[ Problem View ]
Measurements in Electric Circuits Description: Several questions related to a single-loop circuit with one resistor, an ammeter, and a voltmeter. These devices are not necessarily considered ideal. The students are not assumed to know the rules for series and parallel circuits; however, they can make qualitative determinations. Learning Goal: To understand the role of the internal resistance of various devices and the use of the ammeter and the voltmeter.
Consider the circuit shown. All wires are considered ideal; that is, they have zero resistance. We will assume for now that all other elements of the circuit are ideal, too: The value of resistance
is a constant, the internal resistances of the battery (
are zero, and the internal resistance of the voltmeter (
) and the ammeter (
) is infinitely large.
Part A
What is the reading
of the voltmeter?
Express your answer in terms of the EMF . ANSWER:
EMF = EMF
Part B
The voltmeter, as can be seen in the figure, is connected to points 1 and 3. What are the respective voltages between points 1 and 2 and between points 2 and 3? ANSWER:
)
Part C
What is the reading of the ammeter? Express your answer in terms of ANSWER:
and
.
= EMF/R
To make things more interesting, we now assume that the battery has a nonzero internal resistance
(the voltmeter and the ammeter remain ideal).
Part D
What is the reading of the ammeter now? Express your answer in terms of ANSWER:
,
, and
.
= EMF/(R+R_int)
Part E
What is the reading of the voltmeter now? Hint E.1 Hint not displayed
Express your answer in terms of ANSWER:
,
, and
.
= EMF*R/(R+R_int)
Now assume that the ammeter has nonzero resistance internal resistance.
. The battery still has nonzero
Part F
Compared to their values when voltmeter change when
, how would the readings of the ammeter and the ?
Hint F.1 How to approach this part Hint not displayed ANSWER:
The ammeter reading would increase; the voltmeter reading would stay the same. The ammeter reading would decrease; the voltmeter reading would stay the same. The ammeter reading would decrease; the voltmeter reading would increase. The ammeter reading would increase; the voltmeter reading would increase.
Part G
What is the new reading of the ammeter? Express your answer in terms of ANSWER:
,
,
, and
.
= EMF/(R+R_A+R_int)
Now assume that the ammeter again has zero resistance, but the resistance of the voltmeter is less than infinity. The battery still has nonzero internal resistance. Part H
Compared to their values when voltmeter change when
, how would the readings of the ammeter and the
is some large but finite value?
Hint H.1 Consider the voltmeter first
Observe that
, where is the current flowing through the battery. What
happens to the current when the resistance of the voltmeter drops? Hint H.2 The change in the battery current
When the resistance of the voltmeter drops below infinity, it becomes possible for current to flow both through the voltmeter and through the resistor. The overall resistance of the circuit therefore drops, and the current through the battery increases. How would that affect the reading of the voltmeter? Hint H.3 The reading of the ammeter
The ammeter reading is related simply to the voltmeter reading. The current through the ammeter is given by , where is the voltage between points 1 and 3, which is exactly the voltage that the voltmeter reads. ANSWER:
The voltmeter reading would stay the same; the ammeter reading would increase. The voltmeter reading would stay the same; the ammeter reading would decrease. The voltmeter reading would decrease; the ammeter reading would decrease. The voltmeter reading would increase; the ammeter reading would increase.
Suppose now that the piece of ideal wire between points 1 and 2 is removed and replaced by a nonideal wire with a nonzero resistance. Part I
How would this change affect the readings of the ammeter and the voltmeter? Hint I.1 Hint not displayed ANSWER:
The ammeter reading would stay the same; the voltmeter reading would stay the same. The ammeter reading would decrease; the voltmeter reading would decrease. The ammeter reading would increase; the voltmeter reading would decrease. The ammeter reading would decrease; the voltmeter reading would
increase. The ammeter reading would increase; the voltmeter reading would increase. The ammeter reading would stay the same; the voltmeter reading would increase. [ Print ]
2:
0.1270
3: [ Problem View ]
Brightness of Light Bulbs Description: Asks students to rank brightness of light bulbs in amixed series and parallel circuit.
Consider a circuit containing five identical lightbulbs and an ideal battery.
Part A
Rank the brightness of the five bulbs (A through E) from brightest to dimmest. (The more current flowing through a bulb, the brighter it will be.) Hint A.1 Comparing bulb A to bulb B Hint not displayed Hint A.2 Comparing bulb D to bulb E Hint not displayed Hint A.3 Comparing bulb C to bulb D or E Hint not displayed
Hint A.4 Comparing bulb C to bulb A or B Hint not displayed
List the bulbs in order from brightest to dimmest. Between each pair of bulbs, use the symbol > to indicate that the left-hand bulb is brighter than the right-hand bulb, or = to indicate that the bulbs have the same brightness. For example, "B=C=E>A>D" means that bulbs B, C, and E all have the same brightness, and that they are brighter than bulb A, which in turn is brighter than bulb D. ANSWER:
C>A=B>D=E C>A=B>E=D C>B=A>D=E C>B=A>E=D
Now consider what happens when a switch in the circuit is opened.
