Question 1

An inductor produces a back emf in a DC series RL circuit when a switch connecting the battery to the circuit is closed. We can explain this by
A) Lenz's law.
B) increasing magnetic flux within the coils of the inductor.
C) increasing current in the coils of the inductor.
D) all of the above.
E) only (a) and (c) above.

Accepted Answer

Lenz's law explains the generation of back emf as a reaction to changes in magnetic flux to oppose the change in current. The increasing magnetic flux within the coils of the inductor is a direct result of the increasing current in the coils, which together explain the phenomenon of back emf in an inductor in a DC series RL circuit.

Question 2

A circuit contains two inductors of 6.0 mH inductance in parallel placed in series with an inductor of 8.0 mH inductance. After one of the 6.0 mH inductors burns out, the repairman wants to replace all three inductors with one inductor of equivalent inductance. Assuming inductors combine in series and parallel the same way resistors do, what inductance should he use?
A) 3.0 mH
B) 3.4 mH
C) 4.8 mH
D) 11 mH
E) 20 mH

Accepted Answer

D

Question 3

When a switch is closed, completing an LR series circuit, the time needed for the current to reach three-quarters its maximum value is ____ time constants.
A) 0.500
B) 0.693
C) 0.725
D) 1.33
E) 1.39

Accepted Answer

E

Question 4

What is the inductance of a series RL circuit in which R = 1.0 KΩ if the current increases to one-third of its final value in 30 μs?
A) 74 mH
B) 99 mH
C) 49 mH
D) 62 mH
E) none of the above

Accepted Answer

The time constant τ for an RL circuit is given by τ = L/R. The current reaches one-third of its final value in 30 μs, which corresponds to τ = 30 μs / ln(3). Solving for L gives L = τ * R = (30 μs / ln(3)) * 1.0 KΩ = 74 mH.

Question 5

After a switch is thrown to remove the battery from a DC LR circuit, but the circuit is still left complete, the time constant represents
A) the time rate of change of the current in the circuit.
B) the time rate of change of the induced emf in the circuit.
C) the magnitude of the ratio of the current to the time rate of change of the current.
D) all of the above.
E) only (a) and (b) above.

Accepted Answer

The time constant in an LR circuit represents the magnitude of the ratio of the current to the time rate of change of the current. It is a measure of how quickly the current will decay to zero after the battery is removed, not directly related to the time rate of change of the current or induced emf itself.

Question 6

Which of the following are the units of a henry and a farad respectively? A) J ⋅ s²/C² and C²/J B) V ⋅ s/A and V/C C) V/(A ⋅ s) and C/V D) N ⋅ m/A² and 1/J E) None of the above

Accepted Answer

The henry is the unit of inductance and is defined as the electromotive force (emf) induced in a circuit when the current through the circuit changes at a rate of one ampere per second. The farad is the unit of capacitance and is defined as the capacitance of a capacitor that stores one coulomb of charge when one volt is applied across it. The units of the henry are J ⋅ s²/C² and the units of the farad are C²/J.

Question 7

A series LC circuit contains a 100 mH inductor, a 36.0 mF capacitor and a 12 V battery. The period of the electromagnetic oscillations in the circuit is
A) 0.0227 s.
B) 0.377 s.
C) 2.26 s.
D) 105 s.
E) 1750 s.

Accepted Answer

The period of the electromagnetic oscillations in the circuit is given by:
$T = 2\pi\sqrt{LC}$

Substituting the given values, we get:
$T = 2\pi\sqrt{(100 	ext{ mH})(36.0 	ext{ mF})} = 0.377 	ext{ s}$

Therefore, the answer is B.

Question 8

A series LC circuit contains a 100 mH inductor, a 36.0 mF capacitor and a 12 V battery. The frequency of the electromagnetic oscillations in the circuit is A) 5.73 × 10−⁴ Hz. B) 9.55 × 10−³ Hz. C) 0.442 Hz. D) 2.65 Hz. E) 44.0 Hz.

Accepted Answer

The resonant frequency of a series LC circuit can be determined using the formula:
f = 1 / 2π√(LC)

Plugging in the given values, we get:
f = 1 / 2π√(100 × 10^-3 H × 36.0 × 10^-6 F)
f = 2.65 Hz

Therefore, the resonant frequency of the circuit is 2.65 Hz, which is option D.

Question 9

When a switch is closed, completing an LR series circuit, the time needed for the current to reach one half its maximum value is ____ time constants.
A) 0.250
B) 0.500
C) 0.693
D) 1.00
E) 1.44

Accepted Answer

The time constant of an LR circuit is calculated by dividing the inductance (L) by the resistance (R). This is represented by the formula:

τ = L / R

When a switch is closed, the current in the circuit begins to increase. The equation for the current in an LR circuit is:

i = (V / R) * (1 - e^(-t/τ))

Where i is the current, V is the voltage, R is the resistance, t is time, and τ is the time constant.

