PHY 1160C Homework Chapter 19: Direct-Current Circuits Ch 19: 5, 11, 18, 37, 46, 54, 75, 80, 83 19.5 A carbon resistor has a resistance of 50 W when measured at room temperature. What is its resistance when immersed in liquid nitrogen which is at its boiling temperature ( 196 C)? R = Ro(1 + a D T) Take room temperature to be about 25 C. The boiling temperature of liquid nitrogen is 196 C. Therefore D T = Tf Ti = ( 196 C) (25 C) = 221C For carbon, the temperature coefficient a, is 0.0005/C R = Ro(1 + a D T) R = (50 W )(1 + [ 0.0005/C ][- 221C ]) R = (50 W )(1 + 0.1105) R = (50 W )(1.1105) R = 55.5 W For most materials, a reduction in temperature would mean a reduction in resistance. But carbon is the exception; reducing its temperature causes an increase in resistance. 19.11 A resting human neuron has a potential difference of about 75 m across it and the current through it is about 200 µa. How much power is being released in the neuron? P = I P = (200 x 10 6 A)(75 x 10 3 ) P = 1.5 x 10 5 A- Ch 19, p 1
Now, what is an A-? P = 1.5 x 10 5 A- A- = A- C/s A J/C A- = C s J C A- = J s P = 1.5 x 10 5 W W J/s = W 19.18 Ordinarily wire is specified as 10 gauge wire or 12 gauge wire and so on rather than specifying its diameter. The table below gives the correlation between gauge specification and diameter. Notice that the larger the gauge number, the smaller the wire. For each gauge of wire, calculate the resistance of a 10 m piece of copper wire. Gauge number diameter 8 0.3264 cm 10 0.2588 12 0.2053 The resistance R is given by R = r L/A. For copper, the resistivity is r = 1.59 x 10 8 W -m. For each piece of wire, the length is L = 10 m. We need to calculate the cross-section area A = r 2 = (d/2) 2 = d 2 /4. For simple and repeated calculations, I like to use a spreadsheet. gauge diameter radius radius Area Resistance number ( cm ) ( cm ) ( m ) ( m^2 ) ( W ) 8 0.3264 0.1632 1.63e-3 8.37e-6 0.0190 10 0.2588 0.1294 1.29e-3 5.26e-6 0.0302 12 0.2053 0.1027 1.03e-3 3.31e-6 0.0480 For ClarisWorks (and most spreadsheets) the notation 1.63e 3 means 1.63 x 10 3. Ch 19, p 2
19.37 Calculate the power supplied by a 12 volt power supply in the circuit if the resistors have the following values: R1 = 250 W, R2 = 750 W, and R3 = 1000 W. Calculate the power used by each resistor. R 1 R 2 R 3 Figure 19.33 Problem 19.37 These resistors are in series so the equivalent resistor has a resistance Req given by Req = R1 + R2 + R3 Req = 250 W + 750 W + 1000 W = 2000 W The current through this equivalent resistor would be R = /I I = /R = (12 )/(2000 W ) = 0.006 A = 6 ma This is the current supplied by the battery and the power supplied by the battery is P = I P = (0.006 A)(12 ) = 0.072 W = 72 mw P = 72 mw Now, to find the power absorbed by each of the resistors, we could find the voltage across each of the resistors. We know the current through each of the resistors (0.006 A = 6 ma) and we know each resistance so we will use an expression for power that requires only the current and resistance, P = I P = I (IR) = I 2 R P1 = (0.006 A) 2 (250 W ) = 0.009 A 2 W = 0.009 W P2 = (0.006 A) 2 (750 W ) = 0.027 A 2 W = 0.027 W P3 = (0.006 A) 2 (1000 W ) = 0.036 A 2 W = 0.036 W Ch 19, p 3
Ptot = P1 + P2 + P3 = 0.072 W And, of course, that is just the power supplied by the battery to the equivalent resistor. We need to convince ourselves that the units are correct; that is, that units of A 2 W are, indeed, units of watts, W. A 2 W = A 2 (/A) = A- and in problem 19.11 we already found that A- = W that is, that (amp)(volt) = watt 19.46 Calculate the power supplied by the 12 volt power supply in the circuit above if the resistors have the following values: R1 = 250 W, R2 = 750 W, and R3 = 1000 W. Calculate the power used by each resistor. R 1 R 2 R 3 Figure 19.34 Problem 19.46 These resistors are connected in parallel so the equivalent resistance, Req, is given by 1 R eq = 1 R 1 + 1 R 2 + 1 R 3 1 = 1 R eq 250 W + 1 750 W + 1 1000 W 1 = 0.00633 R eq W Req = 158 Ch 19, p 4
