Solution Derivations for Capa #7

Transcription

1 Solution Derivations for Capa #7 1) Consider te beavior of te circuit, wen various values increase or decrease. (Select I-increases, D-decreases, If te first is I and te rest D, enter IDDDD). A) If R1 decreases, te current troug R3. B) If R3 decreases, te current from te battery. C) If R6 decreases, te power output from te battery. D) If R1 decreases, te current troug R6. E) If te voltage V decreases, te total current from te battery. F) If R3 decreases, te current troug R4. A) Increases, Te equivalent resistance of te parallel resistor decreases. From V = IR, te current will increase. B) Increases. Same as (A). C) Increases. P = IV. If te current rises due to a decrease in resistance, te power will also rise. D) Increases. See (A). E) Decreases. Less voltage to pus current troug te circuit. F) Decreases. More will want to flow troug R3 wit a now lower resistance. 2) A battery wit an emf of V as an internal resistance r. Wen connected to a resistor R te terminal voltage is V and te current is 0.11 A. Wat is te value of te external resistor R? Use Om as units of resistance. V E = Given (EMF) V = Given I = Given r =? R =? 1

2 To solve tis problem, you don t ave to worry about te internal resistance. Tus, Om s law applies directly to te given information. V = IR R = V I Note tat te voltage is not equal to te supplied EMF wen te resistor is connected. 3) Wat is te internal resistance r of te battery? Internal resistance of a battery is similar to aving two resistors in series. Tus, te equivalent resistance of te circuit is simply From V = IR, R tot = r + R R tot = r + R = V E I Tus, te internal resistance is r = V E I R Tis time te total voltage supplied by te EMF is used since we are taking into account tat tere is internal resistance. In te previous problem, we used te remaining voltage available after te voltage drop across te internal resistor. 4) Te picture sows a battery connected to two cylindrical resistors in parallel. Bot resistors are made of te same material and are of te same lengt, but te diameter of resistor A is twice te diameter of resistor B. (For eac statement select T True, F False). A) Te resistance of wire B is four times as large as te resistance of wire A. B) Te current troug te battery is five times larger tan te current troug wire B. C) Te power dissipated in wire A is 16 times te power dissipated in wire B. D) Te resistance of wire B is twice as large as te resistance of wire A. E) Te voltage drop across wire B is larger tan te voltage drop across wire A. A) True. Resistance is given by R = ρl A. Tus, if te area increases, te resistance decreases by te square of tat factor (in te cylinder case). 2

3 B) True. Te resistance troug B is four times tat in A. Togeter, tey dissipate all te current. Te total current is ten 5 times te current in A. (Te current in A plus 4 times tat current in B). C) False, P = IV. From (E), tey bot ave te same voltage drop. From (B), A as 4 times te current. Tus, it only dissipates 4 times te power. D) False. See (A) E) False. Te voltage drop across a parallel configuration of resistors is equal. It must be from te definition. Te wires are connected togeter at eac end, so tey must be at te same voltage. Tus, te current must split accordingly. 5) Consider te tree circuits sown above. All te resistors and all te batteries are identical. (For eac statement select T True, F False). A) Te power dissipated in circuit A is twice te power dissipated in circuit B. B) Te total power dissipated in circuit C is twice te total power dissipated in circuit B. C) Te current troug a resistor is te same in circuits A and B. D) Te current troug a resistor is te same in circuits A and C. E) Te voltage across a single resistor in circuit C is twice te voltage across a single resistor in circuit B. A) True. P = V 2 R Te resistance in B is twice tat in A, so A dissipates twice te power. B) False. It is actually 4 times. Te equivalent resistance of C is 1 1 R + 1 R 3 =

