This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: P202 Spring 2006
Ch 24: ac Circuits Homework due at 11:59 pm Feb. 26, 2006
Error allowance on all numerical answers is 3%. 1. The relationship Vms = Vmax M5 is valid only for voltages that vary sinusoidally.
What is the rms voltage for the square wave voltage shown below? V 5.0 V e51) V Solution: The definition of root—mean—square is “square ROOT of the MEAN
value of the SQUARE of a quantity". The square of the voltage sketched above is 25 V‘zfor all values oft (time). The square root of25 is 5. So the rms voltage is 5
V. Solution: 5 2. An ac generator with a frequency of F=[20,25,30 or 35] Hz and an rms voltage of
15 V is connected to a C=[integer between 30 & 40] uF capacitor. Assume that
the generator produces a sinusoidal waveform. (a) What is the maximum current in the circuit? mA (b) What is the current in the circuit when the voltage across the capacitor is 7.5 V
and increasing? (c) What is the current in the circuit when the voltage across the capacitor is 7.5 V
and decreasing? Solution: We know how to find the rms current in the circuit given the rms voltage
and the capacitive reactance. We use the generalized form of Ohm '3 law given in
eqn 24—8 of Walker. The maximum current in the circuit is related to the rms
current by the usual sqrt(2) factor for a sinusoidal voltage (eqn 245 of Walker). General phasor Phasor for part (b) Phasor for part ©
diagram T o solve parts (b) and (c) it helps to draw phasor diagrams for a circuit with a capacitor (above) for various times, t. Note that (at is the angle between the x axis
and the current in all cases even though the angle is only marked on the leftmost ﬁgure for clarity. Remembering that the value of a quantity represented by a
phasor is just the component of the phasor along the y axis, we see from the leftmost diagram that the current is given by I = [max sin cot and the voltage
across the capacitor is given by V = Vmax sin(a)t — 7t / 2). This is just another way
of saying that the voltage across a capacitor lags behind the current. From the second phasor diagram we can see that the voltage is positive (i.e. the y
component of the voltage phasor is positive) and increasing when cat is between 71/2 and trand that in this case the current is positive and decreasing (remember
that both phasors rotate counter clockwise). From the third figure, we see that the voltage is positive and decreasing when cat is between nand 3W2 and that in this
case the current is negative. To ﬁnd the actual values of the current, we use the given voltage ( 7.5 V) and Vmax = sqrt(2) Vrms substituted into the equation given above for V to ﬁnd cat. We then
plug this into the equation above for I and use 1mm found in part (a) to calculate I. The sign ofl is the one we found from the phasor diagrams.
a Full!
vb”? '” Incur : Imps/"QC Viknx’ I'Mﬁ"f2
\E \s" = 1m /(Z1‘.Fc mf») :> I’M”: 0.1333 PC "A
We. tough: that l 2 EM” ; miscl \/ : Vii” sqth x/L) 1)”: J3: :5— Qullwf "ET/z) @ Wt" 60= 7—0300 0‘ Vii30 . (Aka/a, HACAGMuS «L l‘ ‘HJQ I; 17w“, so; HO.;—oa= 0.1333 F: 5.; (10.70“
3}. fa. “n... Us}.Sv 8. alewczwmlj I \n‘ hot. so pm. I: .. entailéW; —4 Solution: (3.) 0.1333 F C; (b) 0.124? F C; (c) — 0.1247 F C 3. The rrns current in an RC circuit is I=[0.6 — 0.8] A. The capacitor in this circuit has a capacitance of 15 pF and the ac generator has a frequency of 150 Hz and an
rms voltage of E=[80, 85, 90, 95, or 100] V. What is the resistance in the
circuit? Ohms Solution: This is just a straightforward application of the expression (eqn 24—] l
of Walker) for the impedance of an RC circuit. We know the capacitance and the frequency so We can find the capacitive reactance. We know the total impedance
of the circuit because we are given both the rms current and voltage and the deﬁnition of impedance is Vrms/Irms. Given Z and X6 we can find R.
E'— = I ‘E :3) "Z ; E/ I i: \/ 241 e). if: 22+ l/(ZITIS'OKHKlOé)?‘ Solution: sqrt((E/l)2 ~ 5003.5) 4. An RC circuit comprises a resistance of R:[3.0 — 4.0] kOhrn, a capacitance of C=[0.2 — 0.4 in steps of 0.05] [LP and a generator with an rms voltage of V=[15,20,25, or 30] V and a frequency ofF=[100, 120, 130, 140 or 150] Hz. (a) What is the power factor of this circuit? (b) Will the power factor increase [radio button], decrease [radio button] or stay
the same [radio buttons] if the frequency of the generator is increased? Solution: We know C and the frequency so we can find Xc (eqn 24—9 of Walker).
