I.E. Irodov Solutions on Electric Oscillations

1: Due to a certain cause the free electrons in a plane copper plate shifted over a small distance x at right angles to its surface. As a result, a surface charge and a corresponding restoring force emerged, giving rise to so-called plasma oscillations. Find the angular frequency of these oscillations if the free electron concentration in copper is n, = 0.85 x10²⁹ m^-1.

If the electrons (charge of each electron = – e) are shifted by a same distance x, a net + ve charge density (per unit area) is induced on the surface. This will result in an electric field E = nex/ε_o in the direction of x and a restoring force on an electron of – ne²x/ε_o

2: An oscillating circuit consisting of a capacitor with capacitance C and a coil of inductance L maintains free undamped oscillations with voltage amplitude across the capacitor equal to V_m. For an arbitrary moment of time find the relation between the current I in the circuit and the voltage V across the capacitor. Solve this problem using Ohm’s law and then the energy conservation law.

Since there are no sources of emf in the circuit, Ohm’s 1 law reads

q/c = -L(dI/dt)

Where q = change on the capacitor, I = dq/dt = current through the coil. Then

d²q/dt² + ω_o²q = 0, ω_o² = 1/LC

The solution of this equation is,

q = q_m cos(ω_ot + α)

We know, v_m = q_m/C then

I = -ω_o CV_m sin(ω_ot + α)

V = V_m cos(ω_ot – α)

V² + I²/(ω_o²C²) = V²_m

V² + L²I²/C = V²_m

By energy conservation,(1/2)LI² + q²/2C = constant

When the P.D. across the capacitor takes its maximum value V_m , the current I must be zero. Thus “constant” = (1/2)V_m²

Hence, ,(1/C)LI² + V² = V_m² .

3: An oscillating circuit consists of a capacitor with capacitance C, a coil of inductance L with negligible resistance, and a switch. With the switch disconnected, the capacitor was charged to a voltage V_m, and then at the moment t =0 the switch was closed. Find:

(a) the current I (t) in the circuit as a function of time;

(b) the emf of self-inductance in the coil at the moments when the electric energy of the capacitor is equal to that of the current in the coil.

After the switch was closed, the circuit satisfies

-L(dI/dt) = q/C

d²q/dt² + ω_o²q = 0 => q = CV_m cos(ω_ot)

Where we have used the fact that when the switch is closed, we must have

V=q/C = V_m, I = dq/dt = 0 at t=0.

(a)I = aq/dt = -CV_m ω_o sin ω_ot = -V_m √(C/L) sin ω_ot

(b)The electrical energy of the capacitor is q²/2C α cos²ω_ot and of the inductor is (1/2)LI² α sin²ω_ot

The two are equal when ω_ot = π/4

At that instant the emf of the self-inductance is -L (di/dt) = V_mcosω_ot = V_m/√2.

4: In an oscillating circuit consisting of a parallel-plate capacitor and an inductance coil with negligible active resistance the oscillations with energy W are sustained. The capacitor plates were slowly drawn apart to increase the oscillation frequency η-fold. What work was done in the process?

In the oscillating circuit, let q = q_m cosωt be the change on the condenser where ω² = (1/LC) and C is the instantaneous capacity of the condenser (S = area of plates)

C = ε_oS/y

y = distance between the plates. Since the oscillation frequency increases η fold, the quantity ω² = y/ ε_oS L changes η² fold and so does y i.e. changes from y₀ initially to η²y₀ finally. Now the P.D. across the condenser is :

V = (q_m/C) cosωt = (yq_m/ε_oS )cosωt

And hence the electric field between the plates is T

E = (q_m/ε_oS )cosωt

Thus, the change on the plate being q_m cosωt , the force on the plate is

F = (q²_m/ε_oS )cos²ωt

Since this force is always positive and the plate is pulled slowly we can use the average force is q_m²/2ε_oS

And work done is A = (η² – 1) q_m²y_o/2ε_oS

But , q_m²y_o/2ε_oS = q_m²/2C_o =W is the initial stored energy.

Therefore, A = (η² – 1)W.

5: In an oscillating circuit shown in Figure below the coil inductance is equal to L = 2.5 mH and the capacitors have capacitances C₁= 2.0 μF and C₂ = 3.0 μF. The capacitors were charged to a voltage V = 180 V, and then the switch S_w was closed. Find: (a) the natural oscillation frequency; (b) the peak value of the current flowing through the coil.

