# Homework Solutions

## Problems 1, 2, 9, 10, 17, 18, 20, 22, 24, 25, 30, 35

Questions:

Q1 What is meant by the statement "the voltage across an inductor leads the current by 90o"?

While the voltage across an inductor and the current through the inductor are both alternating sinusodialy -- meaning they can be written as a sine or cosine function -- they are not in phase. The voltage across the inductor reaches its maximum value before the current through the inductor reaches its maximum. The voltage across the inductor goes to zero before the current through the inductor goes through zero. The two are out of phase by 90o.

Q2 A night watchman is fired by his boss for being wasteful and keeping all the lights on in the building. The night watchman defends himself by claiming that the building is electrically heated, so his boss's claim is unfounded. Who should win the argument if this were to end up in a court of law?

The night watchman is correct. The lights dissipate heat just as the electric heaters do. Using only the heaters requires just as much electricity as using the heaters and the lights.

Q3 Why does a capacitor act as a short circuit at high frequencyies? Why does it act as an open circuit at low frequencies?

At high frequencies, the capacitor only has time to get a very little charge on it which means it has very little "back voltage" or "counter voltage". At low frequencies, there is enough time for a considerable amount of charge to collect on the capacitor. This means there is a large "reverse voltage" on the capacitor so the sum of the capacitor's voltage and the voltage of the ac source is nearly zero. That means the current will be nearly zero.

Q4 Explain how the acronym "ELI the ICEman" can be used to recall whether current leads voltage or voltage leads current in RLC circuits.

In "ELI the ICEman", the "L" in "ELI" describes the inductor (L) and ELI means the voltage -- the Emf -- leads the current I for an inductor (L).

In "ELI the ICEman",the "C" in "ICE" describes the capacitor (C) ICE means the current I leads the voltage -- or the Emf -- for a Capacitor.

Q8 If the frequency is in a series RLC circuit, what happens to the resistance, the inductive reactance, and the capacitive reactance?

The resistance is unchanged.

The inductive reactance is given by so the inductive reactance XL is doubled -- that is, an inductor acts like has twice the "pseudo-resistance" if the frequency is doubled.

The capacitive reactance is given by so the capacitive reactance XC is halved -- that is, a capacitorr acts like has only half the "pseudo-resistance" if the frequency is doubled.

Q12 What is the impedance of an RLC circuit at the resonance frequency?

At the resonance frequency, the impedance equals the resistance; Z = R.

Q14 What is the advantage of transmitting power a high voltages?

Power is given by P = I V so an increase in voltage means a decrease in current. The transmission cable has some resistance Rtrans. The power dissipated by that transmission cable is Ptrans = I2 Rtrans so reducing the current reduces the power -- and drastically since the power is proportional to the square of the current. Consider ordinary household electricity with a voltage of 240 V and compare that to transmitting electricity from a generating station to your city at 100 times that voltage, 24,000 V. By increasing the voltage by a factor of 100, the current is decreased by a factor of 100 and the transmission losses are reduced by a factor of (100)2 = 10,000.

Q16 Why do power lines carry electrical energy at several thousand volts potential but it is always stepped down to 24 V as it enters your home?

33.1, Show that the rms value for the sawtooth voltage shown in Figure P33.1 is Vrms = Vmax/SQRT(3)

We can write this voltage as

v = v(t) = Vmax[ - 1 + (2/T) t]

or

v = v(t) = Vmax[(2/T) t - 1]

for 0< t < T

that is one "cycle" of our sawtooth function.

v2 = V2max[(4/T2)t2 - (4/T)t + 1]

The mean value of v2 is the integral of this from t = 0 to t = T divided by T,

33.2, (a) What is the resistance of a lightbulb that uses an average power of 75 W when connected to a 60-Hz power source having a maximum voltage of 170 V?

(b) What is the resistance of a 100-W bulb?

For power, we need Vrms, not Vmax;

Vrms = 0.707 (170 V)

Vrms = 120 V

P = IV = I2R = V2/R

R = V2/P

(a) for the 75-W bulb,

R75 = (120 V)2/75 W

R75 = 192

(b) for the 100-W bulb,

R100 = (120 V)2/100 W

R100 = 144

33.9, In a purely inductive ac circuit, as in Figure 33.4, Vmax = 100 V.

(a) If the maximum current is 75 A at 50 Hz, calculate the inducatance L.

Just as

R = V/I

for this ac circuit, we have

XL = V/I

XL = 100 V/75 A

XL = 1.33

XL = 2 f L

L = XL/2 f

L = 1.33 /2 (50 Hz)

L = 4.234 mH

(b) At what angular frequency is the maximum current 2.5A?

This time we have

XL = V/I

XL = 100 V/25 A

XL = 4.0

XL = 2 f L

f = XL / 2 L

f = [ 4 /2 (4.234 x 10 - 3) ] Hz

f = 150 Hz

33.10, When a particular inductor is connected to a sinusoidal voltage with a 120-V amplitude, a peak current of 3.0 A appears in the inductor.

(a) What is the maximum current if the frequency of the applied voltage is doubled?

Since the inductive reactance is give by

XL = 2 f L

doubling the frequency will also double the inductive reactance which means the current will be cut in half.

Imax(new) = 1.5 A

(b) What is the inductive reactance at these two frequencies?

Just as, in a dc circuit,

R = V/I

so, in this inductive ac circuit,

XL = V/I

so

XL(a) = 120 V / 3.0 A = 40

XL(b) = 120 V / 1.5 A = 80

33.17, (a) For what linear frequency (f in cycles per second or Hz, rather than the angular freqency in radians per second) does a 22.0-F capacitor have a reactance below 175 ?

