Download Solution to Assignment 4

Solution to Assignment 4
Four resistors and an ideal (zero resistance) battery are laid out in a star-shaped pattern, and connected in three
different ways, as shown above. The resistors have resistances of R, 2R, 3R, and 4R, as labeled on the diagram. We
will give a value for R later, but you don't need to know it to answer the first two parts of this problem.
Solution:
In Circuits A and C, the potential of all the resistors lying on the outer edge of the
pentagon are the same since they are all connected to the negative terminal of the
battery. At the same time, the potential of the other end of the resistors are the same
since they all meet at the center of the pentagon and are connected to the positive
terminal of the battery. By the above argument, Circuits A and C are the equivalent and
can be redrawn as the following:
4R
3R
2R
R
By the same considerations, Circuit B can be redrawn as the following:
4R
2R
R
3R
We see that the loop containing the 2R and 3R resistors is closed with the potential
difference across the 2R and 3R resistors in series is zero. So, no current will flow
through the loop and we can ignore this part of the circuit in calculating its equivalent
resistance. Therefore, the equivalent resistance of circuit A is 24R/50 and that of circuit B
is 4R/5.
(a) Rank the circuits based on the magnitude of the current through the battery, from largest to
smallest. Use only > and/or = signs in your answer, such as C>A=B
Since the current through the battery is inversely proportional to the effective resistance
of the circuit, the order is C=A>B.
(b) In terms of R, what is the equivalent resistance of Circuit A? Just enter the numerical
coefficient in front of the R in the answer box below.
24/50 R
For the rest of this problem, use these numbers. The battery voltage is 24.0 V, and the value of R
is 2.00 ohms.
(c) In Circuit A, calculate the magnitude of the current in the battery.
The current through the battery is the voltage of the battery divided by the equivalent
resistance of the circuit. Therefore the current is 24/(24 *2/50) = 25 A.
(d) In Circuit B, calculate the total power dissipated in the resistors.
The power dissipated by a resistor is i2*R . Since the resistors 4R and 1R are in parallel
connected across the battery, the potential difference across them is the same equal to
24V. This gives the current through the 1R resistor to be 12A. So, the power dissipated
by it is (12A)2*(2Ω) = 288 W. Similarly, the power dissipated by the 4R resistor is 72 Watt.
And the total power dissipated by the circuit is 288 W + 72 W = 360 W.
Circuit a has a battery connected in a circuit with four identical light bulbs (numbered 1 through
4) and three switches (lettered A through C). The switches are shown in their open positions. The
battery is ideal - it has no resistance of its own. We will also assume that the resistance of each
bulb is the same no matter how much current passes through it. Circuit b is exactly the same as
circuit a, except that a fifth light bulb, identical to the other four, has been added to the circuit.
Since the brightness of a bulb is determined by the power dissipated by the bulb, which
in turn is given by Ibulb2Rbulb or Vbulb2/Rbulb. For a given bulb resistance Rbulb, the brightness
increases if either the current through the bulb or the voltage across the bulb is
increased.
(a) In Circuit a, if you want to maximize the brightness of bulb 4, what should you do with the
switches? Select all the correct answers below.
Switch A should be open.
Switch A should be closed.
having it open or closed has no effect on bulb 4.
should be closed.
Switch A doesn't matter -
Switch B should be open.
Switch B
Switch B doesn't matter - having it open or closed has no effect on bulb 4.
Switch C should be open.
Switch C should be closed.
having it open or closed has no effect on bulb 4.
Switch C doesn't matter -
If the current has to flow through bulb 4, clearly switch ‘c’ should be closed. On closing
switch B, bulb 4 is still directly connected across the battery so its voltage is unchanged
and so should its power dissipation. Therefore, switch B doesn’t matter. Similarly,
closing A doesn’t affect the voltage of bulb 4 and hence its power dissipation. So, switch
A also doesn’t matter.
(b) In Circuit b, if you want to maximize the brightness of bulb 4, what should you do with the
switches? Select all the correct answers below.
Switch A should be open.
Switch A should be closed.
having it open or closed has no effect on bulb 4.
should be closed.
Switch A doesn't matter -
Switch B should be open.
Switch B
Switch B doesn't matter - having it open or closed has no effect on bulb 4.
Switch C should be open.
Switch C should be closed.
having it open or closed has no effect on bulb 4.
Switch C doesn't matter -
Switch C has to be closed for the bulb 4 to have any current flowing through it.
Switch A doesn’t matter by the same argument as above.
When switch B is closed, bulbs 3 and 4 become parallel, and together have an equivalent
resistance of R34 = R/2, where R is the resistance of each bulb. Since R34 is in series with
bulb 5, they divide the voltage of the battery in proportion to their resistance value. It
follows that the voltage across R34 or bulb 4 (or bulb 3) is ε/3 (and that across bulb 5 is
2ε/3). When switch B is open, the voltage of the battery is divided equally between bulb 4
and bulb 5. So, the voltage across bulb 4 is ε/3, which is bigger than if switch B is closed.
