 9ECh. 20 - Prob. 10ECh. 20 - Prob. 11ECh. 20 - Prob. 12ECh. 20 - Prob. 13ECh. 20 - Prob. 14ECh. 20 - Prob. 15ECh. 20 - Prob. 16ECh. 20 - Prob. 17ECh. 20 - Prob. 18ECh. 20 - Prob. 19ECh. 20 - Prob. 20ECh. 20 - Prob. 21ECh. 20 - Prob. 22ECh. 20 - Prob. 23ECh. 20 - Prob. 24ECh. 20 - Prob. 25ECh. 20 - Prob. 26ECh. 20 - Prob. 27ECh. 20 - Prob. 28ECh. 20 - Prob. 29ECh. 20 - Prob. 30ECh. 20 - Prob. 31ECh. 20 - Prob. 32ECh. 20 - Prob. 33ECh. 20 - Prob. 34ECh. 20 - Prob. 35ECh. 20 - Prob. 36ECh. 20 - Prob. 37ECh. 20 - Prob. 38ECh. 20 - Prob. 39ECh. 20 - Prob. 40ECh. 20 - Prob. 41ECh. 20 - Prob. 42ECh. 20 - Prob. 43ECh. 20 - Prob. 44ECh. 20 - Prob. 45ECh. 20 - Prob. 46ECh. 20 - Prob. 47ECh. 20 - Prob. 48ECh. 20 - Prob. 49ECh. 20 - Prob. 50ECh. 20 - Prob. 51ECh. 20 - Prob. 52ECh. 20 - Prob. 53ECh. 20 - Prob. 54ECh. 20 - Prob. 55ECh. 20 - Prob. 56ECh. 20 - Prob. 57ECh. 20 - Prob. 58ECh. 20 - Prob. 59ECh. 20 - Prob. 60ECh. 20 - Prob. 61ECh. 20 - Prob. 62ECh. 20 - Prob. 63ECh. 20 - Prob. 64ECh. 20 - Prob. 65ECh. 20 - Prob. 66ECh. 20 - Prob. 67ECh. 20 - Prob. 68ECh. 20 - Prob. 69ECh. 20 - Prob. 70ECh. 20 - Prob. 71ECh. 20 - Prob. 72ECh. 20 - Prob. 73ECh. 20 - Prob. 74ECh. 20 - Prob. 75ECh. 20 - Prob. 76ECh. 20 - Prob. 77ECh. 20 - Prob. 78ECh. 20 - Prob. 79ECh. 20 - Prob. 80ECh. 20 - Prob. 81ECh. 20 - Prob. 82ECh. 20 - Prob. 83ECh. 20 - Prob. 84ECh. 20 - Prob. 85ECh. 20 - Prob. 86ECh. 20 - Prob. 87ECh. 20 - Prob. 88ECh. 20 - Prob. 89ECh. 20 - Prob. 90ECh. 20 - Prob. 91ECh. 20 - Prob. 92ECh. 20 - Prob. 93ECh. 20 - Prob. 94ECh. 20 - Prob. 95E
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