Zeena Husain, Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points If the emf of a battery is 6 V and a current of 69 A is measured when the battery is shorted, what is the internal resistance of the battery? Correct answer: 0.0869565 . Explanation: Given : E = 6 V and I = 69 A . Pa =4 Pb Pa 7. =8 Pb Explanation: 6. Given : R1 = R2 = R = 1 I = 7 A. and
1
When the battery with an emf E is shorted, with current I flowing, the internal resistance Ri is Ri = 6V E = = 0.0869565 . I 69 A
In case a, the two resistors are parallel, so the total resistance is 1 1 R2 + R 1 1 = + = Ra R1 R2 R1 R2 R1 R2 R Ra = = . R1 + R 2 2 In case b, the two resistors are in series, so the total resistance is given by Rb = R1 + R2 = 2 R. Pa = I 2 R a = 1 2 I R 2 Pb = I 2 R b = 2 I 2 R
002 (part 1 of 2) 10 points Consider the following two cases. case a: 7A 1 case b: 7A 1 1 = 7 A Rb Pa . Pb 1 = 7 A Ra
1 2 I R Pa 1 = = 2 2 = . Pb 2I R 4
003 (part 2 of 2) 10 points Find Pa . Correct answer: 24.5 W. Explanation: Pa = 1 2 I R 2 1 = (7 A)2 (1 ) 2 = 24.5 W .
Find the ratio of dissipated powers, 1. 2. 3. 4. 5. Pa Pb Pa Pb Pa Pb Pa Pb Pa Pb 1 8 1 = 2 = =1 =2 = 1 correct 4
004 (part 1 of 1) 10 points Consider the circuit a 22.7 45.4
45 .4
45.4
b
22.7
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban What is the equivalent resistance between the points a and b ? Correct answer: 39.725 . Explanation: Given : R1 R2 R3 R4 R5 = R = 22.7 , = 2 R = 45.4 , = 2 R = 45.4 , = R = 22.7 , and = 2 R = 45.4 . c R1 R2 d 7R 4 7 (22.7 ) = 4 = 39.725 . =
2
a
R3
R5
005 (part 1 of 4) 10 points Twelve resistors, each of equal value R = 6.6 , are connected so that each is along one edge of a cube, as shown in the figure. A current of I = 1.4 A is introduced at point "a". The current will be allowed to exit only along a wire at point "b", as drawn.
f 9 10 7 d 1 2 6 5 4 3 e h 8 g 11 12 b
R4
b The circuit is redrawn below. R2 = 2 R a R1 = R c d R3 = 2 R R4 = R R5 = 2 R
c
b
I
a
Basic Concepts: Rseries = R1 + R2 + R3 + 1 Rparallel =
1 1 1 + + + R1 R2 R3
Solution: This circuit can be analyzed by straight-forward series/parallel analysis: R4 and R5 are in series, so R45 = R + 2 R = 3 R . R2 and R3 are in parallel with R45 , so 1 1 1 8 = + + = R2345 2R 2R 3R 6R 3R R2345 = . 4 R1 and R2345 are in series, so 1 Req = R + 3R 4
What is the current through resistor R2 ? Correct answer: 0.466667 A. Explanation: Consider the figure. The current I is introduced into node "a". Three independent resistors of equal value are accessible via this node, namely, resistor R1 , R2 , and R3 . Because these three resistors are equal and the remainder of the circuit is symmetric, the current will split equally between the three paths. Thus, the current in each of the resistors R1 , R2 , and R3 is I 1.4 A = = 0.466667 A 3 3
I1,2,3 =
006 (part 2 of 4) 10 points What is the current through resistor R7 ? Correct answer: 0.233333 A. Explanation:
