20 resistor as if they were in series:
Req = 20 + 7.5 = 27.5
1
20
1 7.5 27.5
2 2 1
Finding the currents and voltage drops
Now that we have the equivalent resistance, the circuit outside points 1 and
2 looks like
2 + +
27.5

22 Volts
1
By using the loop rule, I get
V
11
= 0 = 22V I
1
(27.5)
which tells me that
I
1
= 22V
27.5 = 0.8 A
Now that we have found the current drawn from the battery, the job of
the equivalent resistor is done. We now want to nd the voltage drop and
the current through each resistor. As we know that I goes through the 20
resistor, we know that the voltage drop is
Vacross 20 = IR = (0.8 A)(20 ) = 16 V
We do not know the current through each of the other two resistors (yet).
All we know is that the current splits into two pieces, and those two pieces
add to give the 0.8 Amps. But we know that the voltage drop across both
resistors has to be the same, and that the left hand side is at
Vlhs = 22 + Vacross 20 = 22 V 16 V = 6 V
while the right-hand side is connected at 0 volts (it is connected to the
negative terminal of the battery). Therefore the potential dierence across
both
the 10 and the 30 resistor must be -6 volts. i.e.
V
10
= V
30
= 6 V
2 From this we can get the currents. We know that
Vacross
10
= I
c
R I
c
= V
10
R
= + 6 V
10 = 0.6 A
Vacross
30
= I
b
R I
b
= V
30
R
= + 6 V
30 = 0.2 A
As promised I
10
+ I
30
add up to give us the full 0.8 Amps.
Power
First let us calculate the power supplied by the battery. Remember, we only
have one real formula for electrical power: P = V I. The battery has a
22 Volt potential dierence, and a current of 0.8 amps owing through it, so
the power it is producing is:
Psupplied by battery = (22 V)(0.8 A) = 17.6 W
Then there is the question of how much power each resistor uses. I could
just apply P = V I to each of them in turn, but I think that it is better
to look at P = (V )
2
/R, so we can directly compare the two resistors in
parallel (they have the same V ). This way we can see immediately that
the 10 resistor will use more power than the 30 resistor without doing a
calculation! Putting in numbers:
P
10
= (V )
2
R
= (6 V)
2
10
= 3.6 W
P
30
= (V )
2
R
= (6 V)
2
30
= 1.2 W.
The minus signs tell us that these resistors are using power, not producing
it.
The nal resistor has 16 Volts across it, and is the most power-hungry:
P
20
= (16 V)
2
20 = 12.8 W
Adding up all the power consumptions, we have
Pused = (12.8 + 3.6 + 1.2) W = 17.6 W,
which is identical to what the battery supplied (as it had to be).
3 Question 2
If we short out resistor three, then our circuit becomes
4 10
+ +
20 0
b c
2
1
22 Volts d
3
30
The voltage drop across the added wire is zero. This tells us that the voltage
drop across the other two resistors in parallel with it are also zero (resistors
in parallel have the same voltage drop). this can only happen if no current
ows down past the 10 or 30 resistors.
Also note that this conclusion does not depend on the actual values of the
two resistances.
Question 3
Answers vary, depends on item chosen. But if the power rating is P , then I
have
P = (V )acrossIthru = (120V )Ithru
So then we get
Ithru = P
120V
Question 5 Determining the resistance of a lightbulb
*** The order of 4 and 5 has been changed ***
This question is rather poorly worded, I am afraid. The physical charac-
teristic of the lightbulb that we control when we make it is the properties of
the lament. For example, the material we make it out of, its length and its
cross-sectional area.
However, all of these dierent physical characteristics give rise to an elec-
trical property of the lightbulb the resistance. We get to control the resist-
ance of a bulb and then can gure out what current would ow through it if
we plugged it into various voltage sources (or conversely what voltage drop
would occur for a given current running through it).
4 When an electrical device claims to be 100W, it means that if you plug it
into a standard voltage outlet that it will dissipate 100W of power. Someone
has chosen the resistance so that this will happen. However, if you plug it
into another voltage source you will get a dierent power, as the resistance
of the device would stay approximately the same.
