AP Physics C – Answers to Free Response Questions involving Inductance and Capacitance (for Practice)

In the post dated 7^th June 2008 two free response questions involving inductance and capacitance were given for your practice. As promised, I give below the answers along with the questions:

(1) In the circuit shown, the battery and the inductor have negligible resistance. The currents through R₁, R₂ and R₃ are i₁, i₂ and i₃ respectively when the switch S is kept closed. The switch S is closed at time t = 0.

(a) Calculate the potential difference across the 8 Ω resistor R₂ immediately after closing the switch S.

(b) If the inductance is replaced by an uncharged 10 μF capacitor what will be the potential difference across the 8 Ω resistor R₂ immediately after closing the switch S? Justify your answer.

(c) After reaching steady state, in case (b) what will be the potential difference across the resistor R₂? Justify your answer.

(d) If a 16 Ω resistor is used instead of the capacitor mentioned in (b), calculate the current through R₂ immediately after closing the switch S.

(e) Show the nature of variation of the potential difference across R₂ with time t in cases (b) and (d).

(a) After closing the switch S, the current I in the inductance branch grows exponentially with time t in accordance with the equation,

I = I₀(1– e^{– Rt /L}) where I₀ is the final steady current.

The current I is therefore zero when t = 0 so that the the potential difference across the 8 Ω resistor R₂ immediately after closing the switch S is zero.

(b) After closing the switch S, the potential difference V across the capacitor (C) will build up exponentially with time t in accordance with the equation,

V = V₀(1– e^–t/RC) where V₀ is the final steady voltage across the capacitor and R is the resistance in series with the capacitor.

The voltage V across the capacitor is therefore zero when t = 0 so that the potential difference across the 8 Ω resistor R₂ immediately after closing the switch S is equal to the charging voltage V₀ which is equal to the voltage across R₃ given by

V₀ = [12 V/(8+8) Ω] ×8 Ω = 6 V.

(c) After reaching steady state in case (b), the potential difference across the resistor R₂ will be zero since the capacitor will be fully charged and there will be no charging current through R₂ to produce a voltage drop across it.

(d) The current through R₂ will rise to the final steady value abruptly if a resistor is used instead of the capacitor mentioned in (b). The current through R₂ immediately after closing the switch S is therefore equal to the current resulting from the division of the main current i₁.

Now, i₁ = 12 V/(8+6) Ω = 6/7 A.

[The 6 Ω resistance substituted above is the parallel combined value of R₃ = 8 Ω and (R₂ + 16 Ω) = 24 Ω].

The above current gets divided between R₃ and the (R₂ + 16 Ω) branch.

The current through R₂ and 16 Ω = Main current×(Resistance of the other branch/Total resistance) = (6/7)×[8/(24+8)] = (3/14) A.

(e) As the potential difference across the capacitor is given by V = V₀(1– e^–t/RC) where V₀ = 6 volts, the potential difference across the resistor R₂ is V₀ – V₀(1– e^–t/RC) = V₀ e^–t/RC.

[This follows from Kirchoff’s voltage law: V_C + V_R2 = V₀]

The nature of variation of the potential difference across R₂ in this case (b) is as shown by curve (b).

When a 16 Ω resistor is used in place of the capacitor, the final current through R₂ is (3/14) A as already shown. The potential difference across R₂ in this case is (3/14)×8 volts = 1.7 volt, nearly. As the current through R₂ will rise abruptly from zero, the potential difference across R₂ will rise abruptly to 1.7 volt as shown by curve (d).

(2) In the above circuit, suppose the switch S was closed for sufficiently long time so that steady state was reached. In this steady state the.switch S is opened at time t = 0.

(a) Will there be any change in the directions of the currents i₂ and i₃ on opening the switch S? Put a tick mark against the correct option among the following:

No change____

Current i₂ alone is reversed____

Current i₃ alone is reversed____

Both i₂ and i₃ are reversed____

Justify your answer.

(b) Calculate the potential difference across resistance R₂ immediately after opening switch S

(c) Calculate the total amount of heat generated in resistance R₂ after switch S is opened.

(d) Calculate the power dissipated in resistance R₂ in the steady state, before opening the switch S.