4 Plugging into we get wic is four times tat of B: C) False. From V = IR, we get P = V 2 R P = 2 V 2 R P = V 2 2R I = V R Since te resistance is greater in B, it as less current. D) True. Te current in C is twice tat in A, since I = V R and te resistance in C is R 2 But, it is split among two equal resistors. So, eac carries alf tat amount of current. Tus, te current troug a resistor in C is te same as te current troug te resistor in A. E) True. Eac resistor in C as te entire voltage drop across it. However, circuit B as two equal resistors in series, so te voltage must drop across eac of tem. Since tey are equal resistance, te voltage drops alf of its full amount across eac one. Remember to answer in te form TFFTT 6) Consider te sections of two circuits illustrated above. All resistances sown ave a finite, non-zero, resistance. (Give ALL correct answers, i.e., B, AC, BCD...) WARNING: You ave only 8 tries for tis problem. A) R cd is always less tan R 3. B) R ab is always less tan R 1. C) After connecting c and d to a battery, te current troug R 3 always equals te current troug R 4. D) After connecting a and b to a battery, te voltage across R 1 always equals te voltage across. A) False, Resistors in series always add to a greater resistance tan eiter one 4

5 individually. B) True, Resistors in parallel always add to a lesser resistance tan eiter one individually. C) True, Te current is te same for all resistors in series as it as no were else to go. D) True, Te voltage is te same for all resistors in parallel because eac of teir ends are connected to eac oter and te voltage must be equal on bot sides. Remember to answer tis one in te form BCD. 7) In te section of circuit below, R 1 = Ω, = Ω and te voltage difference V a V b = V. Te current i = amps. Find te value of R 3. Use Om as your units. R 1 = Given = Given V = Given I = Given We must find te equivalent resistance of te diagram. Tere is one set in parallel and one oter resistor in series. Tus, we can add te two togeter to get 1 R tot = R = R 1 + R 3 R 3 + R 3 From V = IR, R = V I R = R 1 + R 3 = V + R 3 I R 3 = V + R 3 I R 1 ( V R 3 = 1) I R ( + R 3 ) = V I + V I R 3 R 1 R 1 R 3 R 3 + R 1 R 3 V I R 3 = V I R 1 ( R 3 + R 1 V ) = V I I R 1 R 3 = V I R 1 + R 1 V I 8) Wat is te current troug R 3? 5

6 We are given te current and it must all travel troug te diagram. If we find te voltage drop across te first resistor, we know te remaining voltage must drop across te second resistor and can ten find te current. V 1 = IR 1 Since tis is te voltage drop across te first resistor, tere is still V V 1 total voltage to drop over te parallel resistors. Tus, V 3 = V V 1 = IR 3 I = V V 1 R 3 9) Consider te following diagram. (Give ALL correct answers in alpabetical order, i.e., B, AC, BCD...) A) E 1 E 2 = i 1 R 1 + i 2 B) i 1 + i 2 = i 3 C) i 1 = i 2 + i 3 D) E 2 E 3 = i 3 R 3 i 2 E) E 1 E 3 = i 1 R 1 i 3 R 3 F) E 2 E 1 = i 1 R 1 i 2 G) E 3 E 1 = i 3 R 3 i 1 R 1 A) True B) False C) True D) True E) False F) False 6