Question 4 Looking at the resistance of bulbs
*** The order of 4 and 5 has been changed ***
As discussed in question 5, we know that the physical properties of a
lightbulb determine its resistance. Thus, when I say I have a 100W lightbulb
what I actually mean is this lightbulb will produce 100W of power when the
standard voltage of 120 Volts (in the U.S.) is applied across it. If it has some
dierent voltage drop it will generally have a dierent power consumption.
a) Does a 60W lightbulb or a 100W lightbulb have greater current?
This question is really asking about what happens if you plug in a lightbulb
with the same voltage drop across each one. This is not stated explicitly in
the question, but as the normal thing to do with a 100W lightbulb is to have
120 Volt drop across it, so it is not too much of a stretch.
I know that in each case
P = (V )I.
If I use a V = 120 Volts, then I know the power across the 100W lightbulb
is 100W, and the 60W lightbulb is 60W. Rearranging I have
I = P
V .
Thus, the lightbulb with the greater power rating has the greater current.
Finally, although it is not asked for let us put in the numbers. We have
I60W = 60W 120V = 0.5 Amps
while
I100W = 100W 120V = 0.83 Amps
b) Which has the greater resistance?
There are lots of ways of doing this, and I am going to describe two.
5 First of all, there is the way that works directly o our last answer. I know
that I have a current I going through my lightbulb if I have a voltage drop
V across it (that is what we have just done!). Let us choose a point just
before and just after the lightbulb. Then, applying the transport equation,
we have
V = IR.
Rearranging, we have
R = V
I .
The question we just nished informed us that for a given voltage drop the
higher power rating has a higher current. As the current increases, the res-
istance decreases. Hence the higher power rating has lower current.
Placing numbers in, we nd:
R60 W = 120 V
0.5 A = 240
R100 W = 120V
0.833 A = 144 ,
which is as promised.
We can do this a completely dierent way, that does not rely directly on
our current calculation (imagine, if we got part a wrong then we would get
this wrong as well).Recall that the power dissipated by a resistor can also be
written
P = + (V )
2
R
.
Rearranging we have
R = (V )
2
P
.
As the power gets higher (for the same V ) the resistance must decrease.As
promised, this method does not rely on getting the right currents in part a),
and so would be my personal choice for solving this problem.
c) If these two bulbs were connected in series into the same 120V
power supply, which would be brighter?
The key point to remember is that even though the lightbulbs have 100
Watts and 60 Watts written on them, what is actually xed is the resist-
ance of the two bulbs. So our circuit can be envisioned as
6 +
1 = 120 V
R
60W
R
100W
The easy way of getting which one is brighter is to ask what is not changing.
In this case, the current through both resistors is the same and we have a
formula for power dissipated based on currents and resistances:
P = I
2
R.
As the currents are the same, the lightbulb with the greater resistance will
be brighter.
To calculate the power of each lightbulb, we need to work a little harder.
Going around the loop once we have
V = 0 = 120V IR60 W IR100 W = 120V I(R
60
+ R
100
).
This gives us a current of
I =
120V
R
60
+ R
100
=
120V
240 + 144 =
120V
384 = 0.3125 A
The power used by each lightbulb is
P60W = I
2
R60W = (0.3125 A)
2
(240) = 23.44 W
P100W = I
2
R100W = (0.3125 A)
2
(144) = 14.06 W.
As a sanity check, the power supplied by the outlet is
P = (V )I = (120 V)(0.3125 A) = 37.5 W,
exactly the same as the total power used by the two lightbulbs.
Question 6
The thermal energy of an incandescent lightbulb increases. Some energy is
lost to radiation as well (which is kind of the point of a lightbulb !)
1
1
The energy of course is not actually lost, it is more accurate to say that the energy is
transferred to the environment via radiation.
7 Question 7
Trivia: No one in Europe actually operates at 240V. This should read 230V,
which is the EU standard. Areas in the South Pacic (such as New Zealand,
Papua New Guinea and Tonga) operate at 240V.
In any case, let us deal with the 240V as it is twice the US voltage. What
is dissipated by a device plugged into a wall socket? Well, we nd the power
used is
P
= (V )
2
R
So the power used in the U.S.A. would be
P
U SA
= (V
U S
)
2
R
= (120V )
2
R
while in New Zealand (say) it would be
P
N Z
= (240 V )
2
R
= (2V
U S
)
2
R
= 4 (V
U S
)
2
R
= 4 P
U S
.
i.e. power would be dissipated four times quicker than the manufacturer
intended! This extra energy can be turned into heat and will melt the tran-
sistors in cheaper electronics!
8