(a) On opening the switch S, the inductor will try to maintain the current flowing in it because of its inductance (electrical inertia). Therefore, the direction of i₂ will be unchanged. But the direction of i₃ will be reversed since the current now flows because of the voltage induced in the inductance.

(b) Under steady state, the battery delivers a current of 1 A through R₁ since the total resistance across the battery is 8+4 = 12 Ω. [Here 4 Ω is the parallel combined value of R₂ and R₃]. The potential difference across R₂ (and R₃) is therefore 4V. The current in the inductance branch cannot change immediately on opening the switch S so that the potential drop across R₂ will be 4 volt itself immediately after opening the switch S.

(c) Since the voltage across R₂ is 4 V under steady state with the switch S closed, the current through R₂ is 4 V/8 Ω = 0.5 A.The switch is opened at t =0 and the current (I₀) at t = 0 is 0.5 A itself. The current (I ) decays in accordance with the equation,

I = I₀ e^{– R t /L} where R = R₂+R₃ = 8+8 =16 Ω (which is the total resistance in series with L, on opening the switch S).

Therefore, we have I = 0.5 e^–16t/2 = 0.5 e^{– 8 t}

Heat produced per second in the resistance R₂ = I ²R₂

= 0.25 e^{– 16 t} ×8 = 2 e^{– 16 t}

Therefore, total heat (H) produced in R₂ after opening the switch S is given by

H = ₀ ∫^∞ 2e^{– 16 t} = 0.125 joule.

(d) In the steady state before opening the switch S, the current flowing in R₂ is 0.5 A as shown above. The power (P) dissipated in the resistance R₂ is given by

P = I²R₂ = 0.5²×8 = 2 watt.

Note: Even if your theoretical steps, substitutions and calculations are correct, you will lose points if proper units are not supplied. This is especially important in questions involving numerical values.

Work out as many questions as possible to make you strong in facing the AP Physics Exam.

AP Physics C - Free Response Questions involving Inductance and Capacitance (for Practice)

In the post dated 23-12-2007 a free response question on inductance was discussed. You can access that question along with other questions related to inductance by clicking on the label ‘inductance’ below this post. The circuit pertaining to the question referred to above is shown in the figure.

Let us modify the question to facilitate more practice. Here are two questions:

(1) In the circuit shown, the battery and the inductor have negligible resistance. The currents through R₁, R₂ and R₃ are i₁, i₂ and i₃ respectively when the switch S is kept closed. The switch S is closed at time t = 0.

(a) Calculate the potential difference across the 8 Ω resistor R₂ immediately after closing the switch S.

(b) If the inductance is replaced by an uncharged 10 μF capacitor what will be the potential difference across the 8 Ω resistor R₂ immediately after closing the switch S? Justify your answer.

(c) After reaching steady state, in case (b) what will be the potential difference across the resistor R₂? Justify your answer.

(d) If a 16 Ω resistor is used instead of the capacitor mentioned in (b), calculate the current through R₂ immediately after closing the switch S.

(e) Show the nature of variation of the potential difference across R₂ with time t in cases (b) and (d).

(2) In the above circuit, suppose the switch S was closed for sufficiently long time so that steady state was reached. In this steady state the switch S is opened at time t = 0.

(a) Will there be any change in the directions of the currents i₂ and i₃ on opening the switch S? Put a tick mark against the correct option among the following:

No change____

Current i₂ alone is reversed____

Current i₃ alone is reversed____

Both i₂ and i₃ are reversed____

Justify your answer.

(b) Calculate the potential difference across resistance R₂ immediately after opening the switch S.

(c) Calculate the total amount of heat generated in resistance R₂ after switch S is opened.

(d) Calculate the power dissipated in resistance R₂ in the steady state, before opening the switch S.

Try to answer the above two questions which carry 15 points each. You can take about 15 minutes for answering each question.

I’ll be back shortly with model answers for your benefit.

Answer to AP Physics Free Response Question on Inductance

A free response question involving inductance was posted on 23^rd December 2007 for your practice. This was the question:

In the circuit shown in the figure, the battery and the inductor have negligible resistance. The currents through R₁, R₂ and R₃ are i₁, i₂ and i₃ respectively. The switch S is closed at t = 0.