Part B
What happens to bulb A? Hint B.1 How to approach this part
How does the resistance of bulb C alone compare with the resistance of bulb C in parallel with bulbs D and E? How does this affect the resistance and current in the circuit as a whole (as compared to before)? ANSWER:
It gets dimmer. It gets brighter. Its brightness stays the same.
Part C
What is the current
now flowing in bulb C?
Express your answer in terms of the applied voltage bulb.
and
, the resistance of a single
= (2*EMF)/(3*R)
ANSWER: Part D
What happens to bulb C? Part D.1 Current in bulb C earlier
The total resistance of the earlier circuit was What is the current
, where
is the resistance of one bulb.
flowing in bulb C?
Express your answer in terms of
and
.
ANSWER: Answer not displayed ANSWER:
It gets dimmer. It gets brighter. Its brightness stays the same.
This is why appliances in your home are connected only in parallel. Otherwise, turning some on or off would cause the current in others to change, which could damage them (typically in the case of an overload) or prevent them from functioning (if the current is too low). [ Print ] 4. [ Problem View ]
Batteries in Series or Parallel Description: Computation of current through a resistor for parallel and series connected batteries.
You are given two circuits with two batteries of emf
and internal resistance
Circuit A has the batteries connected in series with a resistor of resistance has the batteries connected in parallel to an equivalent resistor. Note that the symbol
should be entered in your answers as EMF.
each. , and circuit B
Part A
In which direction does the current in circuit A flow? Hint A.1 Conventions
Remember that the conventional current flows from a positive to a negative terminal. ANSWER:
clockwisecounterclockwise
Part B
What is the current through the resistor of resistance Hint B.1 Hint not displayed Hint B.2 Hint not displayed
Express the current in terms of ANSWER: Part C
,
, and
.
2*EMF/(2*R_1+R_2) = 2*E/(2*R_1+R_2)
in circuit A?
Calculate the current
through the resistor of resistance
for circuit B.
Hint C.1 Which rule to use Hint not displayed Part C.2 What is the emf for loop 1?
The diagram shows the circuit divided into two loops: branch,
is the current in the branch below it, while
is the current in the topmost is the current in the lowest
branch, which contains . Find an expression for the emf using the voltage drops across the two resistors in loop 1. Express your answer in terms of ANSWER:
,
,
, and
.
= I_3*R_2+I_1*R_1
Part C.3 What is the emf for loop 2? Part not displayed Part C.4 Application of Kirchhoff's junction rule (current rule)
You should now have two equations involving all the variables in the circuit diagram. To solve for , you need a relationship between and . Choose the correct relation by applying Kirchhoff's junction rule to one of the junctions. Recall that Kirchhoff's junction rule states that the algebraic sum of all the currents into a junction is zero:
.
ANSWER:
Now solve the three equations you have obtained for the currents in each branch to obtain an expression for
(
). To do this, you could either add the two
equations other than the one above, or substitute for into this one. Express your answer in terms of ANSWER:
,
, and
and
from the other equations
.
2*EMF/(2*R_2+R_1) = 2*E/(2*R_2+R_1)
Part D
What is the power dissipated by the resistor of resistance ,
, and
for circuit A, given that
?
Hint D.1 What formula to use Hint not displayed
Calculate the power to two significant figures. ANSWER:
-2 = 6.40×10
W
Part E
For what ratio of and would power dissipated by the resistor of resistance same for circuit A and circuit B?
be the
Hint E.1 Hint not displayed Hint E.2 Hint not displayed ANSWER:
= 1.00
Part F
Under which of the following conditions would power dissipated by the resistance circuit A be bigger than that of circuit B? Hint F.1 How to think about the problem
in
Hint not displayed
Some answer choices overlap; choose the most restrictive answer. ANSWER:
[ Print ]
5. zero emf clockwise top plate zero =C*EMF =EMF^2*C =C*EMF*(1-exp(-t/(R*C))) =EMF/R*exp(-t/(R*C)) =9.21*tau =5.530×10-2 =q_0*exp(-t/(R*C)) =-q_0/(R*C)*exp(-t/(R*C)) 6.
=(epsilon_0*r^2/d*(K-1)*V)/DeltaT