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban Now consider what happens as the current continues to flow toward point "b". Look at the current through resistor R2 . As it progresses, it hits the node which joins resistors R6 and R7 . Again, because these resistors are equal and the remainder of the circuit is symmetric, the current I2 must split equally between the two. Hence, I2 I 1.4 A = = = 0.233333 A 2 6 6 Note, in addition, that this process also takes place with currents I1 and I3 , as shown in the figure below. Thus, I7 = I49 = I 1.4 A = 0.233333 A 6 6 I49 = I 6 I 3
3
I1012 =
Note that we could achieve this identical current distribution if we were to arrange the twelve resistors in the following configuration:
1 I 2 I/3 I/3 I/3 3 part (a)
4 I/3 I3 = I9=I/6 5 6 I/3 I2 = I8=I/6 7 8 I/3 I1 = I6=I/6 9 part (b) part (c) I4=I/6 12 I5=I/6 11 I7=I/6 10
007 (part 3 of 4) 10 points What is current through resistor R10 ? Correct answer: 0.466667 A. Explanation: Next, to find the current through resistor R10 , we need to consider the currents flowing into resistor R10 . Carefully consider the original figure. The currents in resistors R7 and R9 are the only currents available to flow into resistor R10 to get to point "b". Consequently, the current there must be I 1.4 A I I = 0.466667 A I10 = I7 +I9 = + = = 6 6 3 3 Again, note the role of symmetry in this process (see the figure below). Two equal currents flow into each of the resistors 10, 11, and 12. Hence, the currents in resistors R11 and R12 equal that through R10 ; i.e. I I10,11,12 = = 0.466667 A 6 008 (part 4 of 4) 10 points Finally, calculate the equivalent resistance between points "a" and "b". Correct answer: 5.5 . Explanation: At last, we wish to calculate the equivalent resistance of our cubic setup. To assist in this endeavor, recall the following facts derived earlier: I I13 = 3
4
c,d,e
1 I a 3 2
5 6 7 8 9 part (d)
f,g,h
10 11 12 b I
I
a
R/3
R/6
R/3
b
I
part (e)
All resistors are equal, thus, from symmetry, the points "c", "d", and "e" are at the same potential and can be connected together. Likewise, the points "f", "g", and "h" are at the same potential and can be connected together. The revised figure is shown above with a stepwise reduction of the circuit. Hence, the equivalent resistance from "a" to "b", Rab , is: Rab = 1
1 R
+
1 R
+
1 R
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban + +
1 R 1 R
4
+
1 R
+
1 R
1 +
If the switch is closed,
1 R
+
1 R
+
1 R
1
1 1 +R+R 1 1 1 = R+ R+ R 3 6 3 5 5 = R = (6.6 ) = 5.5 . 6 6
I=
V r+ R 2 R 2
2
and V2 P =
009 (part 1 of 1) 10 points The power dissipated in the combination of the resistors labeled R in the circuit does not depend on whether the switch is opened or closed. Neglect the internal resistance of the voltage source. R S R E 1.19
R r+ 2
Then V2R = (r + R)2 2 R r+ 2 V2R
2
2 r2 + 2 r R +
R2 = R2 + 2 r R + r 2 2 1 2 R 2
r2 =
Determine the value of R. Correct answer: 1.68291 . Explanation: Given : R R E r r = 1.19 . S
R= 2r = 2 (1.19 ) = 1.68291
010 (part 1 of 2) 10 points A 16 V battery has an internal resistance r. 16 V r 12 and
1A What is the value of r? Correct answer: 13 . Explanation:
If the switch is open, I= V r+R
and the power for the resistor R is P = V2R (r + R)2
Given :
E R2 R3 I1
= 16 V , = 4 , = 12 , = 1 A.
4
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban E and R1 = r I2 = I 1 - I 3 1 =1A- A 4 3 = A 4 = 0.75 A .
