Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 32, Problem 70AP

(a)

To determine

The relation among the currents using Kirchhoff’s law.

(a)

Expert Solution
Check Mark

Answer to Problem 70AP

The relation among the currents using Kirchhoff’s law is I1=I2+I.

Explanation of Solution

Apply Kirchhoff’s junction rule which states that at any junction the sum of the cuurent must be equal to zero.

    I1=I2+I                                                                                                                    (I)

Here, I is the equivalent current in the circuit, I1 is the current through resistor R1 and I2 is the current through resistor R2.

Therefore, the relation among the currents using Kirchhoff’s law is I1=I2+I.

(b)

To determine

The relation using Kirchhoff’s rule in the left loop.

(b)

Expert Solution
Check Mark

Answer to Problem 70AP

The relation using Kirchhoff’s rule in the left loop is I1=ε+IR2R1+R2.

Explanation of Solution

Apply Kirchhoff’s loop rule in the left loop.

    ε=I1R1+I2R2                                                                                                          (II)

Here, ε is the emf across the left loop.

Rearrange equation (I) to get the expression for I2.

    I2=I1I

Substitute I1I for I2 in equation (II).

    ε=I1R1+(I1I)R2I1=ε+IR2R1+R2

Therefore, the relation using Kirchhoff’s rule in the left loop is I1=ε+IR2R1+R2.

(c)

To determine

The relation using Kirchhoff’s rule in the outer loop.

(c)

Expert Solution
Check Mark

Answer to Problem 70AP

The relation using Kirchhoff’s rule in the outer loop is ε=(ε+IR2R1+R2)R1+LdIdt.

Explanation of Solution

Apply Kirchhoff’s rule in the outer loop.

    ε=I1R1+LdIdt                                                                                                        (III)

Substitute ε+IR2R1+R2 for I1 in equation (III).

    ε=(ε+IR2R1+R2)R1+LdIdt                                                                                          (IV)

Therefore, the relation using Kirchhoff’s rule in the outer loop is ε=(ε+IR2R1+R2)R1+LdIdt.

(d)

To determine

The equation involving only current.

(d)

Expert Solution
Check Mark

Answer to Problem 70AP

The equation involving only current is εR2(R1+R2)IR1R2(R1+R2)LdIdt=0.

Explanation of Solution

Multiply and divide left hand side of equation (IV).

    ε(R1+R2)(R1+R2)=(ε+IR2(R1+R2))R1+LdIdtεR1(R1+R2)+εR2(R1+R2)=εR1(R1+R2)+IR1R2(R1+R2)+LdIdtεR2(R1+R2)IR1R2(R1+R2)LdIdt=0                                    (V)

Therefore, the equation involving only current is εR2(R1+R2)IR1R2(R1+R2)LdIdt=0.

(e)

To determine

The given expression.

(e)

Expert Solution
Check Mark

Answer to Problem 70AP

The given equation I=εR1((1e(R(L))(t))) has been derived.

Explanation of Solution

Equation (V) is of the form.

    εIRLdldt=0                                                                                                        (VI)

The solution to equation (V) is

    I=εR(1eRt/L)                                                                                                  (VII)

Compare equation (V), (VI) and (VII) for the value of I.

    I=εR2(R1+R2)R1R2(R1+R2)(1e(R1R2(R1+R2))(tL))=εR1((1e(R1R2(R1+R2))(tL)))                                                                             (VIII)

Substitute R for R1R2R1+R2 in equation (VIII)

    I=εR1((1e(R(L))(t)))

Therefore, the given equation I=εR1((1e(R(L))(t))) has been derived.

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Chapter 32 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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