XC = 1/ 2 f C

f = 1/2 C XC

f = 1/[2 (22.0 x 10 - 6)(175 )]

f = 41.3 Hz

(b) Over this same frequency range, what is the reactance of a 44.0-microFarad capacitor?

XC = 1/2 f C

XC = 1/[2 (41.3)(44.0 x 10 - 6)]

XC = 87.5

But we could have calculated that more efficiently simplly by using

XCa = 1/ 2 f Ca

XCb = 1/ 2 f Cb

XCb/XCa = Ca/Cb

XCb = XCa(Ca/Cb)

XCb = (175 )(22/44)

XCb = 175 /2

XCb = 87.5

Now that I have shown this both ways, I am not sure which is "more efficient"!

33.18, What maximum current is delivered by a 2.2-F capacitor when connected across

(a) a North American outlet having Vrms = 120 V, f = 60 Hz and

XC = 1/2 f C

XC = 1/2 (60 Hz)(2.2-F)

XC = 1/2 (60)(2.2 x 10 - 6)

XC =1 205

I = V/XC

Irms = Vrms/XC

Irms = (120/1,205)A

Irms = 0.0995 A

(b) a European outlet having Vrms = 240 V, f = 50 Hz?

XC = 1/2 f C

XC = 1/2 (50 Hz)(2.2-F)

XC = 1/2 (50)(2.2 x 10 - 6)

XC =1 447

I = V/XC

Irms = Vrms/XC

Irms = (240/1,447)A

Irms = 0.166 A

33.20, A sinusoidal voltage v(t) = Vmax cos t is applied to a capacitor as inFigure P33.20.

(a) Write an expression for the instantaneous charge on the capacitor in terms of Vmax, C, t, and .

v = v(t) = Vmax cos t

C = Q/V = q/v

q = C v

q = q(t) = C v = C Vmax cos t

q(t) = C Vmax cos t

(b) What is the instantaneous current in the circuit?

33.22, A variable-frequency ac generator with Vmax = 18 V is connected across a 9.4 x 10 - 8 F capacitor. At what frequency should the generator be operated to provide a maximum current of 4.0 A?

XC = V/I

XC = 18 V / 4 A = 4.5

XC = 1/ 2 f C

f = 1/2 C XC

f = 1/[2 ( 9.4 x 10 - 8 )( 4.5 )]

f = 376 kHz

33.24, At what frequency does the inductive reactance of a 57-H inductor equal the capacitive reactance of a 57-F capacitor?

XC = 1/ 2 f C

XL = 2 f L

XL = XC

2 f L= 1/ 2 f C

f2 = 1/4 2 L C

f2 = 1/4 2 (57-H)(57-F)

f2 = 1/4 2 (57 x 10 - 6)(57 x 10 - 6)

f2 = 7.8 x 1010

f = 279 kHz

33.25, A series ac circuit contains the following components: R = 150 , L = 250 mH, C = 2.0 F and a generator with Vmax = 210 V operating at 40.0 Hz. Calculate the following:

(a) inductive reactance,

XL = 2 f L

XL = 2 (40)(250 x 10 - 3)

XL = 0.785

(b) capactive reactance,

XC = 1/ 2 f C

XC = 1/[2 (40)(2.0 x 10 - 6)]

XC = 1989

(d) impedance,

Z = SQRT[ R2 + (XL - XC)2]

Z = SQRT[ (150)2 + (1989 - 0.785)2]

Z = 1994

(d) maximum current and

Z = V/I

I = V/Z

I = 210 V/ 1994

I = 0.105 A

(c) phase angle.

= tan - 1 [(XL - XC)/R]

= tan - 1 [(1989 - 0.785)/150]

= tan - 1 [13.25]

= 85.7o

33.30, The voltage source in Figure P33.30 has an output Vrms = (100 V) cos (1000 t).

Vrms = (100 V) cos (1000 t)

Vrms = Vmax cos ( t)

f = /2

f = 159 Hz

Determine the following:

(a) the current in the circuit and

I = V/Z

Z = SQRT[ R2 + (XL - XC)2]

XL = 2 f L = 2 (159)(50 x 10 - 3) = 50

XC = 1/2 f C = 1/[2 (159)(50 x 10 - 6)] = 20

Z = SQRT[ R2 + (XL - XC)2] = SQRT[402 + (50 - 20)2]

Z = 50

I = [100/50] A

I = 2.0 A

(this is Imax);

Imax = 2.0 A

(b) the power supplied by the source.

From Eq33.29, page 976, we know

Pav = Irms Vrms cos

Pav = (1/2) Imax Vmax cos

= tan - 1 [(XL - XC)/R]

= tan - 1 [(50 - 20)/40]

= tan - 1[0.75]

= 37o

cos = 0.8

Pav = (1/2)(2)(100)(0.8)

Pav = 80 W

(c) Show that the power dissipated in the resistor is equal to the power supplied by the source.

P = I2 R

Pav = I2rms R

Pav = (1/2) I2max R

Pav = (1/2) (2.0)2 (40)

Pav = 80 W

33.35 If 100 MW of power at 50 kV is to be transmitted over 100 km with only 1.0 percent loss, copper wire of what diameter should be used? Assume uniform current density in the conductors.

P = Ploss = 0.01(100 x 10 + 6 W) = 1 x 10 + 6 W = 1 MW

Ptot = IV

I = Ptot/V

I = 108/50 x 103

I = 2 000 A

P = I2 R

R = Ploss/I2

R = 1 x 106/(2 000)2

R = 0.25

R = L / A

A = L / R

A = [(1.7 x 10 - 8)(100 x 103)/0.25 ]m2

A = 0.0068 m2

A = r2

r2 = 0.0068 m2/ = 0.00216 m2

r = 0.0465 m = 4.65 cm

d = 9.3 cm