5
I1
I2
R2
Since R2 and R3 are connected parallel, their equivalent resistance R23 is 1 1 1 R3 + R 2 = + = R23 R2 R3 R2 R3 R2 R3 R23 = R2 + R 3 (4 ) (12 ) = 4 + 12 = 3 . Using Ohm's law, we have E = I1 r + I1 R23 E - I1 R23 r= I1 16 V - (1 A) (3 ) = 1A = 16 - 3 = 13 . 011 (part 2 of 2) 10 points Determine the magnitude of the current through the 12 resistor on the right-hand side of the circuit. Correct answer: 0.25 A. Explanation: The potential drop across the 12 resistor on the right-hand side of the circuit is E3 = E - I 1 r = 16 V - (1 A) (13 ) = 16 V - 13 V = 3 V, so the current through the resistor is E3 I3 = r3 3V = 12 1 = A 4 = 0.25 A ,
R3
I3
012 (part 1 of 1) 10 points A length of wire is cut into 4 equal pieces. The 4 pieces are then connected parallel, with the resulting resistance being 5 . What was the resistance r of the original length of wire? Correct answer: 80 . Explanation:
Given : n = 4 and Rp = 5 . The resulting resistance Rp of n equal piece resistors of resistance r connected parallel is Rp = r , n
but when they are in series, the total resistance Rs is Rs = n r = n 2 Rp = 42 (5 ) = 80 .
013 (part 1 of 2) 10 points Four identical light bulbs are connected either in series (circuit 1) or parallel (circuit 2) to a constant voltage battery with negligible internal resistance, as shown.
Husain, Zeena Homework 6 Due: Mar 1 4:00 2004, am Inst: Sonia Paban
6
We can see that the bulbs in circuit 2 are more than 4 times brighter than the bulbs in circuit 1. 014 (part 2 of 2) 10 points If one of the bulbs in circuit 2 is unscrewed and removed from its socket, the remaining 3 bulbs 1. go out. circuit 1 2. become dimmer. 3. become brighter. 4. are unaffected. correct Explanation: Since the bulbs are parallel, after one of the bulbs is unscrewed, the voltage across each remaining bulb is unchanged, and the brightness is unaffected. 015 (part 1 of 3) 10 points 3.5 V 1.7 2.6 5.7 and
circuit 2 Compared to the individual bulbs in circuit 1, the individual bulbs in circuit 2 are 1. the same brightness. 1 2. less than as bright. 4 3. more than 4 times brighter. correct
2.3
I1
I2
4. 4 times brighter. 1 as bright. 4 Explanation: Solution: In circuit 1, the voltage across each light bulb is 5. V =IR= E E R= , 4R 4
8.1 V
In the circuit shown in the figure, find I1 . Correct answer: 1.74386 A. Explanation:
so the power of each bulb in circuit 1 is P1 = V2 E2 = . R 16 R Given : RA RB RC RD E1 E2 E3 = 2.3 , = 1.7 , = 2.6 , = 5.7 , = 8.1 V , = 3.5 V , = 2.2 V .
In circuit 2, the voltage across each bulb is identical; namely E. Hence the power of each bulb in circuit 2 is 1 E2 = P1 . P2 = R 16
2.2 V
I3
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban E2 RB RD
7
Expanding along the first column, the denominator is 1 D = RA + R B 0 =1 0 RC RD RD 1 0 RC -1 RD RD
RA
RC
I1
I2
E1
E3
I3
At a junction (Conservation of Charge) I1 + I 2 - I 3 = 0 . Kirchhoff's law on the large outside loop. (RA + RB ) I1 + RD I3 = E1 + E (2) (1)
1 -1 +0 RC RD = 0 - RC RD - (RA + RB ) (RD + RC ) = (2.6 ) (5.7 ) - (2.3 + 1.7 ) (5.7 + 2.6 ) = -48.02 2 - (RA + RB ) and I1 = -83.74 V D1 = 1.74386 A . = D -48.02 2
Kirchhoff's law on the right-hand small loop. RC I2 + R D I3 = E 3 Using determinants, 0 E1 + E 2 E3 1 0 RC 1 0 RC -1 RD RD -1 RD RD (3)
016 (part 2 of 3) 10 points Find the power supplied by 8.1 V battery on the bottom left-hand side of the circuit. Correct answer: 14.1252 W. Explanation: P1 = I1 E1 = (1.74386 A) (8.1 V) = 14.1252 W . 017 (part 3 of 3) 10 points Find the power supplied to the 2.3 resistor on the left-hand side of the circuit. Correct answer: 6.99438 W. Explanation:
2 PA = I1 RA = (1.74386 A)2 (2.3 ) = 6.99438 W .
I1 =
1 RA + R B 0
Expanding along the first row, the numerator is 0 D1 = E 1 + E 2 E3 =0-1 + (-1) 1 0 RC -1 RD RD RD RD
E1 + E 2 0 E3 RC = - [(E1 + E2 ) RD - E3 RD ] - [RC (E1 + E2 ) - 0] = RD (E3 - E1 - E2 ) - RC (E1 + E2 ) = (5.7 ) (2.2 V - 8.1 V - 3.5 V) - (2.6 ) (8.1 V + 3.5 V) = -83.74 V .