When t = τ, the expression inside the parentheses becomes e^(-1) or approximately 0.37. Therefore, the current is 0.37 times the maximum value at one time constant. Using simple algebra, we can solve for the time required for the current to reach one-half its maximum value:

0.5 = (1 - e^(-t/τ))
e^(-t/τ) = 0.5
-t/τ = ln(0.5)
t = -τ * ln(0.5)

Substituting the value of τ, we get:

t = 0.693 * (L / R)

Therefore, the time needed for the current to reach one half its maximum value in an LR series circuit is 0.693 time constants. The closest option is C, which is the correct answer.

Question 10

There is no current in the circuit shown in the figure below until the switch is closed. The current through the 20-Ω resistor the instant after the switch is closed is either [1] 15 A or [2] 5.0 A, and the current through the 20-Ω resistor after the switch has been closed a long time is either [3] 5.0 A or [4] 15 A. Which combination of the above choices is correct? 
A) [1] and [3]
B) [1] and [4]
C) [2] and [3]
D) [2] and [4]
E) None of these

Accepted Answer

The answer of There is no current in the circuit...

Question 11

Coaxial Cable A has twice the length, twice the radius of the inner solid conductor, and twice the radius of the outer cylindrical conducting shell of coaxial Cable B. What is the ratio of the inductance of Cable A to that of Cable B?
A)

B)

C) 2
D)

E)

Accepted Answer

The inductance of a coaxial cable is proportional to the natural logarithm of the ratio of the outer to inner radii. Since cable A has twice the radius of cable B, the ratio of the outer to inner radii for cable A is also twice that of cable B. Therefore, the natural logarithm of the ratio of the outer to inner radii for cable A is twice that of cable B. The inductance of cable A is therefore twice that of cable B. Therefore, the ratio of the inductance of cable A to that of cable B is 2:1. Hence, the correct answer is C.

Question 12

A series LC circuit contains a 100 mH inductor, a 36.0 mF capacitor and a 12 V battery. The angular frequency of the electromagnetic oscillations in the circuit is A) 36.0 × 10−⁴ rad/s. B) 6.00 × 10−² rad/s. C) 2.78 rad/s. D) 16.7 rad/s. E) 277 rad/s.

Accepted Answer

The angular frequency of the electromagnetic oscillations in a series LC circuit is given by:

$$\omega = \frac{1}{\sqrt{LC}}$$

Substituting the given values, we get:

$$\omega = \frac{1}{\sqrt{(100 	ext{ mH})(36.0 	ext{ mF})}} = 16.7 	ext{ rad/s}$$

Therefore, the correct answer is D.

Question 13

The figure shows an LR circuit with a switch and a 240-volt battery. At the instant the switch is closed the current in the circuit and the potential difference between points a and b, V_ab, are A) 0 A, 0 V B) 0 A, −240 V C) 0 A, +240 V D) 0.024 A, 0 V E) 0.024 A, +240 V

Accepted Answer

The answer of The figure shows an LR circuit with...

Question 14

When a switch is closed to complete a DC series RL circuit,
A) the electric field in the wires increases to a maximum value.
B) the magnetic field outside the wires increases to a maximum value.
C) the rate of change of the electric and magnetic fields is greatest at the instant when the switch is closed.
D) all of the above are true.
E) only (a) and (c) above are true.

Accepted Answer

When a switch is closed in a DC series RL circuit, the electric field in the wires indeed increases to a maximum value as current starts to flow. Simultaneously, the magnetic field outside the wires also increases to a maximum value due to the current. The rate of change of both the electric and magnetic fields is greatest at the instant the switch is closed because that's when the circuit transitions from no current (and thus no fields) to a state where current begins to flow, initiating the build-up of these fields. Therefore, all of the statements are true.

Question 15

A circuit contains two inductors of 6.0 mH inductance in series placed in parallel with an inductor of 8.0 mH inductance. After one of the 6.0 mH inductors burns out, the repair person wants to replace all three inductors with one inductor of equivalent inductance. Assuming inductors combine in series and parallel the same way resistors do, what inductance should she use?
A) 3.0 mH
B) 3.4 mH
C) 4.8 mH
D) 11 mH
E) 20 mH

Accepted Answer

The two 6.0 mH inductors in series have a combined inductance of 12.0 mH. This 12.0 mH is in parallel with the 8.0 mH inductor. The equivalent inductance for inductors in parallel is given by 1/L_eq = 1/L1 + 1/L2. So, 1/L_eq = 1/12 + 1/8, which gives L_eq = 4.8 mH.

Question 16

Before the switch is closed in the figure, the potential across the capacitor is 200 V. At some instant after the switch is closed, the instantaneous current is 0.70 A. What is the energy in the capacitor at this instant? 
A) 49 mJ
B) 31 mJ
C) 80 mJ
D) 0.13 J
E) 62 mJ

Accepted Answer

The answer of Before the switch is closed in the...