The current supplied by the battery to this equivalent resistor would be R = /I I = /R I = 12 /158 W = 0.076 A The power supplied by the battery to this equivalent resistor would be P = I P = (0.076 A)(12 ) = 0.912 W P = 0.912 W To find the power supplied to each individual resistor, we can use an expression that requires the resistance and the voltage since we know that for each resistor, P = I = (/R) = 2 / R For each resistor, = 12, since they are in parallel, P1 = 2 /R1 = (12 ) 2 /250 W = 0.576 W P2 = 2 /R2 = (12 ) 2 /750 W = 0.192 W P3 = 2 /R3 = (12 ) 2 /1000 W = 0.144 W Ptot = P1 + P2 + P3 = 0.912 W And, of course, that is just the power supplied to the equivalent resistor. And we need to convince ourselves that we have, indeed, handled the units correctly. That is, is a watt (W) truely the same as 2 /W? 2 /W = 2 /(/A) = /(1/A) = -A = W as we saw in problem 19.11. Ch 19, p 5
19.54 Find the equivalent resistance in the circuit shown in Figure 19.36 for R1 = 100 W, R2 = 500 W, R3 = 750 W, R4 = 1000 W, R5 = 50 W. If these are connected to a 12 volt battery, what is the power supplied to the equivalent resistor? Find the current through each of the individual resistors and the power supplied to each. R 1 R 2 R R 3 4 R 5 Figure 19.36 Problem 19.54 Resistors R2, R3, and R4 are connected in parallel, so the first thing we will do is find the single resistor that is equivalent to just those three. 1 R eq = 1 R 2 + 1 R 3 + 1 R 4 1 = 1 R eq 500 W + 1 750 W + 1 1000 W 1 = 0.00433 R eq W Req = 231 W We will shortly need to calculate another equivalent resistance, so let s call this one Req,1. Now our circuit looks like this Ch 19, p 6
R 1 R eq,1 R 5 I think it is clear that these three resistors, R1, Req,1, and R5 are in series. If that is not entirely clear, it may help to redraw the circuit diagram as R 1 R eq,1 R 5 For these three resistors in series, the equivalent resistance is given by Req = R1 + Req,1 + R5 and to avoid confusion, we will label this one as Req,2; that is, Req,2 = R1 + Req,1 + R5 Req,2 = 100 W + 231 W + 50 W Req,2 = 381 W That is, our equivalent single-resistor circuit would be R eq,2 with Req,2 = 381 W. The current through this circuit is Ch 19, p 7
R = /I I = /R I = 12 / 381 W I = 0.0315 A The power given out by the battery and absorbed by this single resistor is P = I P = (0.0315 A)(12 ) P = 0.378 W This means the same current of 0.0315 A flows through each of the resistors in the previous all series circuit, R 1 R eq,1 R 5 I = 0.0315 A Now we can find the potential difference across R1, R5, and Req,1 and the power absorbed by each resistor. R = /I = I R 1 = (0.0315 A)(100 W ) = 3.15 5 = (0.0315 A) (50 W ) = 1.575 eq,1 = (0.0315 A) (231 W ) = 7.275 Notice that 1 + eq,1 + 5 equals the 12.0 volts of the battery. Knowing the current through each of these resistors and the voltage across each, it is now easy to calculate the power used by each, P = I P1 = I 1 = (0.0315 A)(3.15 ) = 0.0992 W P5 = I 5 = (0.0315 A)(1.575 ) = 0.0496 W Peq,1 = I eq,1 = (0.0315 A)(7.275 ) = 0.2292 W Ch 19, p 8
As an intermediat check, notice that P1 + Peq,1 + P5 = 0.378 W which is the power supplied to the single equivalent resistor Req,2. That is as it ought to be, of course. Each of the resistors R2, R3, and R4 in the bank of parallel resistors has a voltage of 7.275 across it so the current through each is given by R = /I I = /R I2 = 7.275 / 500 W = 0.01455 A I3 = 7.275 / 750 W = 0.0097 A I4 = 7.275 / 1000 W = 0.007275 A Notice, again, that the sum of these currents, I2 + I3 + I4 = 0.0315 A is the current through resistors R1 and R5 with which this group are in series. The power supplied to each of these resistors is P = I P2 = I2 = (0.01455 A)(7.275 ) = 0.1058 W P3 = I3 = (0.0097 A)(7.275 ) = 0.0706 W P4 = I4 = (0.007275 A)(7.275 ) = 0.0529 W As a final check, look at the power supplied to each individual resistor, Ptot = P1 + P2 + P3 + P4 + P5 Ptot = 0.0992 W + 0.1058 W + 0.0706 W + + 0.0529 W + 0.0496 W = 0.3781 W Ptot = 0.378 W And that is the power supplied by the battery to the single equivalent resistor that we calculated long ago. Ch 19, p 9
19.75 A 12.0 automobile battery (ie, a battery with an emf of 12.0 ) has an internal resistance of 0.15 W. The equivalent resistance of the headlights on the automobile is 6 W. Find the terminal voltage of the battery when the headlights are turned on. The resistance of the starter motor is only 0.1 W. Find the terminal voltage of the battery when the starter is also turned on. A good circuit diagram will be very helpful. R int R E = 12.0 When only the lights are on, the external resistance is R = 6.0 W. The total equivalent resistance of the entire circuit including the internal resistance of the battery then is Rtot = Rint + R Rtot = 0.15 W + 6.0 W = 6.15 W From this we can find the current through the circuit, I = /R I = 12 / 6.15 W = 1.951 A This causes an drop in the terminal voltage due to the internal resistance of int = I Rint = (1.951 A)(0.15 W ) int = 0.293 Therefore the terminal voltage will be trm = E int trm = 12.0 0.293 trm = 11.7 As you can see, there is some small change in the voltage but Ch 19, p 10
