Practice Exam 9

AC Circuits & Electromagnetic Waves

Questions10

Time90 min

Points100

CalculatorAllowed

Use \(\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\,\text{m}^2)\), \(\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\text{m/A}\), \(c = 3.00\times10^8\ \text{m/s}\). Voltages and currents labelled “rms” are RMS values.

1Reactance & phase (conceptual)8 pts

(a) The frequency of a source is doubled. What happens to the reactance of an inductor, and of a capacitor? (b) The current through an unknown single element leads the voltage across it by \(90^\circ\). Which element is it? (c) In a DC circuit long after switching, how do an ideal inductor and capacitor behave? (d) A series RLC circuit is driven above its resonant frequency. Is it inductive or capacitive, and does the current lead or lag the source voltage?

Solution

(a) \(X_L = \omega L\) doubles; \(X_C = 1/(\omega C)\) halves.
(b) A capacitor (ICE: current before voltage). Charge must flow in before voltage builds.
(c) DC is \(\omega = 0\): \(X_L = 0\), so the inductor is a plain wire; \(X_C\to\infty\), so the capacitor is an open circuit.
(d) Above \(\omega_0\), \(X_L > X_C\), so \(\phi > 0\): inductive, and the current lags the voltage.

2A capacitor on the mains8 pts

A \(4.7\ \mu\)F capacitor is connected directly across a 120 V rms, 60 Hz outlet. Find (a) its reactance, (b) the rms and peak currents, (c) the average power it absorbs, and (d) the rms current if the frequency were 120 Hz.

Solution

\[ \text{(a)}\quad X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi(60)(4.7\times10^{-6})} = 564\ \Omega \] \[ \text{(b)}\quad I_{\mathrm{rms}} = \frac{120}{564} = \boxed{0.213\ \text{A}},\qquad I_0 = \sqrt2\,I_{\mathrm{rms}} = \boxed{0.301\ \text{A}} \]

(c) \(\phi = -90^\circ\), \(\cos\phi = 0\): zero average power. It stores energy for a quarter cycle and returns it the next.

(d) Doubling \(f\) halves \(X_C\) to \(282\ \Omega\), so \(I_{\mathrm{rms}} = 0.425\) A — twice as much.

3Series RLC: impedance, phase, element voltages14 pts

A series circuit with \(R = 60\ \Omega\), \(L = 0.40\) H and \(C = 20\ \mu\)F is connected to a 230 V rms, 50 Hz supply. Find (a) \(X_L\), \(X_C\), \(|Z|\) and the phase angle \(\phi\); (b) the rms current; (c) the rms voltage across each element, and show they are consistent with 230 V; (d) the power factor and average power.

Solution

\(\omega = 2\pi(50) = 314.2\) rad/s.

\[ \text{(a)}\quad X_L = (314.2)(0.40) = 125.7\ \Omega,\qquad X_C = \frac{1}{(314.2)(20\times10^{-6})} = 159.2\ \Omega \] \[ |Z| = \sqrt{60^2 + (125.7 - 159.2)^2} = \sqrt{3600 + 1122} = 68.7\ \Omega,\qquad \phi = \tan^{-1}\!\left(\frac{-33.5}{60}\right) = \boxed{-29.2^\circ} \] \[ \text{(b)}\quad I_{\mathrm{rms}} = \frac{230}{68.7} = \boxed{3.35\ \text{A}}\qquad(3.347\ \text{A unrounded, used below}) \] \[ \text{(c)}\quad V_R = (3.347)(60) = 200.8\ \text{V},\quad V_L = (3.347)(125.66) = 420.6\ \text{V},\quad V_C = (3.347)(159.15) = 532.7\ \text{V} \] \[ \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{200.8^2 + (-112.1)^2} = \sqrt{40\,321 + 12\,566} = 230\ \text{V}\ \checkmark \] \[ \text{(d)}\quad \cos\phi = \frac{R}{|Z|} = \frac{60}{68.7} = 0.873\ \text{(leading)},\qquad P = I_{\mathrm{rms}}^2R = (3.347)^2(60) = \boxed{672\ \text{W}} \]

The circuit is capacitive (\(\phi < 0\)): the current leads the voltage by \(29.2^\circ\). The inductor and capacitor voltages both exceed the supply; only their phasor sum with \(V_R\) must equal 230 V. Check (d) another way: \(VI\cos\phi = (230)(3.35)(0.873) = 672\) W.