Therefore switch B should be open.
c. In Circuit b, you close all the switches. Rank the bulbs in circuit b by their brightness, from
brightest to dimmest, with all the switches closed. Use only > and/or = signs in your ranking,
such as 3>2=5>1>4
When all the switches are closed, the voltage across bulb 1 becomes zero since together
with the closed switch A in parallel, the two acquire a zero equivalent resistance. This
leaves the bulb 2 to have the full share of the battery’s voltage, ε. As we discussed in
part b above, the voltages across bulb 3 and 4 are the same and equal to ε/3; the voltage
across bulb 5 is 2*ε/3. Therefore, the order of the brightness, which should be the same
as the order of the voltage across is 2>5>4=3>1.
(d) Starting with all the switches closed in Circuit b, you then open switch B. When you open
switch B, what happens to bulbs 3, 4, and 5? Select all the correct answers below.
Bulb 3 gets brighter.
brighter.
Bulb 3 gets dimmer.
Bulb 4 gets dimmer.
5 gets dimmer.
Bulb 3 stays the same.
Bulb 4 stays the same.
Bulb 4 gets
Bulb 5 gets brighter.
Bulb
Bulb 5 stays the same.
By similar considerations we made in the above parts, when switch B is opened, the
voltage across bulb 5 would go from 2ε/3 to ε/2, so bulb 5 will get dimmer. For bulb 4, the
voltage across it will go from ε/3 to ε/2, so bulb 4 will get brighter. For bulb 3, the current
through it will go from ε/(3R) to zero, so bulb 3 will get dimmer.
(e) Now, all the switches are closed in both circuits. Which of the following statements are true
when all switches are closed?
Bulb 2 in Circuit a is brighter than Bulb 2 in Circuit b.
than Bulb 3 in Circuit b.
Bulb 3 in Circuit a is brighter
Bulb 2 in Circuit b is the same brightness as Bulb 5 in Circuit b.
Circuit a has more power dissipated in it than does Circuit b.
The current through the
battery in Circuit a is larger than the current through the battery in Circuit b.
through Bulb 2 in Circuit a is larger than the current through Bulb 2 in Circuit b.
The current
The
potential difference across Bulb 4 in Circuit a is larger than the potential difference across Bulb 4
in Circuit b.
1. Since the left part of circuit is identical in both the circuits, bulb 2 is equally bright
in both the cases.
2. Since the potential difference across bulb 3 in circuit a, is equal to the voltage of
the battery which is greater than the voltage difference across bulb 3 in circuit b,
bulb 3 is brighter in circuit a.
3. In circuit b, the voltage across bulb 2 is ε and that across bulb 5 is 2ε/3, the two do
not have the same brightness.
4. By calculating the power dissipated by each of the bulb and adding them up, we
find that the total power dissipated by circuit a is 3ε 2/R, and that circuit b is
5ε 2/3R.
5. Adding the currents through the right and left branches in both the circuits, we
find that the current through the battery in circuit a is 3ε/R and that in circuit b is
5ε/(3R).
6. Since left part of the circuit is identical in both circuits, bulb 2 is equally bright in
both the cases.
7. The potential difference across bulb 4 in circuit a, is equal to the voltage of the
battery. This is greater than the voltage difference across bulb 4 in circuit b, which
shares the EMF of the battery with bulb 5.
For the rest of this problem, use these numbers. The battery voltage is 30.0 V, and the
resistance of each light bulb is 6.00 ohms. For convenience, the circuit diagrams are shown
again here, too, with the switches set as in the questions below.
(f) In Circuit b, with switch A open and switches B and C both closed, calculate the equivalent
resistance of the circuit.
Bulbs 3 and 4 are in parallel and the combination of 3&4 are in series with bulb 5.
Therefore the equivalent resistance of the right branch is 3*6/2 = 9 Ω. Bulbs 1 and 2 are
in series, therefore resistance of the left branch is 2 * 6 Ω= 12 Ω. The 12 Ω and 9 Ω are in
parallel. Therefore the equivalent resistance of the circuit is 1/(1/9 + 1/12) = 5.14 Ω.
(g) In Circuit b, with switch A open and switches B and C both closed, calculate the magnitude
of the current in bulb 4.
We had found the current through bulb 4 to be ε/(3R), which becomes 30/(3*6) A=1.67A. (ε
is the EMF of the battery and R is the resistance of each of bulb.)
(h) In Circuit b, with switch A open and switches B and C both closed, calculate the magnitude
of the potential difference across bulb 5.
We had found that the potential difference across bulb 5 is 2ε/3 = 2*30V/3 = 20 V.
(i) Now, we consider Circuit a. In Circuit a, with switch A open and switches B and C both
closed, calculate the equivalent resistance of the circuit.
Since bulbs 3 and 4 are in parallel, the equivalent resistance of the right part of circuit a
is 3 Ω, while that of the left part of circuit a is 12 Ω (as shown in part (e) of this question).