not very much. However, if we try to start the car, now we will notice a large effect. When we try to start the car the starter motor (R1 = 0.10 W ) and the lights (R2 = 6.0 W ) are connected in parallel. Their equivalent resistance is given by 1 R eq = 1 R 1 + 1 R 2 1 = 1 R eq 0.10 W + 1 6.0 W 1 = 10.167 R eq W Req = 0.098 W Notice that this is very nearly the resistance of the starter motor. The fact that the lights are also on has very little effect on the circuit. Now we go through exactly the same reasoning and calculations as before but contrast the results! The total equivalent resistance of the entire circuit including the internal resistance of the battery now is Rtot = Rint + R Rtot = 0.15 W + 0.098 W = 0.248 W From this we can find the current through the circuit, I = /R I = 12 / 0.248 W = 48.32 A (This is a very large current. The wire which carries this current goes directly from the battery to the starter and is very thick; look for it sometime). This causes an drop in the terminal voltage due to the internal resistance of int = I Rint = (48.32 A)(0.15 W ) int = 7.25 Therefore the terminal voltage will be trm = E int trm = 12.0 7.25 Ch 19, p 11
trm = 4.75 Now the terminal voltage is less than half what it was before. You will clearly see the lights dim. 19.80 For a series RC circuit with R = 1.5 kw, C = 25 µf, and bat = 18, find the following: a) the initial current. b) the final charge on the capacitor. c) the time constant of the circuit. d) the current and capacitor charge after two time constants. e) the time at which the current has dropped to onehalf of its initial value. Let s begin with a circuit diagram R = 15 kw C = 25 µf = 18 bat a) the initial current. Initially, before there is any charge on the capacitor, the current is just the net voltage, the voltage of the battery, divided by the resistance, Io = /R = 18 /15,000 W Io = 0.0012 A = 1.2 ma S b) the final charge on the capacitor. Finally, when there is no more current, the potential difference across the capacitor will be the same as the potential difference across the battery (18 ). Ch 19, p 12
What charge must this capacitor have to have a voltage of 18? C = Q/ Q = C Qf = (25 x 10 6 F) (18 ) Qf = 0.000 45 C = 0.45 mc = 450 µc c) the time constant of the circuit. The time constant is given by t = RC t = (1 500 W ) ( 25 x 10 6 F) t = 0.0375 s d) the current and capacitor charge after two time constants. The current, as a function of time is given by i = Io e t/ t For t = 2 t, this is i = (1.2 ma) e 2 i = (1.2 ma)(0.135) i = 0.162 ma The charge on the capacitor, as a function of time, is For t = 2 t, this is q = Qf [ 1 e t/ t ] q = [450 µc] [ 1 e 2 ] q = [450 µc] [ 1 0.135 ] q = [450 µc] [ 0.865 ] q = [450 µc] [ 0.865 ] q = 389 µc e) the time at which the current has dropped to onehalf of its initial value. Ch 19, p 13
As we already know, the current, as a function of time is given by i = Io e t/ t For what time t is i = 0.5 Io? i = 0.5 Io = Io e t/ t 0.5 = e t/ t The natural logrithm (written as ln(x) or labeled on a calculater as ln ) is the inverse of the exponential function. That is ln(e x ) = x ln (0.5) = 0.693 ln (e t/ t ) = t/t Therefore, t/t = 0.693 t = 0.693 t t = (0.693)(0.0375 s) t = 0.026 s 19.83 The capacitor in the circuit sketched below is initially charged with 5 x 10 4 C of charge. At time t = 0 the switch is closed; find the following: a) the initial current. b) the time constant of the circuit. c) the time at which one-half the charge remains on the capacitor. R = 150 W C = 400 µf Q = 500 µc o S Ch 19, p 14
Initially, the potential difference across the capacitor is o = C Qo o = (400 x 10 6 F)(500 x 10 6 C) o = 2 x 10 7 This means the initial current will be I = /R Io = o/r Io = (2 x 10 7 )/(150 W ) Io = 3 x 10 5 A = 30 µa The time constant of this circuit is t = RC t = (150 W)(400 x 10 6 F) = 0.02 s The charge on the capacitor, as a function of time, is q = Qoe t/ t For what value of time t is q = 0.5 Qo? q = 0.5 Qo = Qoe t/ t 0.5 = e t/ t ln(0.5) = ln(e t/ t ) 0.693 = t/t t = 0.693 t t = (0.693)(0.02 s) t = 0.0139 s Ch 19, p 15