4Resonance & Q10 pts

For the circuit of Question 3: (a) find the resonant frequency \(f_0\); (b) find the rms current at resonance; (c) find the quality factor and the bandwidth \(\Delta f\); (d) what rms voltage appears across the inductor at resonance?

Solution

\[ \text{(a)}\quad \omega_0 = \frac{1}{\sqrt{(0.40)(20\times10^{-6})}} = \frac{1}{\sqrt{8.0\times10^{-6}}} = 353.6\ \text{rad/s},\qquad f_0 = \frac{353.6}{2\pi} = \boxed{56.3\ \text{Hz}} \] \[ \text{(b)}\quad I_{\max} = \frac{V}{R} = \frac{230}{60} = \boxed{3.83\ \text{A}} \] \[ \text{(c)}\quad Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{60}\sqrt{\frac{0.40}{20\times10^{-6}}} = \frac{141.4}{60} = 2.36,\qquad \Delta f = \frac{f_0}{Q} = \frac{56.3}{2.36} = 23.9\ \text{Hz} \] \[ \text{(d)}\quad V_L = I_{\max}\,\omega_0L = (3.83)(353.6)(0.40) = 542\ \text{V} = Q\,V\ \checkmark \]

At 50 Hz the circuit was below \(f_0\), consistent with its being capacitive in Question 3. At resonance \(|Z| = R\), \(\phi = 0\), and the inductor and capacitor each carry 542 V that cancel each other.

5RMS & average power10 pts

A space heater is rated 1500 W at 120 V rms. Treat it as a pure resistor. Find (a) its resistance, (b) the rms and peak currents, (c) the maximum instantaneous power, and (d) explain the mistake in “\(P = V_0I_0 = 3000\) W”.

Solution

\[ \text{(a)}\quad R = \frac{V_{\mathrm{rms}}^2}{P} = \frac{120^2}{1500} = \boxed{9.6\ \Omega} \] \[ \text{(b)}\quad I_{\mathrm{rms}} = \frac{P}{V_{\mathrm{rms}}} = 12.5\ \text{A},\qquad I_0 = \sqrt2(12.5) = 17.7\ \text{A} \] \[ \text{(c)}\quad p_{\max} = V_0I_0 = (169.7)(17.7) = 3000\ \text{W} \]

(d) \(V_0I_0\) is the peak of \(p(t) = V_0I_0\cos^2\omega t\), which swings between 0 and 3000 W at 120 Hz. Its average is half that: \(\tfrac12V_0I_0 = V_{\mathrm{rms}}I_{\mathrm{rms}} = 1500\) W. Ratings are averages.

6Power-factor correction · EE application10 pts

A factory load draws 12 kW at power factor 0.75 lagging from a 480 V rms, 60 Hz supply. (a) Find the supply current. (b) What capacitance in parallel raises the power factor to 0.95 lagging? (c) Find the new current and the fractional reduction in line losses.