The peak value of the current is 70 and it is related to the voltage V by the first equation, LI = V –

Or Lω_oI_o cosω_ot = V – (1/(c₁+c₂)

(The P.D. across the inductance is V at t = 0)

= V + 1/(C₁+C₂). I_o/ω_o(cos ω_ot – 1)

Hence, I₀ = (C₁ + C₂) ω_oV = V√[(C₁ + C₂)/L] = 8.05 A.

6: An electric circuit shown in Figure below has a negligibly small active resistance. The left-hand capacitor was charged to a voltage V₀ and then at the moment t = 0 the switch S_w was closed. Find the time dependence of the voltages in left and right capacitors.

Initially q₁ = C V₀ and q₂ = 0. After the switch is closed change flows and we get,

q₁ + q₂ = Cv₀

q₁/C + L(dI/dt) – q₂/C = 0

Also,

Hence,

The solution of this equation subject to

I = 0 at t = 0 is I = I₀ sinω_ot.

On integrating, we have

q₁ = A – (I₀/ω_o) cosω_ot and q₂ = B + (I₀/ω_o) cosω_ot

Therefore, (A-B)/C – (2I₀/ω_oC) cosω_ot + LI₀ω_o cosω_ot = 0

A = B = CV₀/2 and CV₀/2 + I₀/ω₀ = 0

So, q₁ = CV₀/2 (1 + cosω_ot) and q₂ = CV₀/2 (1 – cosω_ot)

7: An oscillating circuit consists of an inductance coil L and a capacitor with capacitance C. The resistance of the coil and the lead wires is negligible. The coil is placed in a permanent magnetic field so that the total flux passing through all the turns of the coil is equal to φ. At the moment t = 0 the magnetic field was switched off. Assuming the switching off time to be negligible compared to the natural oscillation period of the circuit, find the circuit current as a function of time t.

The flux in the coil is

ϕ(t) = ϕ for t < 0 and ϕ(t) = 0 for t > 0

The equation of the current is

…(1)

This shows, LC (d²I/dt²) + I = 0 or with ω₀² = 1/LC

I = I_o sin(ω_ot + α)

Putting values in (1) -LI_o ω₀ cos(ω_ot + α) = -I_o/ω₀C[cos(ω_ot + α) – cos α

This implies, cos α = 0, therefore, I = I_o cos(ω_ot) and , I = – I_o cos(ω_ot)

From Faraday’s law: ε = -dϕ/dt = -L(dI/dt)

Or integrating form, t = – ε to – ε where ε ->0

ϕ = LI₀ with + sign in I or I = (ϕ/L) cos(ω_ot)

8: The free damped oscillations are maintained in a circuit, such that the voltage across the capacitor varies as V = V_mθ^-βt cos ωt.

Find the moments of time when the modulus of the voltage across the capacitor reaches (a) peak values; (b) maximum (extremum) values.

(a) Given The phrase ‘peak values’ is not clear. The answer is obtained on taking |cos ωt| = 1 i.e. t = πn/ω.

(b) For extrema, dV/dt = 0

-β cosωt – ω sinωt = 0

tan ωt = -β/ω

ωt = nπ + tan^-1(-β/ω).

9: A certain oscillating circuit consists of a capacitor with capacitance C, a coil with inductance L and active resistance R, and a switch. When the switch was disconnected, the capacitor was charged; then the switch was closed and oscillations set in. Find the ratio of the voltage across the capacitor to its peak value at the moment immediately after closing the switch.

The equation of the circuit is

10: A circuit with capacitance C and inductance L generates free damped oscillations with current varying with time as I = l_m e^-βt sin ωt. Find the voltage across the capacitor as a function of time, and in particular, at the moment t = 0.

-dQ/dt = I = I_m e^-βt sin ωt

= gm I_m e^{-βt +iωt} (gm means imaginary part)

Then,

11: An oscillating circuit consists of a capacitor with capacitance C = 4.0 μF and a coil with inductance L = 2.0 mH and active resistance R = 10Ω. Find the ratio of the energy of the coil’s magnetic field to that of the capacitor’s electric field at the moment when the current has the maximum value.

