PHYSICS FOR SCI & ENGR W WEBASSIGN
PHYSICS FOR SCI & ENGR W WEBASSIGN
10th Edition
ISBN: 9781337888486
Author: SERWAY
Publisher: CENGAGE L
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Chapter 31, Problem 46AP

At t = 0, the open switch in Figure P31.46 is thrown closed. We wish to find a symbolic expression for the current in the inductor for time t > 0. Let this current be called i and choose it to be downward in the inductor in Figure P31.46. Identify i1 as the current to the right through R1 and i2 as the current downward through R2. (a) Use Kirchhoff’s junction rule to find a relation among the three currents. (b) Use Kirchhoff’s loop rule around the left loop to find another relationship. (c) Use Kirchhoff’s loop rule around the outer loop to find a third relationship. (d) Eliminate i1 and i2 among the three equations to find an equation involving only the current i. (e) Compare the equation in part (d) with Equation 31.6 in the text. Use this comparison to rewrite Equation 31.7 in the text for the situation in this problem and show that

i ( t ) = ε R 1 [ 1 e ( R / L ) t ]

where R′ = R1R2/(R1 + R2).

Figure P31.46

Chapter 31, Problem 46AP, At t = 0, the open switch in Figure P31.46 is thrown closed. We wish to find a symbolic expression

(a)

Expert Solution
Check Mark
To determine
The relation among the three currents by Kirchhoff’s junction rule.

Answer to Problem 46AP

The relation among the three currents by Kirchhoff’s junction rule are i1=i2+i .

Explanation of Solution

Given info: The figure that shows the given circuit is shown below.

PHYSICS FOR SCI & ENGR W WEBASSIGN, Chapter 31, Problem 46AP

Figure (I)

According to Kirchhoff’s junction rule, the total incoming currents are equal to the total outgoing currents at a junction.

From the circuit diagram equating the incoming currents to the outgoing current,

i1=i2+i

Here,

i1 is the current flow through the resistance R1 .

i2 is the current flow through the resistance R2 .

i is the current flow through the inductor.

Conclusion:

Therefore, the relation among three currents by Kirchhoff’s junction rule are i1=i2+i .

(b)

Expert Solution
Check Mark
To determine
The relationship between the given variables around the left loop by Kirchhoff’s loop rule.

Answer to Problem 46AP

The relationship between the given variables around the left loop by Kirchhoff’s loop rule is εi1R1i2R2=0 .

Explanation of Solution

Given info: The figure that shows the given circuit is shown in figure (I).

According to Kirchhoff’s loop rule, the sum of all the voltage across all the elements in a loop must be zero.

From the circuit diagram equating the voltage across the elements in the left loop is equal to zero.

εi1R1i2R2=0 (1)

Here,

ε is the emf of the battery.

Conclusion:

Therefore, the relationship between the given variables around the left loop by Kirchhoff’s loop rule is εi1R1i2R2=0 .

(c)

Expert Solution
Check Mark
To determine
The relationship between the given variables around the outer loop by Kirchhoff’s loop rule.

Answer to Problem 46AP

The relationship between the given variables around the outer loop by Kirchhoff’s loop rule is εi1R1Ldidt=0 .

Explanation of Solution

Given info: The figure that shows the given circuit is shown in figure (I).

According to Kirchhoff’s loop rule, the sum of all the voltage across all the elements in a loop must be zero.

From the circuit diagram equating the voltage across the elements in the outer loop is equal to zero.

εi1R1Ldidt=0 (2)

Conclusion:

Therefore, the relationship between the given variables around the outer loop by Kirchhoff’s loop rule is εi1R1Ldidt=0 .

(d)

Expert Solution
Check Mark
To determine
The equation that involve only current i

Answer to Problem 46AP

The equation that involve only current i is ε'iR'Ldidt=0 .

Explanation of Solution

Given info: The figure that shows the given circuit is shown in figure (I).

From equation (1), the expression for the

εi1R1i2R2=0

Substitute i2+i for i1 in above equation and rearrange for i2 .

ε(i2+i)R1i2R2=0i2R1+iR1+i2R2=εi2(R1+R2)=εiR1i2=εiR1R1+R2 (3)

From equation (2), the expression for the

εi1R1Ldidt=0

Substitute i2+i for i1 in above equation and rearrange for i2 .

ε(i2+i)R1Ldidt=0(i2+i)R1+Ldidt=ε(i2+i)=εLdidtR1i2=εLdidtR1i (4)

Equate equation (3) and equation (4) for i2 .

