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
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 + 
T)
Take "room temperature" to be about 25oC.
The boiling temperature of liquid nitrogen is - 196oC.
Therefore,
For carbon, the temperature coefficient
R = Ro (1 + R = ( 50
R = ( 50
R = (50
R = 55.5
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 mV across it and the current through it is about 200
A. How much power is being released in the neuron?
P = I V P = (200 x 10 - 6 A)(75 x 10 - 3 V)
P = 1.5 x 10 - 5 A - V
Now, what is an A - V (an amp - volt)?
P = 1.5 x 10 - 5 A - V
P = 1.5 x 10 - 5 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 cm 12 0.2053 cm The resistance R is given by R = r L / A. For copper, the resistivity is r = 1.59 x 10 - 8
A = r2 =
For simple and repeated calculations, I like to use a spreadsheet.
For ClarisWorks (and most spreadsheets) the notation 1.63e-3 means 1.63 x 10 - 3.
19.37 Calculate the power supplied by a 12 - volt power supply in the circuit if the resistors have the following values: R1 = 250
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
The current through this equivalent resistor would be
R = V / I I = V / R = (12 V) / (2000
This is the current supplied by the battery and the power supplied by the battery is
P = IV P = (0.006 A)(12 V) = 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 V P = I (IR) = I2 R
P1 = (0.006 A)2 (250
P2 = (0.006 A)2 (750
P3 = (0.006 A)2 (1000
Ptot = P1 + P2 + P3 = 0.072 W
And, of course, that is just the power supplied by the battery to the equivalent resistor.
We ought to convince ourselves that the units are correct; that is, that units of A2 W are, indeed, units of watts, W.
A2 W = A2 (V / A) = A - V and in problem 19.11 we already found that
A - V = 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
Figure 19.34 Problem 19.46
These resistors are connected in parallel so the equivalent resistance, Req, is given by
Req = 158
The current supplied by the battery to this equivalent resistor would be
R = V / I I = V / R
I = 12 V / 158
The power supplied by the battery to this equivalent resistor would be
P = I V P = (0.076 A) (12 V) = 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 V = (V/R) V = V2 / R For each resistor, V = 12 V, since they are in parallel,
P1 = V2 / R1 = (12 V)2 / 250 P2 = V2 / R2 = (12 V)2 / 750
P3 = V2 / R3 = (12 V)2 / 1000
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 V2 /
V2 / as we saw in problem 19.11.
19.54 Find the equivalent resistance in the circuit shown in Figure 19.36 for R1 = 100
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.
Req = 231
We will shortly need to calculate another equivalent resistance, so let's call this one Req,1. Now our circuit looks like this -
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
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
Req,2 = 381
That is, our equivalent single - resistor circuit would be
with Req,2 = 381
R = V / I I = V / R
I = 12 V / 381
I = 0.0315 A
The power given out by the battery and absorbed by this single resistor is
P = I V P = (0.0315 A) (12 V)
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,
Now we can find the potential difference across R1, R5, and Req,1 and the power absorbed by each resistor.
R = V / I V = I R
V1 = (0.0315 A) (100
V5 = (0.0315 A) (50
Veq,1 = (0.0315 A) (231
Notice that V1 + Veq,1 + V5 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 V P1 = I V1 = (0.0315 A) (3.15 V) = 0.0992 W
P5 = I V5 = (0.0315 A) (1.575 V) = 0.0496 W
Peq,1 = I Veq,1 = (0.0315 A) (7.275 V) = 0.2292 W
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 V across it so the current through each is given by
R = V / I I = V / R
I2 = 7.275 V / 500
I3 = 7.275 V / 750
I4 = 7.275 V / 1000
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 V P2 = I2 V = (0.01455 A) (7.275 V) = 0.1058 W
P3 = I3 V = (0.0097 A) (7.275 V) = 0.0706 W
P4 = I4 V = (0.007275 A) (7.275 V) = 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.
19.75 A 12.0 V automobile battery (ie, a battery with an emf of 12.0 V) has an internal resistance of 0.15
A good circuit diagram will be very helpful.
When only the lights are on, the external resistance is R = 6.0
Rtot = Rint + R Rtot = 0.15
From this we can find the current through the circuit,
I = V / R I = 12 V / 6.15
This causes a drop in the terminal voltage due to the internal resistance of
Vint = I Rint = (1.951 A)(0.15 Vint = 0.293 V
Therefore the terminal voltage will be
Vtrm = E - Vint Vtrm = 12.0 V - 0.293 V
Vtrm = 11.7 V
As you can see, there is some small change in the voltage - but 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
Req = 0.098
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
From this we can find the current through the circuit,
I = V / R I = 12 V / 0.248
(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 a drop in the terminal voltage due to the internal resistance of
Vint = I Rint = (48.32 A)(0.15 Vint = 7.25 V
Therefore the terminal voltage will be
Vtrm = E - Vint Vtrm = 12.0 V - 7.25 V
Vtrm = 4.75 V
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 k
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 one - half of its initial value.
Let's begin with a circuit diagram -
a) the initial current.
Always, the current is just the net voltage divided by the resistance,
Initially, before there is any charge on the capacitor, the cnet voltage is just the voltage of the battery. Therefore, initially, before there is any charge on the capacitor, the current is just this net voltage, the voltage of the battery, divided by the resistance,
Io = V / R = 18 V / 15,000 Io = 0.0012 A = 1.2 mA
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 V). This means the net voltage is zero so the current is zero.
What charge must this capacitor have to have a voltage of 18 V?
C = Q / V Q = C V
Qf = (25 x 10 - 6 F) (18 V)
Qf = 0.000 45 C = 0.45 mC = 450
c) the time constant of the circuit.
The time constant is given by
= RC
d) the current and capacitor charge after two time constants.
The current, as a function of time is given by
i = Io For t = 2
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
q = Qf [ ]
For t = 2
q = [450 q = [450
q = [450
q = 389 µC
e) the time at which the current has dropped to one - half of its initial value.
As we already know, the current, as a function of time is given by
i = Io For what time t is i = 0.5 Io?
i = 0.5 Io = Io 
0.5 =
The "natural logrithm" (written as ln(x) or labeled on a calculater as "ln") is the inverse of the exponential function. That is,
ln(ex) = x ln (0.5) = - 0.693
ln (
Therefore,
- t / t = 0.693
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.
Initially, the potential difference across the capacitor is
Vo = C Qo Vo = (400 x 10 - 6 F) (500 x 10 - 6 C)
Vo = 2 x 10 - 7 V
This means the initial current will be
I = V / R Io = Vo / R
Io = (2 x 10 - 7 V) / (150
Io = 3 x 10 - 5 A = 30
The time constant of this circuit is
The charge on the capacitor, as a function of time, is
q = Qo For what value of time t is q = 0.5 Qo?
q = 0.5 Qo = Qo 0.5 =
ln(0.5) = ln(
- 0.693 = - t /
t = 0.693
t = (0.693) (0.02 s)
t = 0.0139 s
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