7 G) True Use Kircoff s rules and follow a loop around te circuit. Only pay attention to te directions tat are labeled, not wic way you feel te current sould flow. Remember to enter tis question in te form ACDG. 10) Wat is te magnitude of te current troug R 1? DATA: R 1 = = R 3 = 16.0 Ω. V 1 = 12.0 Volts. V 2 = 2V 1. Assigning arbitrary directions to te current flow and obtaining 3 equations, V 1 = I 2 + I 1 R 1 (1) V 2 = I 3 R 3 + I 2 (2) I 3 = I 1 + I 2 (3) Now solving for te tree unknowns wit te tree equations, plugging (3) into (2) V 2 = (I 1 + I 2 ) R 3 + I 2 = I 1 R 3 + I 2 R 3 + I 2 (4) Solving (1) for I 2 and plugging into (4), V 1 = I 2 + I 1 R 1 I 2 = I 1R 1 V 1 V 2 = I 1 R 3 + I 2 R 3 + I 2 V 2 = I 1 R 3 + I 1R 1 V 1 R 3 + I 1R 1 V 1 V 2 = I 1 R 3 + I 1R 1 V 1 R 3 + I 1 R 1 V 1 V 2 + V 1 = I 1 R 3 + (I 1 R 1 V 1 ) R 3 + I 1 R 1 = I 1 R 3 + I 1R 1 R 3 V 1 R 3 V 2 + V 1 + V 1 R 3 = I 1 R 3 + I 1R 1 R 3 + I 1 R 1 = I 1 Tus, I 1 = V 2 + V 1 + V 1 R 3 R 3 + R 1R 3 + R 1 7 ( R 3 + R 1R 3 + R 1 ) + I 1 R 1

8 11) Te circuit wic was at position a for a long time is suddenly switced to position b at time t = 0. (For eac statement select T True, F False). WARNING: you ave 8 tries only for tis question. A) In te instant after te switc is trown, te voltage across te capacitor is zero. B) In te instant after te switc is trown te voltage across te resistor is zero. C) Te current troug te resistor equals te current troug te capacitor at all times. D) In te instant after te switc is trown te current troug te capacitor is zero. A) False, te capacitor is currently carged, so tere is a voltage cange. B) False, te capacitor as a voltage cange across its plates so tere will be a voltage across te resistor. C) True, te resistor and capacitor are in series; te current must always be equal. D) False. Carge from te capacitor is moving around te circuit, so tere is a current troug it. 12) Wat is V c at time t = 2.0 ms? DATA: R = 475 Ω, C = 4.0 µf, V b = 2.95 V. R = Given C = Given V b = Given t = Given Te voltage for a discarging capacitor is given by equation 28-8 on page 710 of te textbook. V = V 0 e t/rc Wic equates to our variables as V C = V b e t/rc 8

9 13) Wat is te energy dissipated in te resistor after time t = 0 (tat is, for all time after t = 0, meaning te during te time period from t = 0 to infinity.) One of te energy equations for a capacitor is U = 1 2 CV 2 As time goes to infinity, te entire capacitor will discarge. It was carged to te initial voltage of te battery. 14) A solid truncated cone is made of a material of resistivity 3.50 Ω m. Te cone as eigt = 1.06 m, and radii a = 0.33 m and b = 0.89 m. Wat is te total resistance for tis cone? (Use Om as your units.) ρ = Given a = Given b = Given = Given To solve tis problem we need to treat te frustum like a collection of infinitely many resistors tat ave te same radius over teir lengt. In oter words, we need to integrate. Te equation for resistance is R = ρl A were L is te lengt of te resistor and A is te cross-sectional area. We will be taking small portions of tis resistor so tat te radius is te same over tat lengt. Tus, te lengt will be te differential dx. We will be integrating over te lengt of te resistor taking te left side to be x = 0 and te rigt side to be x =. So far, ρ R = 0 A dx Now, we must relate A to dx. Wen x = 0, r = a and wen x =, r = b. Tus, we can tink of te outer surface of te cone as a line tat must go toug te points (0, a) and (, b). Using point slope form, te slope is b a. Tus, te equation of tis line using point-slope form is y a = b a x y = b a x + a Basically, y is simply a relation of te radius of te resistor to te lengt of it. Tus, it is r. Te area of eac of tese little slices is ( b a A = πr 2 = π x + a 9 ) 2

10 Substituting tis back into te integral, R = 0 ρ π ( b a x + a) 2 dx = ρ π 0 Using a u-substitution of u = b a b a x + a, du = transformed integral R = ρ π (b a) a R = ρ πb (b a) + b du u 2 = dx ( b a x + a) 2 ( ρ 1 π (b a) u ρ πa (b a) dx, dx = du, we get te b a b u=a ) 10