Question 17

The magnetic field in a superconducting solenoid is 3.0 T. How much energy per unit volume is stored in the solenoid, in J/m³? μ₀ = 4π × 10−⁷ T ⋅ A/m

Accepted Answer

The answer of The magnetic field in a superconducting solenoid...

Question 18

For the circuit shown, what is the rate of change of the current in the inductor when the current in the battery is 0.50 A? 
A) 600 A/s
B) 400 A/s
C) 200 A/s
D) 800 A/s
E) 500 A/s

Accepted Answer

We can use the formula for the rate of change of current in an inductor, which is given by:
$$\frac{di_L}{dt}=\frac{1}{L}\cdot V_{	ext{induced}}$$
where $L$ is the inductance of the inductor.
When the current in the battery is 0.50 A, the current through the resistor will also be 0.50 A since they are in series. This means that the voltage drop across the resistor is given by:
$$V_R=IR=0.50	ext{ A}\cdot 2	ext{ }\Omega=1	ext{ V}$$
Since the inductor and resistor are in series, the induced voltage across the inductor must also be equal to 1 V. Therefore, we have:
$$\frac{di_L}{dt}=\frac{1}{L}\cdot V_{	ext{induced}}=\frac{1}{2	ext{ H}}\cdot 1	ext{ V}=0.5	ext{ A/s}$$
which is closest to choice C: 200 A/s.

Question 19

Find the magnetic energy stored in the air gap between two very large magnetic pole pieces, one North, one South, each with an area of 100 cm². Assume the 0.05 T magnetic field is uniform within the 2-cm gap.

Accepted Answer

The answer of Find the magnetic energy stored in the...

Question 20

The switch in the figure is closed at t = 0 when the current I is zero. When I = 15 mA, what is the potential difference across the inductor? 
A) 240 V
B) 60 V
C) 0
D) 180 V
E) 190 V

Accepted Answer

The answer of The switch in the figure is closed...

Quiz 32: Inductance

An inductor produces a back emf in a DC series RL circuit when a switch connecting the battery to the circuit is closed. We can explain this by

When a switch is closed, completing an LR series circuit, the time needed for the current to reach three-quarters its maximum value is ____ time constants.

What is the inductance of a series RL circuit in which R = 1.0 KΩ if the current increases to one-third of its final value in 30 μs?

After a switch is thrown to remove the battery from a DC LR circuit, but the circuit is still left complete, the time constant represents

Which of the following are the units of a henry and a farad respectively?

A series LC circuit contains a 100 mH inductor, a 36.0 mF capacitor and a 12 V battery. The period of the electromagnetic oscillations in the circuit is

A series LC circuit contains a 100 mH inductor, a 36.0 mF capacitor and a 12 V battery. The frequency of the electromagnetic oscillations in the circuit is

When a switch is closed, completing an LR series circuit, the time needed for the current to reach one half its maximum value is ____ time constants.

Coaxial Cable A has twice the length, twice the radius of the inner solid conductor, and twice the radius of the outer cylindrical conducting shell of coaxial Cable B. What is the ratio of the inductance of Cable A to that of Cable B?

A series LC circuit contains a 100 mH inductor, a 36.0 mF capacitor and a 12 V battery. The angular frequency of the electromagnetic oscillations in the circuit is

The figure shows an LR circuit with a switch and a 240-volt battery. At the instant the switch is closed the current in the circuit and the potential difference between points a and b, Vab, are

When a switch is closed to complete a DC series RL circuit,

Before the switch is closed in the figure, the potential across the capacitor is 200 V. At some instant after the switch is closed, the instantaneous current is 0.70 A. What is the energy in the capacitor at this instant?

The magnetic field in a superconducting solenoid is 3.0 T. How much energy per unit volume is stored in the solenoid, in J/m3? μ0 = 4π × 10−7 T ⋅ A/m

For the circuit shown, what is the rate of change of the current in the inductor when the current in the battery is 0.50 A?

Find the magnetic energy stored in the air gap between two very large magnetic pole pieces, one North, one South, each with an area of 100 cm2. Assume the 0.05 T magnetic field is uniform within the 2-cm gap.

The switch in the figure is closed at t = 0 when the current I is zero. When I = 15 mA, what is the potential difference across the inductor?

The figure shows an LR circuit with a switch and a 240-volt battery. At the instant the switch is closed the current in the circuit and the potential difference between points a and b, V_ab, are

The magnetic field in a superconducting solenoid is 3.0 T. How much energy per unit volume is stored in the solenoid, in J/m³? μ₀ = 4π × 10−⁷ T ⋅ A/m

Find the magnetic energy stored in the air gap between two very large magnetic pole pieces, one North, one South, each with an area of 100 cm². Assume the 0.05 T magnetic field is uniform within the 2-cm gap.