Week 7 - Direct Current
I have not failed. I’ve just found 10,000
ways that won’t work.
Thomas A. Edison
Exercise 7.1: Discussion Questions
a) In which 230 V light bulb1 does the filament have greater resistance: a 60 W bulb or a 120 W bulb?
If the two bulbs are connected to a 230 V line in series, through which bulb will there be a greater
voltage drop? What if they are connected in parallel? Explain your reasoning.
Solution: Since P = V 2 /R it is the 60 W bulb that has the higher resistance. If they are connected
in series the same current will go trough both, so V = RI indicate that the higher resistance bulb,
i.e. the 60 W bulb will have the highest voltage drop across it. If they are connected in parallel they
will both have the same voltage drop.
b) You connect a number of identical light bulbs to a flash light battery. How will the brigtness of the
bulbs differ between a parallel connection and a series connection? Will the battery last longer if you
connect them in parallel or in series? Explain.
Solution: Say we have n bulbs. For a series connection the current is determined tough each bulb is
determined by the voltage drop across it divided by the total resistance, I = V /(nR), where we have
used the fact that the resistances are equal, so that the total resistance is just n times the resistance
of each bulb. Now the voltage drop across each bulb is also equal and is given by V = Vtot /n, so the
current trough each bulb is
Vtot
.
n2 R
I=
(1)
Therefore the power each bulb is dissipating in the form of light (and a little heat) is
P =I
V2
Vtot
= 3tot .
n
n R
(2)
So the power of each bulb goes down with the third power of the number of lightbulbs!
Now for a parallel connection the bubls will all have the same voltage Vtot across them. And the
current trough all the bubles will be I = Vtot /R. Therefore the power disapated in each bulb will be
P =
1 230 V
Vtot
R
Vtot =
is the standard voltage in Norwegian households.
1
2
Vtot
R
(3)
The same as with just one bulb in series! Therefore a lot of bulbs in parallel will all shine much
brigher than a lot of bulbs in series. However the power delivered by the battery in the parallel case
is
2
Vtot
R
(4)
2
Vtot
V2
= 2tot
3
n R
n R
(5)
Pbat = n
while it is
Pbat = n
for the series connection. Therefore a battery will survive much longer with a series connection. In
fact we see here that the power delivered by the battery actually decreases in the series connection
when we increase the number of lightbulbs.
Figure 1
Exercise 7.2:
Consider the circuit shown in figure 14. The current trough the 6.00 Ω resistor is 4.00 A, in the direction
shown. What are the currents trough the 25.0 Ω and 20.0 Ω resistors?
Solution: Let’s call the resistor with X Ohms for RX and the current trough it IX . In order to find the
current trough R25 we must find the current trough R8 . This can be found by the loop rule applied to
the loop containing both R6 and R8 . We get
Week 7 – October 17, 2011
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I6 R6 = 4.00 × 6.00 AΩ = I8 R8 = I8 8.00 Ω
(6)
implying
I8 =
4.00 × 6.00
A = 3.0 A.
8.00
(7)
The current trough R25 is then just
I25 = I8 + I6 = (4.00 + 3.00) A = 7.00 A.
(8)
In order to find the current trough R20 we apply the loop rule trough the larger loop containing R20 ,
and the equivalent resistance Req of the three resistors R6 , R8 and R25 . We get
I25 Req = I20 R20 .
(9)
where
R6 R8
+ R25 = 28.4 Ω
R6 + R8
(10)
I20 R20
7.00 × 20.0
A = 0.94 A.
=
Req
28.4
(11)
Req =
implying a current of
I25 =
Answer:
I25 = 7.00 A
(12)
I25 = 0.94 A
(13)
Exercise 7.3:
In the circuit shown in figure 15, find (a) the current in the resistor 3.00 Ω resistor; (b) the unknown
emfs ε1 and ε2 ; (c) the resistance R. Note that three currents are given.
Solution:
Figure 3 shows the assumed direction for the different currents in the circuit. Let’s use the same
convention as in the last exercise and label the resistance with X Ohms and the current going trough it
as RX and IX . From the junction rule we can write down that
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Figure 2
Figure 3
IR
=
I4 + Iε1
Iε1
=
Iε2 + I3
(15)
Iε3
=
I6 + IR .
(16)
(14)
Substituting 16 into 15 and we get
Iε1 = I6 + IR + I3 .
(17)
We can further substitute this result into 14 and obtain
IR = I4 + I6 + IR + I3 ⇒ I3 = −(I4 + I6 ) = −8.00 A
(18)
which means that the current is going in the opposite direction relative to the assumed one.
To find the unknown emfs we apply the loop rule and obtain
ε2 = −I3 R3 + I6 R6 = (−8.00 × 3.00 − 5.00 × 6.00) V = 54.0 V
(19)
ε1 = −I3 R3 + I4 R4 = (8.00 × 3.003.00 × 4.00) V = 36.0 V.
(20)
Applying the looprule to the uppermost rule we get
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IR R = ε2 − ε1 ⇒ R =
ε2 − ε1
54.0 − 36.0
=
Ω = 18.0 Ω.
IR
2.00
(21)
Answer: IX is the current trough the resistor with X Ohms. Note that the sign of the current depends
on which direction one assumes it is going.
I3 = −8.00 A
(22)
ε1 = 36.0 V
(23)
ε2 = 54.0 V
(24)
R = 18.0 Ω
(25)
Figure 4
Exercise 7.4: The Wheatstone Bridge
The circuit shown in figure 4 is known as the Wheatstone bridge.2 It is an arrangement used to determine
the resistance of an unknown resistor R. The three resistors R1 , R2 and R3 can be varied and their
values are always known. With the two switches closed the resistors are varied until the galvanometer
reads zero; the bridge is then said to be balanced. Show that under these conditions the value of the
unkown resistor is R = R1 R3 /R2 .
Solution: Using Kirchhoff’s loop rule for the two loops ’separated by the bridge’ we get
2 The
Wheatstone Bridge allows for extremely accurate measurements of unkonwn resistances.
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I2 R2 = I1 R1
(26)
I3 R3 = IR R.
(27)
and
The last equation we implies that
I3
R3 .
IR
R=
(28)
Now the junction rule, under the condition that the Galvanometer reads zero, implies that
I1 = IR
(29)
I2 = I3 .
(30)
and
Using this information we can replace I3 /IR with I2 /I1 and from the uppermost equation
I2
R1
=
I1
R2
(31)
implying that
R=
R1 R3
.
R2
(32)
Exercise 7.5: Spherical current
Consider for example two concentric metal spherical shells of radius a and b which are seperated by
a weakly conducting material of conductivity σ. (Conductivity is defined as σ ≡ 1/ρ where ρ is the
resistivity.)
a) Assume that at time t = 0 there is a charge Q is on the inner shell and −Q on the outer shell. Find
the current density as a function of position between the spherical shells.
b) What is the total current I(t = 0) flowing from the inner to the outer shell?
c) What is the resistance between the shells? Hint: Find the potential difference first.
d) Find the charge on the inner shell at a time t. How does this compare to the discharging capacitor
(equation 26.16 in the book)?
Week 7 – October 17, 2011
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