(a) Calculate the currents i₁, i₂ and i₃ immediately after closing the switch S (when t = 0).

(b) Calculate the currents i₁, i₂ and i₃ when steady state is reached.

(c) Show the nature of variation of the current through the battery graphically, with time ‘t’ on the X-axis and the current (i₁) on the Y-axis.

(d) What are the values of the slopes of the above graph at t = 0 and at t = ∞.

(e) The switch S is opened and the resistance R₃ is disconnected. If the switch S is now closed, what time is required for the current to attain 63.2% of the final steady value?

As promised, I give below the answer:

(a) Since the parallel branch containing R₃ does not contain any inductance, the current through that branch (i₃) rises to the maximum possible value in this condition immediately on closing the switch S. The same current now flows through R₁ since the initial current i₂ ( at t = 0) through the branch containing the inductance L and the resistance R₂ is zero. [Note that in an LR circuit, the current I at the instant ‘t’ is given by I = I₀(1– e^–Rt/L) where I₀ is the final steady current]

Therefore, when t = 0, i₁ = i₃ = (12 V)/ (8 Ω + 8 Ω ) = 0.75 A and i₂ = 0

(b) Steady state is reached when t = ∞. In this condition, the current is controlled by the resistances only [since dI/dt is zero and no opposing induced voltage L(dI/dt) is developed across the inductance]. The circuit then behaves as made of the parallel combination of R₂ and R₃ (having effective value of 4 Ω) in series with the battery and the resistance R₁. The effective resistance of the circuit is (8 + 4) Ω = 12 Ω and the current delivered by the battery (i₁) is (12 V)/ (12 Ω) = 1 A.

Since this current is equally divided between the two parallel branches, current through each branch is 0.5 A. So, i₂ = i₃ = 0.5 A.

(c) The variation of the current i₁ with time t is shown in the adjoining graph. [The current grows exponentially in the inductive branch and this is why the current delivered by the battery finally rises to the steady value of 1 A as shown].

(d) The slope of the graph when t = o is infinity since the current grows abruptly to 0.75 A because of the purely resistive parallel branch containing R₃. The slope when t = ∞ is zero since the current has settled to the final steady value.

(e) When R₃ is disconnected, We have a series LR circuit with L = 2 H and R = R₁ + R₂ = 16 Ω. The time constant of the circuit is L/R = 2/16 = 0.125 s. The current grows to 63.2 % of the final steady current within this time.

[Time constant is L/R which is the time required for the current to become (1 – 1/e) times the final steady current. This is obtained by putting t = L/R in the equation,

I = I₀(1– e^–Rt/L)].

Merry Christmas!

AP Physics C-Free Response Question on Inductance (for Practice)

In the circuit shown in the figure, the battery and the inductor have negligible resistance. The currents through R₁, R₂ and R₃ are i₁, i₂ and i₃ respectively. The switch S is closed at t = 0.

(a) Calculate the currents i₁, i₂ and i₃ immediately after closing the switch S(when t = 0).

(b) Calculate the currents i₁, i₂ and i₃ when steady state is reached.

(c) Show the nature of variation of the current through the battery graphically, with time ‘t’ on the X-axis and the current (i₁) on the Y-axis.

(d) What are the values of the slopes of the above graph at t = 0 and at t = ∞.

(e) The switch S is opened and the resistance R₃ is disconnected. If the switch S is now closed, what time is required for the current to attain 63.2% of the final steady value?

The above question carries 15 points. The division of points among the sections (a), (b), (c), (d) and (e) can be as 4 + 3 + 4 +2 +2. Try to answer the above question within 15 minutes or less. I’ll be back with the answer shortly.

AP Physics C - Multiple Choice Questions on Inductance

1. An air cored coil of self inductance L has N turns of fine insulated copper wire wound on a former of cross section area A. If the area and number of turns are doubled and the core is a medium of relative permeability 1000, the self inductance of the coil will be

(a) 8000 L (b) 4000 L (c) 8×10^–3 L (d) 4×10^–3 L (e) L

Self inductance of a coil is directly proportional to the area of cross section, relative permeability of the core and the square of the number of turns. The answer therefore is 8000 L.

