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NOTES CHAPTER-21 CURRENT AND DIRECT-CURRENT CIRCUITS _______________________________________________________________________ 1. Current and current density: Current I = Q/t, is the amount of charge flowing past a certain cross sectional area per unit time. The unit for current is the Ampere (A). (1 A = 1C/s) The current can also be expressed as I = nqvdA, where n is the number of charge carriers per unit volume, q is the carrier charge, vd is the drift velocity, and A is the cross sectional area. The current density J = nqvd. The magnitude J = I/A. 2. Ohm's Law: V = IR. 3. Resitivity: = RA/l , where R is the resistance of a conductor of length l and of uniform cross-sectional area A. The resistivity, = RA/l = VA/Il = E/J. The resistance R is a property of a given resistor whereas is characteristic of the material from which the resistor is fabricated. 4. Emf, and terminal voltage Vab of a battery with internal resistance r. = Vab ir, where i is the current flowing through the circuit containing the battery. 5. Power: P = Vab I ( for a general circuit element) = I2R ( power into a resistor) 6. Kirchhoff's junction and loop rules [a] Junction rule: The algebraic sum of the currents into any junction is zero Water pipe analogy for the junction rule: the flow rate of water entering a pipe equals the rate of water leaving. [b] The loop rule: the algebraic sum of all potential differences in a loop must be zero. Remember the following helpful conventions when analyzing dc-circuits using Kirchhoff's junction and loop rules. In the figures below `travel' is the direction in which one imagines going around the loop. This `travel' direction doesn't have to be the direction of the current in a given circuit element in the loop. Potential increase Potential drop 7. Series and parallel connections [a] resistors in series R1 R2 b c R3 d a R = R1 + R2 + R3 R is the net resistance The same current flows through each resistor The potential differences add: Vad = Vab + Vbc + Vcd [b] resistors in parallel 1/R = 1/R1 +1/R2 R is the net resistance The same potential difference Vab = I1R1 = I2R2 exists across the two resistors. The branch currents add: I = I1 + I2 I1 R1 b a I I2 R2 SOLVED EXAMPLES 12 V c d 36 I2 b I1 8 24 I3 6 e 12 a 24 V f 1. Consider the circuit shown in the diagram. [a] Write down Kirchhoff's junction rule at `b' and loop equations for the loops abefa and bcdeb Solution: At junction `b', - I3 + I1 - I2 = 0 I3 = I1 - I2 For loop abefa, - 8I1 - 6I3 -24I1 + 24 = 0 For loop bcdeb - 12I2 12- 36I2 +6I3- = 0 - 54I2+6I1- 12 = 0 or substituting for I3 from [1] ........[3] or substituting for I3 from [1] - 38I1+6I2+24 = 0 ........[2] ...[1] [b] Solve equations obtained in part [a] to determine currents I1 , I2. and I3. Multiply eq. [2] by 9 and add to eq. [2] and solve the resulting equation for I1 to get, I1 = 204/336 = 0.61A Substitute this value of I1 in eq.[2] to get, I2 = - 0.15A I3 = I1 - I2 = 0.61 + 0.15 = 0.76 A [c] Determine the power dissipated in the 6.0- resistor. P = 6.0 (I1 - I2)2 = (0.76)2 6.0 = 3.47W 2. Consider the circuit shown to the right. [a] Write down I1 in terms of I2 and I3. 8V c d 2 I2 b I3 I1 2 4 e 8V 2 1 = 2 - 3 ..........[1] [b] Write down the Kirchhoff's loop equations for: (i) loop abefa 2 21 - 43 - 8 + 2 1+4 = 0 1 - 3 - 1 = 0 ...........[2] (ii) loop bcdeb 22 - 8 + 2 2+ 8 +43 = 0 2 = - 3 ........[3] a 4V f [c] Solve the three equations above to determine I1, I2, and I3. From [1] and [3] 1 = -3- 3 = -2 3 ........