This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: "‘ ......m. .../5:??4? Keg/ .. Instructions: Please read each problem carefully. Circle the correct answer choice, or fill in the blanks as asked for in the pro em , A—iH D 4 g .1. 4. .L a, J.— Usetheﬁgureatthe rightforproblems 13. c1 c2 c, '16 “I" 3‘3"? “'9‘ Forthccireuit consisting of air ﬁlled, parallel plate 3  capuitm. c. = 1.50 W, c,= 3.80 pr, c,= 5.20 pr, and C. = 7.60 21F, connected as shown at the right, C. .1
\[8] #1. Determinetheequivalent capacitance between pointsAandB. ’ Answer: (a) 1.03x10'pF. (b) 1.50x10‘pr. (c) 1.39:: 10' pF. (d) 1.76x 10' 14F. (e) 1.55x10‘pr. @ I K/OF \ [8] #2. A potential of21.6 volts is now applied across points D and B. What charge ' owbe on C4?
Answer: (a) 3.70 x 10* c. (b) 5.50 x 10' c. (c) 9.05 x 10*5 c. (a) 1.37 x 10“ c. 1.64 x 10* c. (t) [5].#3. A piece of glass (dielectric constant 12.!) is placed so as to completely fill between the plates of capacitor C.. What will
he the new value of the capacitance of C4? Answer: (a)3.55 x10'5 F. (b) 4.26 x 10" r. (c) 7.02 x10" F @20 x 10'5 F. (e) 2.33 x 104 r. (1) \ [8] #4. A 12.0 m long section of a certain wire (resistivity =5 8.52 x 10" Qm at room temperature), has a circular cross section of
radius=5.0x10"rn. Determine the value ofits ‘ atroomtempenture. Answer: (a) 1.15 x 10'2 o. (b) 2.77 x 10‘2 o. .30 x 10‘2 n. (a) 2.58 x 10'2 a. (e) 1.30 x 10'2 o. (1) \ [8] #5. An electric current of l5.0 mA exists when an electric ﬁeld of 50.0 V/m is setup in a wire. If the wire has a cross section
area of 9. 10 x 10“ 111’, what is its conductivity?
wer: (a) 2.98 x 10“ (9.111)". (b) 1.81 x 10“ (0.111)". (c) 2.77 x 10‘l (flm)". (a) 1.31 x 10" (tim)".
.30 x 10" (Q~m)". (t) [I #6. Circle each of the true statements
a) When a capacitor is fully charged the electric current that moves from one plate to the other through the capacitor is a
maximum.
If a point charge Q is placed in an inﬁnitely large material of dielectric constant K, the magnitude of the electric ﬁeld at a
1 ’te distance r ﬁom the point charge is E = (l/41t1ccoXQ/rz).
(G, e . Law tells us that the current is directly proportional to the potential difference that is applied across the device.
((93 n a resistance the kinetic energy of the charge carrier is tramformed to heat energy. “1 terrain? a. dielectric wuh 16> Kw on a charged airﬁlled capacitnr'has the affect of reducing the electric ﬁeld between the
. 3 acitor plates. \ [8] #7. For the circuit shown , R. = 2.00 9, R1 = 3.00 0, R3 = 4.009, and R. = 5.00 I) . Determine the equivalent resistance
between a and b. Answer: (a) 8.92 o. @722 o. (c) 6.95 x 102 o.
(d) 6.24 n. (e) 1.1 19. (t) [20] #8. For the circuit shown:
Write three loop rule equations. and a useful junction rule equation
inVOIVing 11, i3 , and i]. 32.0 0 1/ PHYS 222 EXAM #2 Spring 2000 Covering Chapter 25  27 Name S
For #9 & #10, use the Circuit shown at the right. E = 4.50 KV, C == 6.80 uF, + l R
andR=970xlo’n E _
C ‘5)”. Whatisﬂieminimumtimethatshouldpassaﬁertheswitchisclosed, to consider that the capacitor“ is fully ed?
Answer. (3)2. as x10"s. (b) 2.95 x 10 s. (c)3.39x 10“ [email protected] 10“ s. (e) 5.21 x rod 5. \
[5] #10. Consider that the capacitor is initially uncharged before the switch S is closed. What is the current in the resistance R immediatelyaﬁerswitc ' closed?
Answer: (a) 455 mA. @ ' v mA. (c) 555 mA. (d) 632 mA. (e) 645 mA. (0 \[22] #11. The ﬁgure attherightisoftwoconcentric cylindricalooudmtingshellsofradiusa and b, withb>a. The ~ ~
between the shells is ﬁlled with air (dielectric constant = 1.000). The length of each cylinders is L.
Using the steps below, which are based on those given in Chapter 25, determine an expresion
for the capacitance. [4] Step #1: Using Q to represent the m of the charge write on the ﬁgure to show the
charges that you assume on the shells. (Don't forget the signs to indicate the polarity of the charges.) [ 6] Step #2: Choose the correct expression for the electric ﬁeld in the space between the shells:
(Note: Win and N/C are the same.) pa (a). E(a<r<b)=2£l’r (V/m) (b)E(a<r<b)=2w ”(V/m) @Ewmmzﬁo (V/m)
0
— 9": = VI
2““, . (e) E (a < r < b) 27:5021} 0( m) . (f) [6] Step #3: Determine the electric potential difference between the two cylinders.
in) v =(Q/21teoa)[1n(b/L)] (V). = (Q/2neoL)[n(b/a)] (V). (c) v =(Q2/21woab)[lﬂ(L)] (V). (d) V = (Qz/ZnoLlllnw/aﬂ (V) (e) V = (Q/4eob)[ln(Ua)] (V) (f) V = [6] Step #4: Complete the problem by using the above information to determine the capacitance. (a) C = 41r$oL[1n(b/a)]'l (1") = ZMUHWIIH" (F) (C) C = 4Sob[n(L/a)] (F).
(d) C = 21:30a[ln(b/L)]" (F) (e) C = ZNanbUMLH" (F). (D C = \[18] #12 The circuit shows meters to measure currents in and voltages Va.
Match the letter in the meter with what it is measuring. (For example, Va; means the potential difference across R2, and in means the current in R2)
Meter # Measure Meter # Measure Meter # Measure in f in 2 Via
in ( le Vs: g0 ...
View
Full Document
 Spring '11
 Henry

Click to edit the document details