2. A straight air cored solenoid has length 1 m, area of cross section 10 cm² and total number of turns 2000. If a current of 1 A flowing in it is reversed in 0.1 s, the average emf induced in it will be nearly

(a) 100V (b) 10 V (c) 1 V (d) 0.5 V (e) 0.1 V

The emf induced is given by

ε = –L(dI/dt),

where dI is the change in current in the time dt and L = μ₀n²Aℓ where ‘ℓ’ is the length of the solenoid, ‘A’ is its cross section area, ‘n’ is the number of turns per metre and μ₀ is the permeability of air (or free space) which you must remember as 4π×10^–7.

Ignoring the negative sign which is because of Lenz’s law, we have

ε = μ₀n²Aℓ (dI/dt)

Substituting, ε = 4π×10^–7×(2000)²×(10×10^–4)×1×[1– (–1)]/(0.1) volt.

Note that the change in the current is 2 A ( from 1 A to –1 A) since the current is reversed.

Therefore, ε = 0.1 (nearly).

3. In the circuit shown, the switch S is kept closed so that the current flowing in the circuit is the final steady value. When the switch S is opened, the time constant for the decay of current in the inductance is

(a) L/R

(b) L/5R

(c) 5L/4R

(d) 4L/5R

(e) L/3R

The current decays through the resistors R and 4R in series with L and therefore the time constant is L/(R+4R) = L/5R.

4. In the above question, suppose the resistance 4R is disconnected and is then connected between the inductance L and the negative terminal of the battery. The switch S is kept closed. If the switch is opened, the time constant for the decay of current in L is

(a) L/R

(b) infinity

(c) zero

(d) L/5R

(e) 5L/4R

This is a simple case which may however generate confusion in your mind. Note that when you open the switch, the resistance in series with L is infinite and the time constant will be L/∞ = 0.

Physically, the current becomes zero abruptly since there is no conducting path for the current to flow.

[Discharging of a charged capacitor through a resistor is an easy thing since the capacitor can retain the charge during the small time during which the capacitor is disconnected from the charging battery and then connected to the resistor, using a charge-discharge key. In the case of the LR circuit, for studying the current decay pattern, after establishing the steady final current through the circuit, you have to short circuit the leads connected to the battery and immediately disconnect the battery so that the battery is not damaged].

5. A small plane circular coil of radius ‘r’ having ‘n’ turns is placed at a distance ‘d’ (d >> r) from a straight vertical conductor. The coil and the straight conductor are contained in the same vertical plane. The mutual inductance between the coil and the straight conductor is

(a) μ₀ nr²/2πd

(b) π μ₀ nr²/2d

(c) μ₀ nr²/2

(d) μ₀ nr²/2π²d

(e) μ₀ nr²/2d

The mutual inductance is given by

M = Ф/I where I is the current in the straight conductor and Ф is the magnetic flux through the coil.

Since the coil is small, the magnetic field produced at the coil by the current in the straight conductor can be assumed to be constant over the entire area of the coil so that

Ф =nBA = n(μ₀I/2πd)(πr²)

Therefore, M = μ₀ nr²/2d

You will find similar multiple choice questions at physicsplus: Multiple Choice Questions (MCQ) involving Inductance

AP Physics C: Inductance- Equations to be remembered

The following equations are to be remembered in this section:

1. Self inductance (coefficient of self induction) ‘L’ of a circuit is defined in terms of magnetic flux linkage as

L = Ф/I where Ф is the magnetic flux (in weber) linked with the circuit when a current I ampere flows in the circuit.

L is defined in terms of the self induced voltage ‘ε’ as

L =ε /(dI/dt) where dI/dt is the rate of change of current in the circuit.

The induced voltage opposes the change in the current in the circuit in accordance with Lenz’s law and hence the above equation has to be strictly

L = – ε/(dI/dt). But remember that self inductance is a positive quantity. It is the electromagnetic analogue of mass (which has the property of inertia) in mechanics.

[Induced voltage (emf) is represented by the symbol V also, instead of ε].

