A 3 m long solenoid with a diameter of 0.05 m has 3000 turns of wire with total
resistance of 1 Ω uniformly wound along its length. It has an air
core.

(a) Find the approximate self-inductance of this solenoid.

(b) If the solenoid and a 9 Ω resistor are
connected in parallel across the terminals of a 12 V battery, find the final
steady state current in the solenoid.

(c) If the battery is disconnected at time 0 leaving the circuit of the
solenoid and the resistor intact, find the current in the resistor as a function
of time.

(d) Show that the I^{2}R losses in the resistor after
t = 0
is equal to the energy stored in the solenoid at t = 0.

Solution:

- Concepts:

Self inductance, Kirchhoff's rules, RL circuits - Reasoning:

The coil, resistor and battery form an RL circuit, which can be analyzed using Kirchhoff's rules. - Details of the calculation:

(a) B = μ_{0}nI = magnitude of B inside the solenoid.

F = NBπR^{2}= LI = flux. L = Nμ_{0}nπR^{2}= μ_{0}n^{2}πR^{2}l, l = length of solenoid, n = N/l.

L = (4π*10^{-7 }N/A^{2})*(1000 m^{-1})^{2}*π(0.025 m)^{2}*3m = 7.4*10^{-3 }H.

(b) I_{s}= 12 V/1 Ω = 12 A.

(c) LdI/dt + RI = 0. R = 10 Ω.

I = I_{0}exp(-Rt/L) = 12 A exp(-10 Ωt/(7.4*10^{-3 }H)) = 12 A exp(-1351 t/s)

The current through the solenoid is decaying exponentially.

(d) P = I^{2}R = 144 A^{2}exp(-2703 t/s)*10 Ω.

Energy dissipated = ∫_{0}^{∞ }P dt = 0.53 J.

Energy stored in solenoid at t = 0: E = ½LI^{2}= ½(7.4*10^{-3 }H)(12 A)^{2}= 0.53 J.

A solenoid of length l
= 1 m and cross-sectional area A = 1 cm^{2} has N = 10000 turns of wire
with a total resistance of 1 ohm. At t = 0 the solenoid is
connected to a V = 100 V power supply. Find the total work done by
the power supply in the time interval it takes for the current to increase from
zero to (1 - 1/e) = 0.6321 times its steady state value.
Give a numerical answer.

Solution:

- Concepts:

Induced emf, transient - Reasoning:

As the current changes, an emf = -LdI/dt is induced. Here L is the self-inductance of the solenoid. - Details of the calculation:

Kirchhoff's loop rule yields

V = LdI/dt + IR, and I(t) = (V/R)(1 – exp(-(R/L)t).

The time it takes for the current to increase from zero to (1 - 1/e) times its steady state value is found from

(1 - 1/e) = 1 – exp(-(R/L)t), t = L/R.

The power supplied by the battery is

P = VI = dW/dt.

The total work done in the time interval from t = 0 to t = L/R is

W = (V^{2}/R) ∫_{0}^{L/R}(1 – exp(-(R/L)t)dt.

∫_{0}^{L/R}(1 – exp(-(R/L)t)dt = L/R – (L/R)(1 – 1/e).

W = LV^{2}/R^{2}(1/e) = 0.3678 LV^{2}/R^{2}.

We can also write W = ½ LI^{2}(t = L/R) + ∫_{0}^{L/R}I^{2}(t) R dt =

(L/2)(1 - 1/e)^{2}(V^{2}/R^{2}) + (V^{2}/R) ∫_{0}^{L/R}(1 – exp(-(R/L)t)^{2}dt,

since the battery has to supply the energy dissipated in the resistance and the energy stored in the magnetic field of the coil.

∫_{0}^{L/R}(1 – exp(-(R/L)t)^{2}dt = ∫_{0}^{L/R}(1 – 2exp(-(R/L)t + exp(-(2R/L)t)dt

= L/R – 2(L/R)(1 – 1/e) + (L/2R)(1 – 1/e^{2}) = (L/R)(2/e – 1/(2e^{2}) -½)

= (L/R)*0.168

W = 0.1998*LV^{2}/R^{2}+ 0.168*LV^{2}/R^{2}= 0.3678 * LV^{2}/R^{2}.

The self-inductance of a solenoid is L = μ_{0}N^{2}A/l .

For the given solenoid we therefore have L = 1.26*10^{-2}J/A^{2}, and V^{2}/R^{2}= 10^{4}A^{2}.

We therefore have for the work W = 46.4 J.

The circuit above is composed of a constant voltage source V, an inductor L, a
resistor R and two switches S_{1} and S_{2}. Prior to time t =
0 both switches S_{1} and S_{2} are open.

(a) What is the current flowing through the inductor L immediately after S_{1}
is closed at t = 0?

(b) What is the steady-state current flowing through the inductor L at a much
later time t_{1}?

(c) At time t_{2} > t_{1} S_{2} is closed and S_{1}
is opened at the same time. Write a differential equation and solve it to find
the current flowing through the inductor as a function of time .

(d) Sketch
the current flowing through the in the inductor L, from time t = 0 until a time
much later than t_{2}.

Solution:

- Concepts:

RL transient circuits - Reasoning:

We are asked to analyze the transient behavior of an RL circuit. - Details of the calculation:

(a) Immediately after S_{1}is closed the current through the inductor L is zero.

V – LdI/dt – RI = 0. dI/dt = -(R/L)I +V/L, I(t) = (V/R) (1 - exp(-Rt/L)).

(b) The steady state current through the inductor is I = V/R.

(c) LdI/dt + RI = 0

dI/I = (-R/L)dt, ∫ dI/I = -∫ (R/L)dt, ln(I) = (-R/L)t + C

At t = t_{2}, the current is V/R, so ln(V/R) = (-R/L)t_{2}+ C.

ln(I) = (-R/L)t + ln(V/R) + (R/L)t_{2}

ln(RI/V) = (-R/L)(t - t_{2})

RI/V = e^{(-R/L)(t - t}_{2}^{)}

I = (V/R)e^{-(t - t}_{2}^{)R/L}for any t > t_{2}

(d)

**I**n the circuit shown in the figure, switch S is closed at time t = 0.

(a) Find the current reading of each meter just after S is closed.

(b) What does each meter read long after S is closed?

Solution:

- Concepts:

Transient in RL circuits -
Reasoning:

At t = 0, the inductor acts like an open circuit, a long time after the switch is closed the inductor acts like a short circuit. -
Details of the calculation:

(a) A_{2}an A_{3}read zero, A_{1}an A_{4 }read I = (25 V)/(55 Ω) = 0.45 A.

(b) R_{eff }= (40 + 30/11) Ω = (470/11) Ω. I_{total}= V/R_{eff}= 55/94 A = 0.585 A.

The voltage across the 5 Ω, 10 Ω, and 15 Ω resistors in parallel is 1.6 V.

I_{2}= 0.319 A, I_{3}= 0.16 A, I_{4}= 0.106 A,

A_{1}reads 0.585 A, A_{2}reads 0.319 A, A_{3}reads 0.16 A, A_{4}reads 0.106 A.

(a) Suppose a capacitor is charged by a voltage source, and then switched to a resistor for discharging. Would a larger capacitance value result in a slower discharge, or a faster discharge? How about a larger resistance value?

(b) Now consider an inductor, "charged" by a current source and then switched to a resistor for discharging. Would a larger inductance value result in a slower discharge, or a faster discharge? How about a larger resistance value?