Download Week 7 - Direct Current

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.
Answer: 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.
Answer: 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 its resistance, I = V /R. Now the voltage drop
across each bulb is also equal and is given by V = Vtot /n, so the current trough each bulb is
I=
Vtot
.
nR
(1)
Therefore the power each bulb is dissipating in the form of light (and a little heat) is
P = IV = I
Vtot
V2
= 2tot .
n
n R
(2)
So the power of each bulb goes down with the second 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
= tot
2
n R
nR
(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 1. 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?
Answer:
I25 = 7.00 A
Week 7 – October 17, 2011
2
(6)
compiled October 3, 2012
I20 = 9.94 A
(7)
Figure 2
Exercise 7.3:
In the circuit shown in figure 2, 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.
Answer:
Figure 3
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.
Week 7 – October 17, 2011
I3 = −8.00 A
(8)
ε1 = 36.0 V
(9)
ε2 = 54.0 V
(10)
R = 9.0 Ω
(11)
3
compiled October 3, 2012
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 .
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.
Answer:
J=
σQ
r̂.
4πε0 r2
(12)
b) What is the total current I(t = 0) flowing from the inner to the outer shell?
Answer:
I=
σQ
ε0
(13)
c) What is the resistance between the shells?
2 The
Wheatstone Bridge allows for extremely accurate measurements of unkonwn resistances.
Week 7 – October 17, 2011
4
compiled October 3, 2012
Answer:
R=
∆V
1 b−a
=
.
I
4πσ ba
(14)
d) Find the charge on the inner shell at a time t. Hint: How does this situation compare to that of a
discharing capacitor?
Answer:
Q(t) =
σQ − t
e τ
ε0
(15)
with time constant
τ = RC = 4πε0 R
Week 7 – October 17, 2011
5
ba
.
b−a
(16)
compiled October 3, 2012