E1 + E 2 E3
018 (part 1 of 3) 10 points Given: Three batteries and four resistors are connected in a loop. 25 V 1902 719 24 V a 28 V b
2335
1427
What is the magnitude of the current in the loop? Correct answer: 0.00328999 A.
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban Explanation: Vab = E2 - I R2 - I R1 - E1 = 24 V - (0.00328999 A) (2335 ) - (0.00328999 A) (1427 ) - 28 V = -16.3769 V .
8
Given :
R1 R2 R3 R4 E1 E2 E3 I
= 1427 , = 2335 , = 1902 , = 719 , = 28 V , = 24 V , and = 25 V . E3 R4 E1 b
020 (part 3 of 3) 10 points What is the voltage drop across the top, left resistor? Correct answer: 6.25756 V. Explanation: V3 = I R 3 = (0.00328999 A) (1902 ) = 6.25756 V . 021 (part 1 of 4) 10 points In the circuit shown, the capacitor is initially uncharged. At t1 = 0, the switch S is moved to position "a". R2 R1 C E S b a
R3 E2 a
R2
R1
In this problem, three emf are connected in series, so Etotal = E2 + E3 - E1 = 24 V + 25 V - 28 V = 21 V
RT = R 1 + R 2 + R 3 + R 4 = 1427 + 2335 + 1902 + 719 = 6383 . The current in the circuit is counterclockwise, because E2 + E3 > E1 , and I= Etotal RT 21 V = 6383 = 0.00328999 A ,
Find VR1 , the voltage drop across R1 , as a function of time t1 . 1. VR1 = V0 e-t1 /(R1 C) correct 2. VR1 = V0 1 - e-t1 /(R2 C) 3. VR1 = V0 1 - e-t1 /(R1 C) 4. VR1 = V0 e-t1 /[(R1 +R2 ) C]
019 (part 2 of 3) 10 points Caution: Include the sign in your answer. What is the potential Vab from point a to point b ? Correct answer: -16.3769 V. Explanation:
5. VR1 = V0 e-t1 /(R2 C) 6. VR1 = V0 1 - e-(R1 +R2 ) t1 /(R1 R2 C) 7. VR1 = V0 1 - e-t1 /[(R1 +R2 ) C]
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban 8. VR1 = V0 e
-(R1 +R2 ) t1 /(R1 R2 C)
9
Find the voltage drop, VR1 , across R1 , as a function of time t2 . 1. VR1 = V0 2. VR1 3. VR 1 = V 0 4. VR1 rect R1 e-t2 /(R1 C) R1 + R 2 R1 = V0 1 - e-t2 /(R2 C) R1 + R 2 R1 1 - e-(R1 +R2 )t2 /(R1 R2 C) R1 + R 2 R1 e-t2 /[(R1 +R2 ) C] cor= V0 R1 + R 2
Explanation: Solution: For an "RC" circuit, I = I0 e-t1 /(RC) V0 -t1 /(R1 C) = e . R1 Since I R1 = VR1 , VR1 = V0 e-t1 /(R1 C) . 022 (part 2 of 4) 10 points Find VC , the voltage across C, as a function of time t1 . 1. VC = V0 1 - e-(R1 +R2 ) t1 /(R1 R2 C) 2. VC = V0 1 - e-t1 /[(R1 +R2 ) C] 3. VC = V0 1 - e-t1 /(R2 C) 4. VC = V0 e-t1 /(R1 C) 5. VC = V0 e-t1 /(R2 C) 6. VC = V0 1 - e-t1 /(R1 C) correct 7. VC = V0 e-(R1 +R2 ) t1 /(R1 R2 C) 8. VC = V0 e
-t1 /[(R1 +R2 ) C]
5. VR1 = V0 6. VR1
R1 e-(R1 +R2 ) t2 /(R1 R2 C) R1 + R 2 R1 = V0 1 - e-t2 /[(R1 +R2 ) C] R1 + R 2
R1 1 - e-t2 /(R1 C) R1 + R 2 R1 e-t2 /(R2 C) 8. VR1 = V0 R1 + R 2 Explanation: Now the switch moves to position "b", thereby excluding the battery from the circuit. Note: The equivalent resistance of the circuit is Req = R1 + R2 , because R1 and R2 are in series. 7. VR1 = V0 I = I0 e-t2 /(RC) V0 = e-t2 /(R1 +R2 )C , R1 + R 2 because the capacitor has an initial potential across it of V0 . Thus VR 1 = I R 1 = V0 R1 R1 + R 2 e-t2 /(R1 +R2 )C .