[4] Substitute [4] in [2] to get - 23- 3 = 1 or 3 = - 1/3 which gives 2 = 1/3 and from [1] 1 = 1/3 +1/3 = 2/3 3. At t = 0 s the switch, s in the diagram is thrown up (to a) to close the RC circuit and start charging the capacitor C = 20F. After 160 sec. the charge on the capacitor is 8.0x10-3C. [a] What is the value of the resistance R ? During charging of C. Q = Q0(1 exp[ - t/RC] 8*10-3 = 500*20*10-6(1 exp[ - 160/(R+Ro)C] 0.2 = exp[ - 160/(R+Ro)C] or 5 = exp[160/(R+Ro)C] (R+Ro)C ln 5 = 160 (R+Ro)20*10-6*1.61 = 160 , solve it for R = 4*106 V = 500V 1.0 M C=20.0F R a b s [b] What is the potential difference across C at t = 120 s.? Q = Q0(1 exp[ - t/RC] ...[1] Divide both sides of [1] by C and note that V = Q/C to rewrite [1] as: V(t) = V0(1 exp[ - t/RC], and thus V(t=120s) = 500(1 exp[ - 120/(R+Ro)C] = 500(1 exp[ -120/100]) = 349 V [c] After C is fully charged, the switch s is thrown to b to start discharging the capacitor. How much energy is lost as heat during the first 100 sec. after the capacitor started discharging? The voltage across C at t = 100s is V = 500 exp[ - 100/100] = 184 V The energy lost as heat due to resistive heating is equal to the change in the stored energy in the capacitor. Energy lost = C[(500)2 (184)2] = 2.16W 4. At t = 0.0 s the switch, s in the diagram is flipped up (to a) to close the RC circuit and start charging the capacitor C = 20F. After 160 sec. the charge on the capacitor is 8.0x10-3C. [a] What is the value of the resistance R ? V = 500V 15.0 M R C=20.0F a b s Q = Q0(1 exp[ - t/RC] ( Note Q0 = CV) ...[1] 8*10-3 = 500*20*10-6(1 exp[ - 160/RTC] simplifying [2] gives, 0.2 = exp[ - 160/RTC] or 5 = exp[160/RTC] ...[2] (where 1/RT = 1/R +1/15*106 or RT = R(15*106)/ R+15*106) RTC ln 5 = 160 or RT = 160/[ C ln 5] = 5*106 = R(15*106)/( R+15*106) it Solve to get R = 7.5*106 [b] What is the potential difference across C at t = 120 s.? V(t=120s) = 500(1 exp[ - 120/RTC] = 500(1 exp[ -120/100]) = 349 V [c] After C is fully charged, the switch s is thrown to b to start discharging the capacitor. How much energy is lost as heat during the first 100 sec. after the capacitor started discharging? The voltage across C at t = 100s is V = 500 exp[ - 100/100] = 184 V The energy lost as heat due to resistive heating is equal to the change in the stored energy in the capacitor. Energy lost = C[(500)2 (184)2] = 2.16W Lecture Problems _______________________________ 21.2 The period of revolution for the sphere is T = q q = 2 T 2 , and the average current represented by this revolving charge is I = 21.20 The total clock power is . (270 10 From e = 6 J s 3 600 s 12 clocks 2.50 = 2.43 10 J h . clock 1 h ) W out , the power input to the generating plants must be: Q in Q in W out t 2.43 1012 J h = = = 9.72 1012 J h t e 0.250 and the rate of coal consumption is Rate = (9.72 1012 J h ) 21.34 1.00 kg coal = 2.95 105 kg coal h = 295 m etric ton h 33.0 106 J . +15.0 - (7.00 )I1 - (2.00 )(5.00) = 0 5.00 = 7.00I1 so I1 = 0.714 A I1 + I 2 - 2.00 A = 0 0.714 + I 2 = 2.00 so I 2 = 1.29 A + - 2.00 (1.29 ) - 5.00 ( 2.00 ) = 0 = 12.6 V 21.59 (a) q = CV 1 - e-t ( RC ) -10.0 2.00 106 1.00 10-6 q = 1.00 10-6 F (10.0 V ) 1 - e ( ) ( )( ) = 9.93 C (b) I= dq V -t = e dt R RC 10.0 V -5.00 e = 3.37 10-8 A = 33.7 nA I= 2.00 106 (c) d 1 q2 q dq q dU = = I = dt 2 C C dt C dt dU 9.93 10-6 C = 3.37 10-8 A = 3.34 10-7 W = 334 nW dt 1.00 10-6 C V ( ) (d) Pbattery = I = 3.37 10-8 A (10.0 V ) = 3.37 10-7 W = 337 nW ( ) ANSWERS TO QUESTIONS: The end-of-Chapter Questions are a good source to clarify the