εLdidtR1i=εiR1R1+R2(εiR1R1+R2+i)R1=εLdidt(εiR1+(R1+R2)iR1+R2)R1=εLdidt(ε+iR2R1+R2)R1=εLdidt

Further solve the above equation,

ε(ε+iR2R1+R2)R1=Ldidtε(R1+R2)(ε+iR2)R1R1+R2=LdidtεR2iR1R2R1+R2=Ldidtε(R2R1+R2)i(R1R2R1+R2)Ldidt=0

Assume ε'=ε(R2R1+R2) and R'=(R1R2R1+R2) .

Substitute ε' for ε(R2R1+R2) and R' for (R1R2R1+R2) in equation ().

ε'iR'Ldidt=0

Thus, the require equation in term of current i is ε'iR'Ldidt=0 .

Conclusion:

Therefore, the equation that involve only current i is ε'iR'Ldidt=0 .

(e)

Expert Solution
Check Mark
To determine
The comparison of equation for part (d) with the equation 31.6 in the text use this comparison rewrite the equation 31.7 in the text for the text and show that i(t)=εR1[1e(R'/L)t]

Answer to Problem 46AP

The equation 31.6 in the text and the equation in part (d) are similar also the equation 31.7 can be rewrite for this situation is i(t)=ε'R'[1e(R'/L)t] and it showed that i(t)=εR1[1e(R'/L)t] .

Explanation of Solution

From the textbook the equation 31.6 is given as,

εiRLdidt=0

From the part (d), the equation is given as,

ε'iR'Ldidt=0

Since both the equation shown above are same therefore their solution are also same.

The solution of the equation 31.6 is given in the text as equation 31.7 that is,

i(t)=εR1[1e(R'/L)t]

Similarly rewrite the equation 31.7 for the part (d) equation.

i(t)=ε'R'[1e(R'/L)t]

Substitute ε' for ε(R2R1+R2) and R' for (R1R2R1+R2) in equation ().

i(t)=(ε(R2R1+R2))((R1R2R1+R2))[1e(R'/L)t]i(t)=εR1[1e(R'/L)t]

Conclusion:

Therefore, the equation 31.6 in the text and the equation in part (d) are similar also the equation 31.7 can be rewrite for this situation is i(t)=ε'R'[1e(R'/L)t] and it showed that i(t)=εR1[1e(R'/L)t] .

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Chapter 31 Solutions

PHYSICS FOR SCI & ENGR W WEBASSIGN

Ch. 31 - A toroid has a major radius R and a minor radius r...Ch. 31 - Prob. 7PCh. 31 - Prob. 8PCh. 31 - Prob. 9PCh. 31 - Prob. 10PCh. 31 - Prob. 11PCh. 31 - Show that i = Iiet/ is a solution of the...Ch. 31 - Prob. 13PCh. 31 - You are working as a demonstration assistant for a...Ch. 31 - Prob. 15PCh. 31 - The switch in Figure P31.15 is open for t 0 and...Ch. 31 - Prob. 17PCh. 31 - Two ideal inductors, L1 and L2, have zero internal...Ch. 31 - Prob. 19PCh. 31 - Prob. 20PCh. 31 - Prob. 21PCh. 31 - Complete the calculation in Example 31.3 by...Ch. 31 - Prob. 23PCh. 31 - A flat coil of wire has an inductance of 40.0 mH...Ch. 31 - Prob. 25PCh. 31 - Prob. 26PCh. 31 - Prob. 27PCh. 31 - Prob. 28PCh. 31 - In the circuit of Figure P31.29, the battery emf...Ch. 31 - Prob. 30PCh. 31 - An LC circuit consists of a 20.0-mH inductor and a...Ch. 31 - Prob. 32PCh. 31 - In Figure 31.15, let R = 7.60 , L = 2.20 mH, and C...Ch. 31 - Prob. 34PCh. 31 - Electrical oscillations are initiated in a series...Ch. 31 - Review. Consider a capacitor with vacuum between...Ch. 31 - A capacitor in a series LC circuit has an initial...Ch. 31 - Prob. 38APCh. 31 - Prob. 39APCh. 31 - At the moment t = 0, a 24.0-V battery is connected...Ch. 31 - Prob. 41APCh. 31 - You are working on an LC circuit for an experiment...Ch. 31 - Prob. 43APCh. 31 - Prob. 44APCh. 31 - Prob. 45APCh. 31 - At t = 0, the open switch in Figure P31.46 is...Ch. 31 - Review. The use of superconductors has been...Ch. 31 - Review. A fundamental property of a type 1...Ch. 31 - Prob. 49APCh. 31 - In earlier times when many households received...Ch. 31 - Assume the magnitude of the magnetic field outside...Ch. 31 - Prob. 52CPCh. 31 - Prob. 53CP
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