PHYS 3 Chapter Notes - Chapter 25.4-25.6: Electromotive Force, Internal Resistance, Ammeter

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19 Nov 2017

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Source of emf = a circuit device that provides emf. Not a force, but an energy-per-unit-charge quantity must be the same at the end of the round trip as at the beginning. Electromotive force (emf) = the influence that makes current flow from lower to higher potential. There are parts of the circuit where the potential energy increases and parts where it decreases. Complete circuit = a path that forms a closed loop that allows a conductor to have a steady current. There is no steady motion of charge in an incomplete circuit. If a charge goes around a complete circuit and returns to its starting point, the potential energy. Equation 25. 13-25. 14: for an ideal source of emf: Internal resistance = the resistance that a charge encounters when it moves through the. Equation 25. 15: terminal voltage, source with internal resistance: Is the emf of the source, is the current through the source, is the internal resistance of.

Related Questions

In a circuit containing only resistors, the basic (though notnecessarily explicit) assumption is that the current reaches itssteady-state value instantly. This is not the case for a circuitcontaining inductors. Due to a fundamental property of an inductorto mitigate any "externally imposed" change in current, the currentin such a circuit changes gradually when a switch is closed oropened.
Consider a series circuit containing a resistor of resistance andan inductor of inductance connected to a source of emf andnegligible internal resistance. The wires (including the ones thatmake up the inductor) are also assumed to have negligibleresistance.
Let us start by analyzing the process that takes place after switchis closed (switch remains open). In our further analysis, lowercaseletters will denote the instantaneous values of various quantities,whereas capital letters will denote the maximum values of therespective quantities.
Note that at any time during the process, Kirchhoff's loop ruleholds and is, indeed, helpful:
.
Part A
Immediately after the switch is closed, what is the current in thecircuit?
Hint A.1
How to approach the problem
Hint not displayed
ANSWER:

zero
Answer not displayed
Part B
Immediately after the switch is closed, what is the voltage acrossthe resistor?
Hint B.1
Ohm's law
Hint not displayed
ANSWER:

zero
Answer not displayed
Part C
Immediately after the switch is closed, what is the voltage acrossthe inductor?
Hint C.1
A formula for voltage across an inductor
Hint not displayed
ANSWER:

zero
Answer not displayed
Part D
Shortly after the switch is closed, what is the direction of thecurrent in the circuit?
ANSWER:
clockwise
counterclockwise
There is no current because the inductor does not allow the currentto increase from its initial zero value.
Answer not displayed
Part E
Shortly after the switch is closed, what is the direction of theinduced EMF in the inductor?
ANSWER:
clockwise
counterclockwise
There is no induced EMF because the initial value of the current iszero.
Answer not displayed
Part F
Eventually, the process approaches a steady state. What is thecurrent in the circuit in the steady state?
ANSWER:

Answer not displayed
Part G
What is the voltage across the inductor in the steady state?
Hint G.1
Induced EMF and current
Hint not displayed
ANSWER:
zero

Answer not displayed
Part H
What is the voltage across the resistor in the steady state?
ANSWER:
zero

Answer not displayed
Part I
Now that we have a feel for the state of the circuit in its steadystate, let us obtain the expression for the current in the circuitas a function of time. Note that we can use the loop rule (goingaround counterclockwise):
.
Note as well that and . Using these equations, we can get, aftersome rearranging of the variables and making the subsitution,
.
Integrating both sides of this equation yields
.
Use this last expression to obtain an expression for . Rememberthat and that .
Express your answer in terms of , , and . You may or may not needall these variables. Use the notation exp(x) for .
ANSWER:
=
Answer not displayed
Part J
Just as in the case of R-C circuits, the steady state here is neveractually reached: The exponential functions approach their limitsasymptotically as . However, it usually does not take very long forthe value of to get very close to its presumed limiting value. Thenext several questions illustrate this point.
Note that the quantity has dimensions of time and is called thetime constant (you may recall similar terminology applied to R-Ccircuits). The time constant is often denoted by . Using , one canwrite the expression

as
.
Find the ratio of the current at time to the maximum current.
Express your answer numerically, using three significantfigures.
ANSWER:
=
Answer not displayed
Part K
Find the time it takes the current to reach 99.999% of its maximumvalue.
Express your answer numerically, in units of . Use threesignificant figures.
ANSWER:
=
Answer not displayed

