HW 6-solutions - Lecture notes 6 PDF

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al hasani (laa2832) – HW 6 – coker – (56000) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment covers the material from text Ch. 25, on direct current circuits, as presented in class lectures. Some of the questions have several parts, to cover more concepts. 001 10.0 points Consider the circuit

1

Consider the circuit below. 11.5 V 6.03 V 0.29 Ω 0.23 Ω X 1.12 Ω

R

1.9 A

Y Find the resistance R. Correct answer: 1.23895 Ω. Explanation: R1

E1

R2

E2 X

R3

I

R

Y

The current in the 5.0 Ω resistor is: 1. 1.5 A correct 2. 3.0 A 3. 2.4 A

Let : E1 = 11.5 V , E2 = 6.03 V , R1 = 0.29 Ω , R2 = 1.12 Ω , R3 = 0.23 Ω , I = 1.9 A .

and

From Ohm’s law, the total resistance of the circuit is

4. 0.67 A

V E1 − E2 11.5 V − 6.03 V = = 1.9 A I I = 2.87895 Ω ,

Rtotal =

5. 0.42 A Explanation: The entire voltage of the battery lies across the lower parallel branch in which the 5 Ω resistor lies. Since the current through the series combination is the same, we get that the current through this resistor is: V Req 12 V = 8Ω = 1.5 A

so the resistance R is R = Rtotal − R1 − R2 − R3 = 2.87895 Ω − 0.29 Ω − 1.12 Ω − 0.23 Ω = 1.23895 Ω .

I=

003 (part 2 of 2) 10.0 points Find the potential difference VXY = VX − VY between points X and Y . Correct answer: 8.821 V.

002 (part 1 of 2) 10.0 points

Explanation:

al hasani (laa2832) – HW 6 – coker – (56000) The current in the circuit goes counterclockwise, so the potential difference between X and Y is

2

What happens when switch S is moved to position c, leaving R2 and R3 parallel? 1. The current through R3 increases.

VXY = E2 + R3 I + R I = 6.03 V + (0.23 Ω + 1.23895 Ω) (1.9 A) = 8.821 V . 004 (part 1 of 2) 10.0 points In the figure below the switch S is initially in position a .

2. The current through R2 remains the same as when R1 was in the circuit. 3. The current through R2 is half what it was with R1 in the circuit. 4. The current through R2 and R3 are now the same. correct 5. The current through R3 decreases.

R1 V

b

a

c S

R3

R2

Explanation: Since R2 and R3 have the same terminal voltage and resistance, the current through R2 and R3 must now be the same. 006 (part 1 of 2) 10.0 points

What happens to the current through R3 when the switch is moved to the open position b? R1 = R2 = R3. Neglect the internal resistance of the battery. 1. The current through R3 increases to twice its original value.

The currents are flowing in the directions indicated by the arrows. E1

R1 I1 R2

I2

2. The current through R3 decreases to twothirds its original value.

E2 I3

3. The current through R3 increases to three-halves its original value. 4. The current through R3 is reduced to one-half its original value. 5. The current through R3 remains the same. correct Explanation: The voltage across R3 is the E of the battery, and is unchanged. The current through E . R3 remains the same, R3 005 (part 2 of 2) 10.0 points

Consider going clockwise around the outside loop (meaning all the way around the outer edge, ignoring the segment in the middle). Which expression is the correct loop equation based on Kirchhoff’s laws? 1. E1 − I1R1 + E2 = 0 2. E1 + I1R1 − E2 = 0 3. −E1 + I1 R1 − E2 = 0 correct 4. −E1 − I1 R1 − E2 = 0

al hasani (laa2832) – HW 6 – coker – (56000) Explanation: A general rule to keep in mind when writing down loop equations is that, if you are going in the direction of current flow, voltage sources are positive and other components (like resistors) are negative. So, going clockwise around the circuit’s outer loop, starting with E1 (we can start anywhere on the loop), we are going against current flow, so the voltage source must be written as a negative quantity, −E1 . Then, since we are still going against current flow, the voltage drop across the resistor is positive, +I1 R1. Finally, on the bottom branch of the loop, we just have the voltage source E2, although since we are again opposing current flow, this must be a negative quantity. Setting these terms equal to zero, we obtain −E1 + I1 R1 − E2 = 0. 007 (part 2 of 2) 10.0 points Considering the same circuit from part 1, which of the following expressions would be the correct loop equation if you went counterclockwise around the upper loop? 1. −E1 − I2R2 − I1 R1 = 0 2. E1 + I2R2 − I1R1 = 0 3. E1 − I2R2 − I1R1 = 0 correct 4. −E 1 + I2R2 + I1 R1 = 0 Explanation: Again, we can start anywhere, so let’s start with the voltage source. Since we are going counterclockwise, we are actually going in the direction of current flow, so the voltage term is positive, +E1. Each of the voltage drops across the resistors are negative, because, again, we are following the flow of current. So we end up with E1 − I2R2 − I1 R1 = 0.