concepts introduced in a particular chapter. In the following we have discussed answers to some selected Questions. For the statements of Questions, please consult your Text. Q21.1 Consider all of the eastbound lanes on a highway, taken together. Individual vehicles correspond to bits of charge. The number of vehicles that pass a certain milepost, divided by the time during which they go past, corresponds to the value of the current. Q21.2 Geometry and resistivity. In turn, the resistivity of the material depends on the temperature. Q21.3 The radius of wire B is 3 times the radius of wire A, to make its cross sectional area 3 times larger. Q21.7 Because there are so many electrons in a conductor (approximately 1028 electrons m 3 ) the average velocity of charges is very slow. When you connect a wire to a potential difference, you establish an electric field everywhere in the wire nearly instantaneously, to make electrons start drifting everywhere all at once. Q21.8 The 25 W bulb has a higher resistance. The 100 W bulb carries more current. Q21.9 One ampere hour is 3 600 coulombs. The ampere hour rating is the quantity of charge that the battery can lift though its nominal potential difference. Q21.10 In series, the current is the same through each resistor. Without knowing individual resistances, nothing can be determined about potential differences or power. Q21.11 In parallel, the potential difference is the same across each resistor. Without knowing individual resistances, nothing can be determined about current or power. Q21.14 A wire or cable in a transmission line is thick and made of material with very low resistivity. Only when its length is very large does its resistance become significant. To transmit power over a long distance it is most efficient to use low current at high voltage, minimizing the I 2 R power loss in the transmission line. Alternating current, as opposed to the direct current we study first, can be stepped up in voltage and then down again, with high-efficiency transformers at both ends of the power line. Q21.17 Kirchhoff's junction rule expresses conservation of electric charge. If the total current into a point were different from the total current out, then charge would be continuously created or annihilated at that point. Kirchhoff's loop rule expresses conservation of energy. For a single-loop circle with two resistors, the loop rule reads +e - IR1 - IR 2 = 0 . This is algebraically equivalent to qe = qIR1 + qIR 2 , where q = It is the charge passing a point in the loop in the time interval t . The equivalent equation states that the power supply injects energy into the circuit equal in amount to that which the resistors degrade into internal energy. Suppose e = 12 V and each lamp has R = 2 . Before the switch is closed 12 V = 2 A . The potential difference across each lamp is (2 A )(2 ) = 4 V . the current is 6 The power of each lamp is (2 A )(4 V ) = 8 W , totaling 24 W for the circuit. Closing the Q21.21 switch makes the switch and the wires connected to it a zero-resistance branch. All of the current through A and B will go through the switch and (b) lamp C goes out, with zero voltage across it. With less total resistance, the (c) current in the battery 12 V =3A 4 becomes larger than before and (a) lamps A and B get brighter. (d) The voltage across each of A and B is (3 A )(2 ) = 6 V , larger than before. Each converts power (3 A )(6 V ) = 18 W , totaling 36 W, which is (e) an increase.

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