Part L
Find the time it takes the current to reach 99.999% of its maximumvalue. Assume that ohms and millihenrys.
Express your answer in seconds, using three significantfigures.
ANSWER:
=
Answer not displayed
seconds
Part M
Find the time it takes the current to reach 99.999% of its maximumvalue. Assume that ohms and henrys.
Express your answer in seconds, using three significantfigures.
ANSWER:
=
Answer not displayed
seconds
Part N
What fraction of the maximum value will be reached by the currentone minute after the switch is closed? Again, assume that ohms andhenrys.
Use three significant figures in your answer.
ANSWER:
=
Answer not displayed
Now consider a different situation. After switch has been closedfor a long time, it is opened; simultaneously, switch is closed, asshown in the figure. This effectively removes the battery from thecircuit.
The questions below refer to the time immediately after switch isopened and switch is closed.
Part O
What is the direction of the current in the circuit?
ANSWER:
clockwise
counterclockwise
The current is zero because there is no EMF in the circuit.
Answer not displayed
Part P
What is happening to the magnitude of the current?
ANSWER:
The current is increasing.
The current is decreasing.
The current remains constant.
Answer not displayed
Part Q
What is the direction of the EMF in the inductor?
ANSWER:
clockwise
counterclockwise
The EMF is zero because the current is zero.
The EMF is zero because the current is constant.
Answer not displayed
Part R
Which end of the inductor has higher voltage (i.e., to which end ofthe inductor should the positive terminal of a voltmeter beconnected in order to yield a positive reading)?
Hint R.1
How to approach the problem
Hint not displayed
ANSWER:
left
right
The potentials of both ends are the same.
The answer depends on the magnitude of the time constant.
Answer not displayed
Part S
For this circuit, Kirchhoff's loop rule gives
.
Note that , since the current is decreasing. Use this equation toobtain an expression for .
Hint S.1
Initial and final conditions
Hint not displayed
Hint S.2
Let us cheat a bit
Hint not displayed
Express your answer in terms of , , and . Use exp(x) for .
ANSWER:
=
Answer not displayed

Learning Goal: To work through astraightforward application of Faraday's law to find the EMF andthe electric field surrounding a region of increasing flux

Faraday's law describes how electric fields and electromotiveforces are generated from changing magnetic fields. This problem isa prototypical example in which an increasing magnetic fluxgenerates a finite line integral of the electric field around aclosed loop that surrounds the changing magnetic flux through asurface bounded by that loop. A cylindrical iron rod withcross-sectional area is oriented with its symmetry axis coincidentwith the z axis of a cylindrical coordinate system as shown. It hasa uniform magnetic field inside that varies according to . In otherwords, the magentic field is always in the positive z direction,and it has no other components.

For your convenience, we restate Faraday's law here: , where is theline integral of the electric field, and the magnetic flux is givenby , where is the angle between the magnetic field and the localnormal to the surface bounded by the closed loop.
Direction: The line integral and surface integral reverse theirsigns if the reference direction of or is reversed. The right-handrule applies here: If the thumb of your right hand is taken along ,then the fingers point along . You are free to take the loopanywhere you choose, although usually it makes sense to choose itto lie along the path of the circuit you are considering.

Find , the electromotive force (EMF) around a loop that is atdistance from the z axis, where is restricted to the region outsidethe iron rod as shown. Take the direction shown in the figure aspositive.
Express in terms of , , , , and any needed constants such as , ,and .
Due to the cylindrical symmetry of this problem, the inducedelectric field can depend only on the distance from the z axis,where is restricted to the region outside the iron rod. Find thisfield.
Express in terms of quantities given in the introduction (andconstants), using the unit vectors in the cylindrical coordinatesystem,r^ ,theta^ , and z^ .

PHYS 3 Chapter Notes - Chapter 25.4-25.6: Electromotive Force, Internal Resistance, Ammeter

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