Explanation: By Kirchhoff's law, the sum of voltage around a closed circuit must be zero. Consequently, moving clockwise around the circuit as drawn V0 - V R 1 - V C = 0 . Therefore, VC = V0 - VR1 = V0 1 - e-t1 /(R1 C) . 023 (part 3 of 4) 10 points Much later (t1 ), at some time t2 = 0 (the clock is restarted at t2 = 0), the switch is moved from position "a" to position "b".
024 (part 4 of 4) 10 points Find VC as a function of time t2 . 1. VC = V0 1 - e-t2 /(R1 C) 2. VC = V0 e-t2 /[(R1 +R2 ) C] correct
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban 3. VC = V0 1 - e-t2 /(R2 C) 4. VC = V0 e-t2 /(R1 C) 5. VC = V0 e-t2 /(R2 C) 6. VC = V0 1 - e 7. VC = V0
-(R1 +R2 ) t2 /(R1 R2 C)
10
026 (part 1 of 1) 10 points If R = 14 k , C = 5 mF, E = 17 V, Q = 27 mC, and I = 3 mA, what is the potential difference, Vdif f = Vb - Va ? 17 V
a
1 - e[-t2 /(R1 +R2 ) C]
Q 14 k + - + - 3 mA 5 mF
b
8. VC = V0 e-(R1 +R2 ) t2 /(R1 R2 C) Explanation: Just as before, apply Kirchhoff's law of voltage to find VC = VR1 + VR2 = V0 e-t2 /[(R1 +R2 )C] . 025 (part 1 of 1) 10 points A fully charged capacitor stores 14.3 J of energy. How much energy remains when its charge has decreased to half its original value? Correct answer: 3.575 J. Explanation: Let : U = 14.3 J . 1 CV2 2 1 Q2 . = 2 C Q2 2 C Q2 4 C 1 Q2 2 C
Caution: Be careful with the signs. Correct answer: 19.6 V. Explanation:
Given : R = 14 k = 14000 , C = 5 mF = 5 10-6 F , E = 17 V , Q = 27 mC = 2.7 10-5 C , I = 3 mA = 0.003 A . Starting from the right, across the resistor: VR = IR across the capacitor: VC = -Q/C across the battery: VB = -E Hence, Vb - Va = -E - Q + IR C 2.7 10-5 C = -17 V - 5 10-6 F + (0.003 A)(14000 ) = 19.6 V
and
Q = C V , so the energy is U=
If Q is halved, U falls to one-quarter, as shown 1 2
Uf =
1 = 2 1 = 4 1 = U 4 1 = (14.3 J) 4 = 3.575 J .
027 (part 1 of 1) 10 points Consider a galvanometer with an internal resistance of 85 . If it deflects full-scale when it carries a current of 0.7 mA, what is the value of the series resistance that must be connected to it if this combination is to be used as a voltmeter having a full-scale deflection for a potential difference of 0.7 V? Correct answer: 915 . Explanation:
Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban
11
Given :
V = 0.7 V , I = 0.7 mA = 0.0007 A , r = 85 .
and
The voltage drop across the terminals is given by V =IR+Ir = I (R + r) So R= V -r I 0.7 V = - 85 0.0007 A = 915