Solution

\[ \text{(a)}\quad I_1 = \frac{P}{V\cos\phi_1} = \frac{12\,000}{(480)(0.75)} = \boxed{33.3\ \text{A}} \] \[ \phi_1 = \cos^{-1}0.75 = 41.4^\circ,\quad Q_{\mathrm r1} = P\tan\phi_1 = (12\,000)(0.882) = 10.58\ \text{kvar} \] \[ \phi_2 = \cos^{-1}0.95 = 18.2^\circ,\quad Q_{\mathrm r2} = P\tan\phi_2 = (12\,000)(0.329) = 3.94\ \text{kvar} \]

(b) The capacitor must supply the difference, \(\Delta Q_{\mathrm r} = 6.64\) kvar, and a capacitor across \(V\) supplies \(V^2\omega C\):

\[ C = \frac{\Delta Q_{\mathrm r}}{\omega V^2} = \frac{6639}{(377)(480)^2} = \frac{6639}{8.69\times10^7} = \boxed{76\ \mu\text{F}} \] \[ \text{(c)}\quad I_2 = \frac{12\,000}{(480)(0.95)} = 26.3\ \text{A},\qquad \frac{P_{\mathrm{loss},2}}{P_{\mathrm{loss},1}} = \left(\frac{26.3}{33.3}\right)^2 = 0.62 \]

Line losses fall by about 38% while the load still gets its 12 kW. The capacitor does not change the real power at all — it only cancels reactive current.

7Transformers10 pts

An ideal transformer powers a 6.0 V rms device from a 120 V rms outlet. The primary has 1000 turns. (a) How many secondary turns? (b) The device draws 3.0 A rms. What is the primary current? (c) If the device is a \(2.0\ \Omega\) resistor, what resistance does the outlet “see”? (d) Why would this transformer not work from a 120 V battery?

Solution

\[ \text{(a)}\quad N_s = N_p\frac{V_s}{V_p} = 1000\cdot\frac{6.0}{120} = \boxed{50} \] \[ \text{(b)}\quad I_p = I_s\frac{N_s}{N_p} = (3.0)\frac{50}{1000} = \boxed{0.15\ \text{A}}\qquad(\text{check: } 120\times0.15 = 6.0\times3.0 = 18\ \text{W}) \] \[ \text{(c)}\quad R_{\mathrm{eff}} = \left(\frac{N_p}{N_s}\right)^2R = (20)^2(2.0) = 800\ \Omega = \frac{120\ \text{V}}{0.15\ \text{A}}\ \checkmark \]

(d) Induction needs a changing flux (Faraday). Steady DC makes a constant flux, so the secondary emf is zero — and the primary, now just a coil of low resistance, would draw a large current and overheat.

8Displacement current10 pts

A parallel-plate capacitor has circular plates of radius 4.0 cm, 2.0 mm apart, in air. The voltage across it is rising at \(1.0\times10^6\) V/s. Find (a) the capacitance, (b) the displacement current between the plates, and (c) the magnetic field between the plates at 2.0 cm from the axis. (d) Why did no one notice this effect before Maxwell?

Solution

\[ \text{(a)}\quad A = \pi(0.040)^2 = 5.03\times10^{-3}\ \text{m}^2,\qquad C = \frac{\varepsilon_0A}{d} = \frac{(8.85\times10^{-12})(5.03\times10^{-3})}{2.0\times10^{-3}} = 22.2\ \text{pF} \] \[ \text{(b)}\quad I_d = \varepsilon_0\frac{d\Phi_E}{dt} = \varepsilon_0A\frac{1}{d}\frac{dV}{dt} = C\frac{dV}{dt} = (22.2\times10^{-12})(1.0\times10^6) = \boxed{22.2\ \mu\text{A}} \]

It equals the conduction current in the wires, \(dq/dt = C\,dV/dt\), as it must.

\[ \text{(c)}\quad B(2\pi r) = \mu_0I_d\frac{r^2}{R^2}\ \Longrightarrow\ B = \frac{\mu_0I_dr}{2\pi R^2} = \frac{(4\pi\times10^{-7})(22.2\times10^{-6})(0.020)}{2\pi(0.040)^2} = \boxed{5.6\times10^{-11}\ \text{T}} \]

(d) The field is about a millionth of the Earth's (\(\sim5\times10^{-5}\) T). The displacement current is only large where fields change very fast — which is exactly the regime of radio waves and light.

