THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 122 Assignment # 10 SOLUTIONS:
 
AC CIRCUITS & EM WAVES
 
Fri. 15 Mar. 2002 - finish by Fri. 22 Mar.

1.

A typical ``light dimmer" switch consists of a variable inductor L connected in series with the light bulb B as shown in the diagram on the right. The power supply produces 120 V (rms) at 60.0 Hz and the light bulb is marked ``120 V, 100 W." Assume that the light bulb is a simple resistor whose resistance does not depend on its temperature.

(a)

What maximum inductance L is required if the power in the light bulb is to be varied by a factor of five?   ANSWER:  Since we are looking for a given power ratio, the answer will be independent of the rms driving voltage (see below). The frequency is $\omega = 2\pi$(60 Hz) = 377 s^-1. The light bulb draws $P = {\cal E}^2/R = 100$ W when the voltage drop across it is ${\cal E} = 120$ V, so $R = {\cal E}^2/P = (120)^2/100 = 144$ $\Omega$. Now, for an LR circuit the impedance is just $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + \omega^2 L^2}$   (i.e. we just omit the capacitive reactance term)¹ so that $I_m = {\cal E}_m/Z$ and the average power dissipated in the resistance (light bulb) is $\overline{P} = {1\over2} I_m^2 R = {\displaystyle {R \over 2} \cdot {{\cal E}_m^2 \over Z^2} = {R \over 2} \cdot {{\cal E}_m^2 \over R^2 + \omega^2 L^2} }$. When $L \to 0$ we get a maximum value of $\overline{P}({\rm max}) = {\displaystyle {\cal E}_m^2 \over 2R}$ and when $L \to L_{\rm max}$ we get a minimum value of $\overline{P}({\rm min}) = {\displaystyle {R \over 2} \cdot {{\cal E}_m^2 \over R^2 + \omega^2 L_{\rm max}^2}}$. We want the ratio of these to be five: ${\displaystyle {\overline{P}({\rm max}) \over \overline{P}({\rm min})} = {R^2 . . . . . . m max}^2 = 5 R^2 \quad \Longrightarrow \quad \omega^2 L_{\rm max}^2 = 4 R^2 }$ or ${\displaystyle \omega L_{\rm max} = 2 R }$ giving ${\displaystyle L_{\rm max} = {2 R \over \omega} = {2 \times 144 \over 377} }$   or    .
 

(b)

Could one use a variable resistor in place of the variable inductor? If so, what maximum resistance R would be required? Why isn't this done? (Variable resistors [rheostats] are generally cheaper than variable inductors.)   ANSWER:  The circuit would be cheaper to build using a rheostat in place of the variable inductance and it would work fine: the two resistances in series would just add up to one net resistance $R_{\rm tot} = R_{\rm light} + R_{\rm var}$ so that $I_{rms} = {\cal E}_{rms}/R_{\rm tot}$ and $\overline{P}_{\rm light} = I_{rms} R_{\rm light}^2 = {\cal E}_{rms} R_{\rm light}^2/(R_{\rm light} + R_{\rm var})$, which can be varied from a minimum of ${\cal E}_{rms} R_{\rm light}^2/(R_{\rm light} + R_{\rm max})$ to a maximum of ${\cal E}_{rms} R_{\rm light}$, the ratio of which $[R_{\rm light}/(R_{\rm light} + R_{\rm max})]$ can easily be made equal to 5. The problem with this simpler design is that when the light is dimmed ( $\overline{P}_{\rm light}$ minimized), considerable power $[I_{rms} R_{\rm var}^2]$ is being uselessly dissipated in the rheostat! A pure inductance, by contrast, does not draw any power; it merely shifts the phase of the voltage and current.

 

2.

LCR CIRCUIT TIME-DEPENDENCE:   In the circuit shown, the battery has negligible internal resistance. The switch S is closed for a long time, then opened. Describe qualitatively what happens in the circuit after the switch is closed and then after it is opened again, for two cases:
(a)   $R/2L > 1/\sqrt{LC}$  and
(b)   $R/2L < 1/\sqrt{LC}$.

ANSWER:  This problem has a tricky part, namely the behaviour when the switch is first closed. In this situation $\sum \Delta {\cal E} =0$ around the leftmost loop requires that a current $I_1 = {\cal E}/R$ immediately start flowing through R and continue that way forever. So, for all practical purposes, R simply serves to drain the battery and has no effect at all on the voltages in the outer loop, which consists of a battery driving an undamped LC circuit - which will therefore at its resonant frequency $\omega_0 = 1/\sqrt{LC}$ without any damping at all, regardless of the size of R!
 
When the switch is opened again, we do get different behaviour depending on the ratio of R to L. Specifically, if we define $\lambda = R/2L$ and $\omega_0 = 1/\sqrt{LC}$, for we have an ``overdamped'' oscillator in which the current will decay away exponentially , whereas for we will see at the frequency whose with a time constant $\tau = 1/\lambda$.

 

3.

ELECTROMAGNETIC WAVE:   The electric field associated with a plane electromagnetic wave is given by

$$E_x = 0, \quad E_y = 0 \quad \hbox{\rm and} \quad E_z = E_0 \sin[k(x - ct)] ,$$

where $E_0 = 2.34 \times 10^{-4}$ V/m and $k = 9.72 \times 10^6$ m^-1. The wave is propagating in the +x direction.

(a)

Write expressions for all three components of the magnetic field associated with the wave.
ANSWER:  The wave propagates in the direction given by $\Vec{E} \times \Vec{B}$ (see textbook's section on the POYNTING VECTOR) and so, since $\Vec{E}$ points in the $\kH$ direction and the wave propagates in the $\iH$ direction, we must have $\iH = \kH \times \hat{B}$ where $\hat{B}$ is the unit vector in the $\Vec{B}$ direction. The only unit vector satisfying this criterion is $\hat{B} = - \jH$ and so $\Vec{B}$ is in the negative y direction when $\Vec{E}$ is in the positive z direction. Both must oscillate sinusoidally with the same wavelength and frequency, so they share the same propagation speed c, the same wavenumber k and the same angular frequency $\omega = ck$. This allows us to write down the answer:    .
Having established this, we need only find the relationship between the amplitudes of the E and B fields. Their magnitudes must satisfy $\left\vert\partial E \over \partial x \right\vert = \left\vert\partial B \over \partial t \right\vert$, giving $k E_0 = \omega B_0$ or $B_0 = {k \over \omega} E_0$ or  .   (This also follows from E=cB [always valid].)
 

(b)

Find the wavelength of the wave.   ANSWER:  All you need for this part is the universal relationship

$$k \equiv {2\pi \over \lambda} \quad \Longrightarrow \quad \lambda = {2\pi \over k} = {2\pi \over 9.72 \times 10^6}$$

or    .

NOTE:   This was a rather trivial problem, designed partly to remind you of the properties of waves, our subject for most of the rest of the course. I have been rather longwinded in the solutions; you may be much more economical with words, but be sure you always make your basic reasoning clear!