I = I_m e^-βt sin ωt

Β = R/2L, ω₀ = √(1/LC) and ω = √(ω₀² – β²)

I =

Then, q = I_m e^-βt [sin (ωt+ δ)/√(ω² + β²)] and tan δ = ω/β

Thus, W_M = (1/2)L I_m² e^-2βt sin² ωt

Current is maximum when, d/dt(e^-βt) sinωt = 0

-β sin ωt + ω cos ωt = 0

or tan ωt= ω/β = tan δ

that is, ωt = n π + δ

Hence,

= 1/(4 β²/ω₀²) = ω₀²/4 β² = 1/LC x L²/R² = 5

(W_M is the magnetic energy of the inductance coil and W_E is the electric energy to the capacitor.)

12: An oscillating circuit consists of two coils connected in series whose inductances are L1 and L2, active resistances are R1 and R2, and mutual inductance is negligible. These coils are to be replaced by one, keeping the frequency and the quality factor of the circuit constant. Find the inductance and the active resistance of such a coil.

L = L₁ + L₂, R = R₁ + R₂.

13: How soon does the current amplitude in an oscillating circuit with quality factor Q = 5000 decrease η = 2.0 times if the oscillation frequency is v = 2.2 MHz?

Q = π/βT or β = π/QT

Now, βt = ln η

so, t = (ln η/π) QT = Q ln η/πv = 0.5 ms

14: An oscillating circuit consists of capacitance C = 10μF, inductance L = 25 mH, and active resistance R = 1.0 Ω. How many oscillation periods does it take the current amplitude to decrease e-fold?

Current decreases e fold in time

t = 1/βt = 2L/R sec = 2L/RT oscillations

= 2L/R (ω/2 π)

=

= 15.9 oscillations

15: How much (in percent) does the free oscillation frequency ω of a circuit with quality factor Q = 5.0 differ from the natural oscillation frequency ω0 of that circuit?

16: In a circuit shown in figure below the battery emf is equal to 2.0 V, its internal resistance is r = 9.0Ω, the capacitance of the capacitor is C = 10μF, the coil inductance is L = 100 mH, and the resistance is R = 1.0 Ω. At a certain moment the switch S_w was disconnected. Find the energy of oscillations in the circuit

(a) immediately after the switch was disconnected;

(b) t = 0.30 s after the switch was disconnected.

At t = 0 current through the coil = ε/(R+r)

P.D. across the condenser = ε/(R+r)

(a) At t = 0, energy stored = W₀

(b) The current and the change stored decrease as e^-tR/2L so energy decreases as e^-tR/L

W = W₀ e^-tR/L = 0.10 mJ

17: Damped oscillations are induced in a circuit whose quality factor is Q = 50 and natural oscillation frequency is v₀ = 5.5 kHz. How soon will the energy stored in the circuit decrease η = 2.0 times?

18: An oscillating circuit incorporates a leaking capacitor. Its capacitance is equal to C and active resistance to R. The coil inductance is L. The resistance of the coil and the wires is negligible. Find:

(a) The damped oscillation frequency of such a circuit;

(b) its quality factor.

In a leaky condenser

dq/dt = I-I’ where I’ = V/R = leak current

Now, V = q/c = -L (dI/dt) = -L (d/dt)(dq/dt + V/R)

= -L(d²q/dt²) – L/RC (dq/dt)

Or

Then, q = q_me^-βtsin(ωt +α)

(a) β = 1/2RC, ω₀² = 1/LC and ω = √(ω₀² – β²)

= √(1/LC – 1/4R²C²)

(b) Q = ω/2 β = RC √(1/LC – 1/4R²C²)

= (1/2) √(4R²C/L – 1)

19: Find the quality factor of a circuit with capacitance C = 2.0 μF and inductance L = 5.0 mH if the maintenance of undamped oscillations in the circuit with the voltage amplitude across the capacitor being equal to V_m = 1.0 V requires a power (P) 0.10 mW. The damping of oscillations is sufficiently low.

Given, V = V_me^-βtsin(ωt), ω = ω₀βT << 1

Power loss = (energy loss per cycle)/T

= (1/2) CV_m² x 2β

(energy decreases as W₀ e^-2βt so loss per cycle is W₀ x 2βT)

Thus, <p> = (1/2) CV_m² x R/L

Or R = 2<p>/ V_m² x L/C

Hence, Q = (1/R) √(L/C) = √(C/L) V_m²/2<p> = 100 (on substituting values)

20: What mean power should be fed to an oscillating circuit with active resistance R = 0.45Ω to maintain undamped harmonic oscillations with current amplitude I_m = 30 mA?

Energy is lost across the resistance and the mean power lass is

= R<I²> = (1/2)RI_m² = 0.2 mW.

This power should be fed to the circuit to maintain undamped oscillations.