9EM wave: intensity, fields, pressure10 pts

A 5.0 mW laser pointer makes a beam 2.0 mm in diameter (assume uniform). Find (a) the intensity, (b) \(E_0\) and \(B_0\), and (c) the force the beam exerts on a black surface that absorbs it completely.

Solution

\[ \text{(a)}\quad I = \frac{P}{\pi r^2} = \frac{5.0\times10^{-3}}{\pi(1.0\times10^{-3})^2} = \boxed{1.59\times10^3\ \text{W/m}^2} \] \[ \text{(b)}\quad E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} = \sqrt{\frac{2(1592)}{(3.00\times10^8)(8.85\times10^{-12})}} = \boxed{1.09\times10^3\ \text{V/m}},\qquad B_0 = \frac{E_0}{c} = \boxed{3.65\ \mu\text{T}} \] \[ \text{(c)}\quad F = p_{\mathrm{rad}}A = \frac{I}{c}A = \frac{P}{c} = \frac{5.0\times10^{-3}}{3.00\times10^8} = \boxed{1.7\times10^{-11}\ \text{N}} \]

The beam is more intense than sunlight (about 1000 W/m²), but the force is utterly negligible — radiation pressure matters only for huge areas or tiny masses. A mirror would feel twice the force.

10Maxwell's equations & wave direction10 pts

(a) Compute \(1/\sqrt{\mu_0\varepsilon_0}\) and say why the result was historic. (b) Name the Maxwell equation behind each: (i) an AC generator; (ii) magnetic field lines never begin or end; (iii) the field outside a charged sphere falls as \(1/r^2\); (iv) a magnetic field circles the gap of a charging capacitor. (c) A plane wave travels in the \(+x\) direction with \(\mathbf E\) along \(+\hat{\mathbf z}\) at some instant. Which way does \(\mathbf B\) point there?

Solution

\[ \text{(a)}\quad \frac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}} = \frac{1}{\sqrt{1.112\times10^{-17}}} = \frac{1}{3.335\times10^{-9}} = \boxed{3.00\times10^8\ \text{m/s}} \]

Constants measured with capacitors and current-carrying wires give the measured speed of light — so light is an electromagnetic wave (Maxwell, 1865).

(b) (i) Faraday's law; (ii) Gauss's law for magnetism, \(\oint\mathbf B\cdot d\mathbf A = 0\); (iii) Gauss's law for \(\mathbf E\); (iv) the Ampère–Maxwell law, through its displacement-current term.

(c) The travel direction is \(\mathbf E\times\mathbf B\). We need \(\hat{\mathbf z}\times\hat{\mathbf B} = +\hat{\mathbf x}\). Since \(\hat{\mathbf z}\times\hat{\mathbf y} = -\hat{\mathbf x}\), we get \(\hat{\mathbf z}\times(-\hat{\mathbf y}) = +\hat{\mathbf x}\): \(\mathbf B\) points along \(-\hat{\mathbf y}\), with magnitude \(E/c\).

Scoring

Score	Where you are
90–100	Solid. You have finished Physics II — the whole course fits on one page.
75–89	Concepts fine; slips with \(f\) versus \(\omega\), peak versus RMS, or a missing \(\tfrac12\) in the intensity. Re-check Section 5 and Section 7.
60–74	Phasor sums still shaky (Q3) or resonance unclear (Q4). Re-read Sections 3–4 with the RLC lab open.
< 60	Start from Section 1: phasors and reactance (Section 2) must be automatic before impedance, power or waves make sense.

Where the points usually go

Adding element voltages as magnitudes (Q3c); using \(f\) instead of \(\omega\) in \(X_L\) and \(X_C\) (Q2, Q3); quoting \(V_0I_0\) as average power (Q5); computing the capacitor for power factor 1 instead of 0.95 (Q6); forgetting that the Amperian circle inside the plates encloses only \(r^2/R^2\) of the displacement current (Q8); and the order in \(\mathbf E\times\mathbf B\) (Q10c).