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9 Nov 2019

Consider the depicted circuit.The voltage source is a 5 V power supply. The solenoid isconsidered to be ideal so that it has no internal resistance. Bothresistors are the same: 5.0 Ω.

**Assume the switch had been closed for a long time. UseKirchhoff's rules to determine the voltage drop acrossR2.

**At the instant just after the switch was opened,what is the current though the solenoid

**At the instant just after the switch was opened, what is thecurrent though R2?

**At the instant just after the switch was opened, what thecurrent though R1?

**At the instant just after the switch was opened, what themagnitude of the voltage drop across R1?

**Suppose R1 were replaced with a 5 MΩ resistor and the switch had beenclosed for a long time. What would the magnitude of the voltagedrop across R1 be just after the instant in which theswitch was opened?

*5V

*.5A

*.5A
*.5A

*2.5 V
*5V

(what I thought...)

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9 Nov 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A circuit is constructed with four resistors, one capacitor,one battery and a switch as shown. The values for the resistorsare: R1 = R2 = 65 O, R3 = 115 O and R4 = 136 O. The capacitance isC = 79 µF and the battery voltage is V = 24 V.
1) The switch has been open for a long timewhen at time t = 0, the switch is closed. What is I1(0), themagnitude of the current through the resistor R1 just after theswitch is closed?
___A
2) What is I1(8), the magnitude of the current thatflows through the resistor R1 a very long time after the switch hasbeen closed?
___A
3) What is Q(8), the charge on the capacitorafter the switch has been closed for a very long time?
___µC
4) What is I2(8), the magnitude of the current throughresistor R2, after the switch has been closed for a very longtime?
___A
5)
What is IC(0), the magnitude of the currentthrough the capacitor just after the switch is closed?
___A

A circuit is constructed with four resistors, one capacitor,one battery and a switch as shown. The values for the resistorsare: R1 = R2 = 48 O, R3 = 74 O and R4 = 158 O. The capacitance is C= 45 µF and the battery voltage is V = 12 V. The positive terminalof the battery is indicated with a + sign.
1) The switch has been open for a long timewhen at time t = 0, the switch is closed. What is I4(0), themagnitude of the current through the resistor R4 just after theswitch is closed?
What is Q(8), the charge on the capacitor after theswitch has been closed for a very long time?
After the switch has been closed for a very long time,it is then opened. What is Q(topen), the charge on the capacitor ata time topen = 705 µs after the switch was opened?
What is IC,max(closed), the current that flows throughthe capacitor whose magnitude is maximum during the time when theswitch is closed? A positive value for the current is defined to bein the direction of the arrow shown.
What is IC,max(open), the current that flows through thecapacitor whose magnitude is maximum during the time when theswitch is open? A positive value for the current is defined to bein the direction of the arrow shown.

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

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