Since the left and right branches of the circuit are in parallel, the equivalent resistance of
the circuit is 2.4 Ω.
(j) In Circuit a, with switch A open and switches B and C both closed, calculate the power
dissipated in Bulb 1.
The current through bulb 1 is equal to the current through the bulb 2 which is equal to ε /
(2R) = 30/(2*6) = 2.5 A. Therefore the power dissipated by bulb 1 is i2*R = 2.5*2.5*6 =
37.5W
3. (a) In this simulation you can choose between seven possible circuits by clicking on one of the small circuit
diagrams shown at the upper right. Given the different circuits available to you, and taking into account the limits set
by the sliders in the simulation, determine the following. Express your answers to three significant figures.
(i) What is the maximum possible total current that can be obtained in this simulation?
Among the seven circuits, the one with all three resistors connected in parallel would
have the biggest current since it has the smallest equivalent resistance. As for the
voltage and resistance settings, we know that the current is the largest when the
potential of the battery is the highest and the resistance is the least. Therefore, when the
voltage across the battery is set to 10V and all the resistance of all the resistors are set
to 1Ω, we obtain the largest total current of 30A.
(ii) What is the minimum possible total current that can be obtained in this simulation? Note: for this answer in
particular, you should calculate the current yourself from the voltage and the equivalent resistance - the simulation
does not show enough significant figures for the minimum current.
We know that the current is the minimum when the voltage is the lowest and the
resistance is the biggest. On setting each of the resistors at 10 Ω and the battery voltage
at 1V, we see that the lowest current is found in the circuit with the three resistors in
parallel (which has an equivalent resistance of 30 Ω). By calculating the current of this
circuit given the above values of the voltage and resistance, we find the value of the
lowest current to be 0.033 A.
(b) Now move the sliders so the resistances of the three resistors are equal to one another. With the battery voltage
fixed, rank the circuits according to the following criteria. Use principles of physics to find the answers, and then verify
the answers using the simulation.
If each of the resistors has a value of R then the equivalent resistances of the circuits
from 1 to 7 are R, 2R, 3R, R/2, R/3, 3R/2 and 2R/3 respectively.
(i) Rank the circuits based on the total current, from largest to smallest. Use only > and/or = signs in your ranking
(e.g.,4>3>6>1=7>5>2).
The total current is given by the ratio of battery voltage to that of the equivalent
resistance, the total current is the largest in the circuit with the lowest equivalent
resistance and vice-versa. Therefore the order is
5>4>7>1>6>2>3
(ii) Now rank circuits 2 to 7 only based on the current through resistor R2, from largest to smallest. Use only > and/or
= signs in your ranking (e.g., 4>3>6=7>5>2).
From the simulation, it can be seen that the order is 4=5>2=7>6=3.
4. If the potential difference across the 18 Ω resistor in the figure above is 15.0 V, what is the
emf of the battery? Note that you should be able to do this problem without a calculator.
Solution:
Since the potential across the 18 Ω resistor is 15V, the current through it is 15V/18 Ω
=0.833 A, which is equal to the current through the 6 Ω resistor. It follows that the
potential difference across the 6 Ω and 18 Ω resistor together is 0.833*(18+6) = 20V. This
voltage should be the same as that across the 8 Ω resistor, so the current through the 8
Ω resistor is 20V/8Ω = 2.5A. The total current through the 30 Ω resistor is equal to the
sum of the currents through the 8 Ω resistor and the 6 and 18 Ω resistors, which is
0.833A + 2.5A = 3.333A. Therefore the voltage across the 30 Ω resistor is 30*3.333 = 100
V. The emf of the battery is equal to the sum of the voltages across the 30 Ω and the 8 Ω
resistors, which is 20V + 100V = 120 V.
5.Four resistors, of resistance R, 2R, 3R, and 6R, are connected together in a circuit with an ideal
battery, as shown above. If the magnitude of the current through the 3R resistor in the figure
above is 12.0 A, what is the magnitude of the current through the battery? Note that you should
be able to do this problem without a calculator.
Solution:
Since the current through the 3R resistor is 12 A, the voltage across it is V = I R = 12 *
3R= 36R. Since the 6R and 3R resistors are in parallel, the voltage across the 6R resistor
is also 36R. It follows that the current through the 6R resistor is 36R/6R=6A. The current
that flows through 2R is the sum of those through the 3R and 6R resistors, which is 12A
+ 6A = 18A. This is the current flowing through the right branch of the circuit. From the
current through the 2R resistor, we can determine that the potential difference across it
is 18A*2R=36R. The potential difference across the battery is equal to sum of the
potential differences across the 2R and 3R resistors = 36R + 36R = 72 R. Since the R
resistor on the left branch is connected directly across the battery, its potential
difference is also 72R, which gives the current flowing through it to be 72R/R =72A. The
total current flowing through the battery is equal to the sum of the currents flowing
through the right and left branches of the circuit, which is equal to 72A + 18A = 90A.