Consider the circuit E1 B E2 C

3

r1 i1 r2

A

D

i2 F

R

E

i What equation does the loop DCFED yield? 1. −E1 + i2 r2 − iR = 0 2. E1 − i2 r2 − iR = 0 3. −E2 − i2 r2 + 2iR = 0 4. E2 − i2 r2 + iR = 0 5. −E2 + i2 r2 − 2iR = 0 6. E2 − i2 r2 − iR = 0 correct 7. −E2 − i2 r2 − 2 iR = 0 8. −E1 − i2 r2 − iR = 0 9. E2 + i2 r2 − iR = 0 10. E2 + i2 r2 + iR = 0 Explanation: Recall that Kirchhoff’s loop rule states that the sum of the potential differences across all the elements around a closed circuit loop is zero. If a resistor is traversed in the direction of the current, the change in potential is −IR. If an emf source is traversed from the − to + terminals, the change in potential is +E . Apply the opposite sign for traversing the elements in the opposite direction. Thus by inspection, DCFED : E2 − iR − i2r2 = 0

008 (part 1 of 2) 10.0 points 009 (part 2 of 2) 10.0 points

al hasani (laa2832) – HW 6 – coker – (56000) Find the current i. Symmetry is applicable here. Let E1 = E2 = E = 12 V , r1 = r2 = r = 2.1 Ω , and R = 1.3 Ω .

4

Find the current through the 6 Ω resistor and the 3.4 V battery at the top of the circuit. Correct answer: 3.4 A.

Correct answer: 5.10638 A.

Explanation:

Explanation: Let : E1 = E2 = E = 12 V , r1 = r2 = r = 2.1 Ω , R = 1.3 Ω .

and

Let : R1 = 6 Ω , R2 = 23.6 Ω , E1 = 3.4 V , and E2 = 17 V .

E1 = E2 and r1 = r2. This implies that i1 = i2 . (Loops DCFED and ABFEA have identical loop equations.) Hence the junction rule yields

E1

I1

i 1 + i 2 = 2i 2 = i i i2 = 2

I2

i=

12 V r2 = 2.1 Ω R+ 1.3 Ω + 2 2

010 (part 1 of 2) 10.0 points The currents are flowing in the direction indicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. Assume the batteries have zero internal resistance. 3.4 V 6Ω I1

I1 − I2 − I3 = 0 .

(1)

Pay attention to the sign of the battery and the direction of the current in the figure. Using the lower circuit in the figure, E2 + I 2 R 2 = 0 (2) −E2 −17 V I2 = = = −0.720339 A . R2 23.6 Ω Then, for the upper circuit E1 − I 2 R 2 − I 1 R 1 = 0 (3) E1 + E2 − I 1 R 1 = 0 E1 + E2 3.4 V + 17 V I1 = = 6Ω R1 = 3.4 A .

23.6 Ω Alternate Solution: Using the outside loop 17 V

I3

I3 At nodes,

E2

= 5.10638 A .

I2

R2 E2

Substituting this into the loop equation DCFED, i E2 − iR − r2 = 0 2

R1

−E1 − E2 + I1 R1 = 0 E1 + E2 I1 = . R1

(4)

al hasani (laa2832) – HW 6 – coker – (56000) 011 (part 2 of 2) 10.0 points Find the current through the 23.6 Ω resistor in the center of the circuit.

From the junction rule, I = i1 + i2 . Applying Kirchhoff’s loop rule, we obtain two equations: E1 = i1 r1 + I R E2 = i2 r2 + I R = (I − i1 ) r2 + I R = −i1 r2 + I (R + r2) ,

Correct answer: −0.720339 A. Explanation: From Eq. (2) I2 = −

E2 17 V = −0.720339 A . =− R2 23.6 Ω 012

12.9 V

(1)

(2)

Multiplying Eq. (1) by r2, Eq. (2) by r1 , E1 r2 = i1 r1 r2 + r2 I R E2 r1 = −i1 r1 r2 + I r1 (R + r2 )

10.0 points

25.8 V

5

Adding, 6Ω

E1 r2 + E2 r1 = I [r2 R + r1 (R + r2)] E1 r2 + E2 r1 r2 R + r1 (R + r2 ) (25.8 V) (5 Ω) + (12.9 V) (6 Ω) = (5 Ω) (22 Ω) + (6 Ω) (22 Ω + 5 Ω)

5Ω

I=

22 Ω

= 0.758824 A . Find the current through the 22 Ω lowerright resistor.