From Xc and R we can ﬁnd the impedance Z (eqn 2411 of Walker). We know that
the power factor is just R/Z (eqn 24—12 of Walker) so this is the answer to part (a).
If the frequency of the generator is increased, the capacitive reactance Xc
decreases (eqn 249), thus Z also decreases (can 2411). Since Z decreases and R stays the same, R/Z increases, i.e. the power factor increases. WM 2
a .I.  _L_.  L . R
VC— (Ac —_ erFCxio‘C‘ f) %—\/ﬂwo + ﬂZWRtioéil POWEI tact—07L (59¢: ice/(ZWFC)L Solution: (a) R/sqrt(R2 + 105/(62832 F (3)2) ; (b) increase 5. An RL circuit consists of a resistance of R=[65  70] Ohms, and inductor of
L=[3035] mH and an ac generator with an rrns voltage of 120 V.
(a) At what frequency is the rms current equal to 1.5 A? Hz
(b) What is the rms voltage across the resistor at this
current? V
(c) What is the rrns voltage across the inductor at this
current? V Solution: Since Vrms = [rm Z, We can ﬁnd Z given the rms values of voltage and
current. The expression for Z involves R (given), L (given) and to (unknown) (see
Walker eqn 24—15). We can solvefor wand hencefrequencyfmi = 2 75]). The
rms voltage across the resistor is given by [m R, so this answers part (b). The rms voltage across the inductor is [m X ,5. Since we know wand L we can ﬁnd XL (eqn
2414 of Walker), so this answers part (c) V‘s“; : Irma ‘t $7 2‘ 2. Ila/I’M! L I'M,  lx‘ éqoo Ivy,— (Zrcl. L.)1Kl0
L5 4 : 8%.” \létmo— I91
L.
vfm I,“ «9 me
me : EM )(L; 1.: to L: L; 2M! Lxmhg
VJ“ : ifJSllooeﬂ Solution: (3) 159.15 sqrt(6400 — R2)/L ; (b) 1.5 R ; (c) 1.5 sqrt(6400 — R2) . An RLC circuit has a resistance of R=[100 7 110] Ohms, an inductance of L:[90 100] mH and a capacitance of C=[5, 10, 15 or 20] HF. (a) What is the power factor when this circuit is connected to a 125 Hz ac
generator? (b) Will the power factor increase [radio button], decrease [radio button] or stay
the same [radio button] if the resistance is increased? (0) Calculate the power factor if the resistance is increased to [ R2 = 500, 525,
550, or 575] Ohms Solution: T he power factor is given by R/Z ( Walker eqn 2412). We know R, L, C andf(and hence to) so we can ﬁnd Z (Walker eqn 24—16) and calculate R/Z as the
power factor for part (a). To see what happens to the power factor when R
increases, substitute eqn 24—] 6 of Walker into the deﬁnition of power factor
(Walker eqn 2412) and divide the top and bottom ofthe eqn by R. We now see
that the term on the bottom of the fraction decreases when R increases. Thus the
power factor will increase. This answers part (b). To actually calculate the power factor for a new value of the resistance, we just substitute the new value into the
deﬁnition of the power factor. sen/eh (wL Vase): “[WLAHJLQ‘...
to; 15?»; cm; 2 “' Jﬁeat (ZTClZ§La<tOg '/271:1LTC»<10'(’>IJ be. '2: dE€z+ (anaemia — HERZak. )1] .. _..r..,. .... f1 . ._ _..__.'. Cos gé : ﬁ/Vﬂt+P(oﬂ?ECHL_ MyefzurﬂCEL
QWLLQIQu‘ m¢ : : Q/J234ygn : Tim... 1 1*? t as“); are“ 1‘ ‘ ._ l' l o
6n» e4  PM ,i t/ :22 + (Marta. — 12%. zit/c }‘ Solution: (a)R/sqrt(R2+[0.7854L1273.2/C]2) (b) increase
(0) R2 / sqrt(R22 + [0.7854 L  1273.2 / (:12) 7. An RLC circuit has a resonance frequency of F=[2.0 — 2.5] kHz. If the
capacitance is C=[40~50] 14F, what is the inductance? rnH Solution: This is a straightforward application of the deﬁnition of resonant
frequency given by eqn 2418 of Walker was: ‘/\/LC.