It will be better to remember the above equation as

ε = –L(dI/dt).

2. An inductance L carrying a current I possesses energy U_L given by

U_L = ½ LI², which is stored in the magnetic field established by the current.

3. Self inductance of a infinitely long straight air-cored solenoid is given by

L = μ₀n²Aℓ where ‘ℓ’ is the length of the solenoid, ‘A’ is its cross section area, ‘n’ is the number of turns per metre and μ₀ is the permeability of air (or free space).

[By the term infinitely long solenoid, we mean a solenoid with radius negligibly small compared to the length]

Note that the self inductance is proportional to the square of the number of turns.

4. Mutual inductance (coefficient of mutial induction) is defined in terms of magnetic flux linkage as

M = Ф/I where Ф is the magnetic flux (in weber) linked with the secondary circuit when a current I ampere flows in the primary circuit.

M is defined in terms of the induced voltage ‘ε’ in the secondary circuit as

M = ε /(dI/dt) where dI/dt is the rate of change of current in the primary circuit.

The induced voltage opposes the change in the current in the primary circuit in accordance with Lenz’s law and hence the above equation has to be strictly

M = – ε /(dI/dt). But remember that mutual inductance is a positive quantity.

It will be better to remember the above equation as

ε = –M(dI/dt).

5. Mutual inductance between an infinitely long straight solenoid and a short secondary coil wound round it (outside) at the middle is given by

M = μ₀nNA where 'n' and A are respectively number of turns per metre and the cross section area of the solenoid (primary) and N is the total number of turns in the secondary coil.

[If the secondary coil is inside the solenoid, the cross section area of the secondary coil is to be used in place of A].

6. Mutual inductance M is related to the primary and secondary self inductances L₁ and L₂ as

M = K√(L₁L₂) where K is the coupling coefficient which can have a maximum value of one. This happens when the entire magnetic flux produced by the primary is linked with the secondary.

Note that self inductance and mutual inductance are directly proportional to the permeability of the core material. If a coil is wound on a core of relative permeability μ_r, its inductance will be μ_r times the value with air core. Further, the self inductance of a given coil (with a given core) is a constant where as the mutual inductance between two given coils depends on the relative disposition of the coils.

7. Growth of current in a series LR circuit connected in series with a direct voltage is exponential and is given by

I = I₀(1– e^–Rt/L) where e is the base of natural logarithms and I₀ is the final current (at infinite time) given by I₀ = V/R, where V is the direct supply voltage.

8. Decay of current in the LR circuit is exponential and is given by

I = I₀e^–Rt/L

The growth and decay of current are shown in the adjoining figure.

9. Time constant of LR circuit = L/R

[You must remember the time constant of a CR circuit also, which is CR. The growth of charge Q on a capacitor connected in series with a direct voltage source of emf V volt is given by

Q= Q₀(1– e^–t/RC) where Q₀ is the final (maximum) charge equal to CV which is attained at infinite time.

When the capacitor having charge Q₀ is allowed to discharge through a resistance R, the decay of charge on the capacitor is given by

Q = Q₀e^–t/RC

The growth and decay of charge are shown in the adjoining figure.

10. Frequency of oscillations generated when a charged capacitor of capacitance C is discharged through an inductor of inductance L is given by

f = 1/2π√(LC)

The above equation can be written in terms of the angular frequency as

ω = 1/√(LC)

[The above equation is obtained by solving Kirchoff’s voltage equation for the LC circuit:

L(dI/dt) + Q/C =0, which can be rewritten as

L(d²Q/dt² ) + Q/C =0.

Therefore, d²Q/dt² = – Q/LC, which is similar to the equation of the simple harmonic motion of a mass ‘m’ attached to a spring of force constant ‘k’, written as

m (d²x/dt²) = – kx

or, d²x/dt² = –ω²x, where ω = √(k/m) is the angular frequency of mechanical oscillations of the spring-mass system.

The angular frequency of oscillations of the LC circuit (angular frequency of variation of charge on the capacitor) is similarly given by

ω = 1/√(LC)]

In the next post, questions (involving inductance) of the type you can expect in AP Physics C Exam will be discussed.