013 10.0 points Consider the circuit

Correct answer: 0.758824 A. E

Explanation:

B E1

B E2 C

r1 i1 r2

F

− −

q+

A

+ +

R

The potential difference is D

1. VB − VA = − iR − 2. VB − VA = − iR −

E

3. VB − VA = − iR +

I 4. VB − VA = + iR + Let : E1 = 25.8 V , E2 = 12.9 V , r1 = 6 Ω , r2 = 5 Ω , and R = 22 Ω .

i

+

A

i2 R

− −

5. VB − VA = + iR − 6. VB − VA = − iR + 7. VB − VA = + iR −

q C q C q C q C q C q C q C

+E − E correct −E −E +E +E −E

al hasani (laa2832) – HW 6 – coker – (56000) 8. VB − VA = + iR +

q +E C

Explanation: Applying Kirchhoff’s law gives q VB − VA = −iR − − E. C

015 (part 2 of 2) 10.0 points A long time after the switch has been closed, the current I∞ supplied by the battery is

2. I∞

The switch has been open for a long period of time.

3. I∞

4. I∞ = 0 .

R2

5. I∞ =

C V

S

Immediately after the switch is closed, the current supplied by the battery is V . correct R1 V (R1 + R2 ) 2. I0 = . R1 R2 1. I0 =

3. I0 = 0 . V . R1 + R2 V . 5. I0 = R2 Explanation: Before the switch is closed, there is no charge on the capacitor, so the voltage is zero across the capacitor at this time. Because it is not possible to change the charge on the capacitor like a step function (or the current should be infinitely large), immediately after the switch is closed, the voltage across the capacitor (and R2 ) is still zero. Therefore, the voltage across R1 is V ; i.e., think of the capacitor as being a short-circuit for this instant of time. So the current supplied by the battery, which is the same as the current going through V R1, is I0 = . R1 4. I0 =

V (R1 + R2) . R1 R2 V . = R2 V . correct = R1 + R2

1. I∞ =

014 (part 1 of 2) 10.0 points

R1

6

V . R1

Explanation: After a long time, the capacitor has been charged and remained stable. That means the current going through R1 is the same as the current going through R2 ; i.e., think of the capacitor as being a open-circuit for this time. So we can write down the equation V = I∞ R 1 + I∞ R 2 , which gives the current I∞ as I∞ =

016

V . R1 + R2 10.0 points

At t = 0 the switch S is closed with the capacitor uncharged. 69 µF 4 kΩ 25 V

S

i

What is the charge on the capacitor when I = 2 mA? Correct answer: 0.001173 C. Explanation:

al hasani (laa2832) – HW 6 – coker – (56000) C

R E S

i

Let : C = 69 µF = 6.9 × 10−5 F , E = 25 V , R = 4 kΩ = 4000 Ω , and I = 2 mA = 0.002 A . Applying Kirchhoff’s law E−

Q −IR = 0 C

Q = C (E − I R) = (6.9 × 10−5 F) × [25 V − (0.002 A)(4000 Ω)] = 0.001173 C . 017 10.0 points A resistor and a capacitor are in series. There is no charge or current in the circuit. At t = 0 a battery is put into series with the other two components, current flows, and the capacitor charges. If the resistance is 2 Ohms, the capacitance is 1 F, and the battery voltage is 6 V, how much positive charge is on the + capacitor plate after 3 seconds? 1. 3.33 C 2. 4.66 C correct 3. 1.12 C 4. 2.59 C 5. 0.76 C 6. 6.0 C 7. 5.55 C Explanation:

7

In this case, as shown in class and in the text, the current falls off in the circuit as the capacitor is charged, according to I(t) = (V /R) exp(−t/RC). In this case V /R = 3 Amps, and RC = 2 sec, so at 3 sec, I = (3) exp(−3/2) = 0.67 Amps. Now, using Kirchhoff’s 2nd rule around the circuit we have V − IR − Q/C = 0 and we can easily solve this equation for Q at 3 sec. Q = C(V − IR) = 4.66 C. Note that the maximum charge that eventually appears on the capacitor is given by Q = CV = 6 Coulombs....