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 52CP

(a)

To determine

The emf across L immediately after t=0 .

(a)

Expert Solution
Check Mark

Answer to Problem 52CP

The emf across L immediately after t=0 is 96V .

Explanation of Solution

Given info: The battery emf is 18.0V , inductance is 0.500H , first resistance is 2.00 and second resistance is 6.00 .

Initially switch s is closed so the total current distribute in two current one is flow in the right loop and another current flow in the left loop.

Write the expression for the total current by Kirchhoff junction rule.

i=i1+i2 (1)

Here,

i is the total current.

i1 is the current in the right loop.

i2 is the current in the left loop.

Write the expression for the current in the right loop.

i1=εR1(1eR2t/L)

Here,

ε is the emf of the battery.

R1 is the resistance of the first resistor.

R2 is the resistance of the second resistor.

l is the inductance of the inductor.

Since for the steady state condition t=0 so,

Substitute 0 for t in equation (2).

i1=εR1(1eR2(0)/L)=εR1

Substitute 18.0V for ε and 2.00 for R1 to find i1 .

i1=18.0V2.00×103Ω1kΩ=9×103A×103mA1A=9mA

Write the expression for the current in the left loop.

i2=εR2

Substitute 18.0V for ε , 6.00 for R2 to find i2 .

i2=18.0V6.00×103Ω1kΩ=3×103A×103mA1A=3mA

Substitute 9mA for i1 and 3mA for i2 in equation (1).

i=9mA+3mA=12mA

Write the expression for the voltage across the inductor by Kirchhoff loop rule after t=0 .

i(R1+R2)+ΔVL=0

Rearrange the term for ΔVL .

ΔVL=i(R1+R2)

Substitute 18.0V for ε and, 2.00 for R1 , 6.00 for R2 , 12mA for i to find ΔVL

ΔVL=12mA(2.00+6.00)ΔVL=96V

Conclusion:

Therefore, the emf across L immediately after t=0 is 96V .

(b)

To determine

The point of the coil that have higher potential.

(b)

Expert Solution
Check Mark

Answer to Problem 52CP

The point a is at high potential than point b .

Explanation of Solution

Given info: The battery emf is 18.0V , inductance is 0.500H , first resistance is 2.00 and second resistance is 6.00 .

Since the current flow upwards to downward direction in the inductor coil due to some reactance of the coil the voltage drop across the inductor coil hence the point b  at lower potential in compare to the point a of the coil.

Write the expression for the voltage drop across the inductor coil.

ΔVL=Ldidt

Here,

L is the inductance of the coil.

Conclusion:

Therefore, the point a is at high potential than point b .

(c)

To determine

To draw: The graph of currents in R1 and R2 as a function of time and show values before and after t=0 .

(c)

Expert Solution
Check Mark

Answer to Problem 52CP

The graph of currents in R1 and R2 as a function of time and show values before and after t=0 is shown below.

Physics for Scientists and Engineers, Chapter 31, Problem 52CP , additional homework tip  1Physics for Scientists and Engineers, Chapter 31, Problem 52CP , additional homework tip  2

Explanation of Solution

Given info: The battery emf is 18.0V , inductance is 0.500H , first resistance is 2.00 and second resistance is 6.00 .

The currents in R1 and R2 are shown below. After t=0 , the current in R1 decreases from an initial values of 9mA according to i=I1eRt/L . Take the original current direction as positive in each resistor the current decreases from 9mA to zero.

In R2 the currents jumps from 3.00mA to 9.00mA upwards and then decrease in magnitude to zero. The time constant of each decay is 1/e=36.8% of the original value.

Physics for Scientists and Engineers, Chapter 31, Problem 52CP , additional homework tip  3Physics for Scientists and Engineers, Chapter 31, Problem 52CP , additional homework tip  4

Figure (I)

(d)

To determine

The time at which the value of current in R2 become 2.00mA .

(d)

Expert Solution
Check Mark

Answer to Problem 52CP

The time at which the value of current in R2 become 2.00mA is 75.2μs .

Explanation of Solution

Given info: The battery emf is 18.0V , inductance is 0.500H , first resistance is 2.00 and second resistance is 6.00 .

Formula to calculate the current in the circuit after t=0 is,

i=i1e((R1+R2)t/L)

Substitute 2.00 for R1 , 6.00 for R2 , 9mA for i1 , 2.00mA for i ,  and 0.500H for L to find i

2.00mA=(9mA)e((2.00+6.00)t/0.500H)e((8.00)t/0.500H)=2.00mA9mA

Take log and solve the equation further,

t=7.52×105s×106μs1s=75.2μs

Conclusion:

Therefore, the time at which the value of current in R2 become 2.00mA is 75.2μs .

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Chapter 31 Solutions

Physics for Scientists and Engineers

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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