3 a_ ,/ 3 —6
m Hca. cows. (2nF'xtO ) n L¥t0 cht0 1% L = 2933/cm Solution: 25.33 i (FZC) 8. An RLC circuit has a capacitance of C=[O.250.3] uF.
(a) What inductance will produce a resonant frequency of 95 M2? pH (b) If the impedance at resonance is 1/5 of the impedance at 10 kHz, what value
of R should be chosen? Ohm Solution: For part (a), simply use the deﬁnition of resonant frequency (eqn 24—18
of Walker). For part (2)), note that the impedance at resonance is just R. To
calculate the impedance at a frequency of 10 kHz, use the deﬁnition given by
Walker eqn (24—16). All of the quantities required to calculate Z are known except
for R. We are given the relationship between the impedance at resonance and the
impedance at 10 kHz so we simply write an equation that impedance at resonance
(R) is equal to 1/5 of the impedance at 10 kHz (an expression involving R as an
unknown). We solve the equation for R. queue.le we: (2n‘lfxla‘)L— l/(cxlo‘ﬁK Lg) islet: La; Mamas
' : "12‘ ‘5} = tiiﬁtxlo'” H = 2‘906?‘ pH
C X 3'512‘lx '0 c. T
TMpeAMu M "Lao \S'Kowua r \e. R: é‘r/th (2“ 'Oul— * ‘/2r:to”c:lo“’°)1
Zuni: (Zr—lo” 2.a;oarxlo“L— t.S_'.Ci\6/C)Z ZQR“: (l.?£xl0_% _(§al.é/C)2
R : 3.7.481/C
Solution: (a) 2.8067 / C ; (b) 3.2487 / C 9. An RC circuit consists of a resistor of R=[30—35] Ohms, a capacitance of C=[20
25] uF and an ac generator with an nns voltage of 120 V. (a) At what frequency will the rms current in the circuit be 2.9 A? Hz
(b) For this frequency, what is the rms voltage across the resistor? V
(c) For this frequency, what is the rrns voltage across the capacitor? V Solution: From the deﬁnition of impedance Z, we know that Vms = In”, Z. So,
given In,” and Vm we can calculate Z. Z is deﬁned in terms of R, C and a) ( Walker eqn 2416). We know R and C so we can calculate a). This answers part
(a). F or part (b) we note that the rms value of the voltage across a resistor is
given by Vm : In,“ R. Since we know 1m and R we can ﬁnd Vrms. For part (c),
we note that the rms voltage across a capacitor is given by Vrms = [ms XC. We know In,” and can calculate X: from its definition knowing C and a). Vrms: J—m: % ILA, If“! 2' 2'0! A i : ills/2'6]: Lil‘ag 2= JR2+ ’/(2nFc»<io'5)'“
' 1?tzixlo3= 91+ 1/(37?l:f‘»<'O£)z
L2WFCx106): l/(l3I1tho3")€2) r: = t‘s—mswo" c. \ll'}127kt03_‘33.1 it _ r
Wyn“ Arms“: Solution: (a) 1.5915 105 / (C sqrt(1.7122 103 — R2)) ; (b) 2.9 R;
(c) 2.9 sqrt(l.7122 103 9R2) 10. A radio tuning circuit contains an RLC circuit with a resistor of R=[4,5, or 6] Ohms, and an inductor of L=[2.5 — 3.0] pH. (3) What capacitance is needed to produce a resonant frequency of 95
MHZ? pF (b) If the capacitance is increased above the value found in part (a), will the
impedance of the circuit increase [radio button], decrease [radio button] or
stay the same [radio button]? (0) Find the impedance of the circuit at resonance. Ohms Solution: Part (a) is just an application of the deﬁnition of the frequency of a
resonant circuit (eqn 24—18 of Walker). Note we don ’I need to know R for this. We
know that the minimum impedance for a circuit occurs at resonance, so the
answer to part (b) is that the impedance will increase (whether we increase or
decrease C). The impedance of a circuit at resonance is R so this is the answer to part (c). u; :14;ch “.0 mﬁot —_' .
7 H LC.
.‘ Ihn ’ __ — 
v‘ w_;..;. (ME '—' '1 (chiofl C n; Ht
3.~>_€2o1x10 = Io'g/Lc __;, C: amen/[ﬂ Solution: 2.8067/ L; (b) increase (c)R ...
View
Full
Document
This note was uploaded on 07/02/2011 for the course PHYS 202 taught by Professor Warren during the Spring '09 term at Indiana.
 Spring '09
 WARREN
 Physics

Click to edit the document details