what to do if i get 2.5v on and outlet
Electrical Circuits Review
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Office D: Qualitative Relationships Between Variables
lx. A resistor with a resistance of R is connected to a bombardment with a voltage of V to produce a electric current of I. What would be the new current (in terms of I) if ...
a. ... the resistance is doubled and the same voltage is used?b. ... the voltage is doubled and the same resistance is used?
c. ... the voltage is tripled and the resistance is doubled?
d. ... the voltage is doubled and the resistance is halved?
e. ... the voltage is halved and the resistance is doubled?
f. ... five times the voltage and i-third the resistance is used?
g. ... one-fifth the voltage and one-fourth the resistance is used?
Answer: See answers below.
This question tests your understanding of the current-voltage-resistance relationship. The current is directly proportional to the voltage and inversely proportional to the resistance. Any alteration in the voltage will upshot in the same alteration of the current. So doubling or tripling the voltage will cause the electric current to exist doubled or tripled. On the other hand, any alteration in the resistance volition effect in the opposite or changed alteration of the electric current. So doubling or tripling the resistance will cause the electric current to be one-half or one-third the original value.
a. The new current will exist 0.5 • I.b. The new current will exist 2 • I.
c. The new current volition be i.5 • I.
d. The new current will be iv • I.
due east. The new electric current will be 0.25 • I.
f. The new electric current will be xv • I.
thousand. The new current will exist 0.8 • I.
61. A wire of length L and cross-sectional area A is used in a circuit. The overall resistance of the wire is R. What would be the new resistance (in terms of R) if ...
a. ... the length of the wire is doubled?b. ... the cross-exclusive expanse of the wire is doubled?
c. ... the length of the wire is doubled and the cross-sectional area of the wire is doubled?
d. ... the length of the wire is tripled and the cantankerous-sectional area of the wire is doubled?
e. ... the length of the wire is halved and the cantankerous-sectional surface area of the wire is tripled?
f. ... the length of the wire is tripled and the cross-exclusive area of the wire is halved?
g. ... the length of the wire is tripled and the diameter of the wire is halved?
h. ... the length of the wire is tripled and the diameter of the wire is doubled?
Reply: See answers below.
This question tests your understanding of the variables which consequence the resistance of a wire. The resistance of a wire expressed past the equation R = Rho • Fifty / A (where Rho is the resistivity of the cloth, L is length of wire, and A is cross-sectional area of the wire). The resistance is straight proportional to the resistivity, directly proportional to the wire length, and inversely proportional to the cross-sectional expanse. Any alteration in the resistivity or the length will event in the same amending in the resistance of the wire. And any alteration in the cross-sectional area of the wire will result in the opposite or inverse alteration in the resistance of the wire.
a. The new resistance volition be 2•R.b. The new resistance volition be 0.five•R.
c. The new resistance will still be R.
d. The new resistance will exist 1.5•R.
e. The new resistance volition be (1/6)•R.
f. The new resistance will be vi•R.
m. The new resistance volition exist 12•R. (Halving the diameter volition make the surface area one-4th the size since surface area is directly proportional to the foursquare of the diameter.)
h. The new resistance will be (iii/4)•R. (Doubling the bore will make the area four times the size since area is directly proportional to the foursquare of the diameter.)
62. An electric apparatus with a current of I and a resistance of R converts energy to other forms at a rate of P when continued to a 120-Volt outlet. What would be the new power rating (in terms of P) if ...
a. ... the current is doubled (and the same 120-Volt outlet is used)?b. ... the electric current is halved (and the same 120-Volt outlet is used)?
c. ... the resistance is doubled (and the same 120-Volt outlet is used)?
d. ... the resistance is halved (and the same 120-Volt outlet is used)?
e. ... the current is tripled (and the same 120-Volt outlet is used)?
f. ... the resistance is tripled (and the same 120-Volt outlet is used)?
grand. ... the same apparatus is powered by a 12-Volt supply?
h. ... the aforementioned appliance is powered by a 240-Volt supply?
Answer: Come across answers beneath.
This question tests your understanding of the mathematical relationship between ability, electric current, voltage and resistance. In that location are three equations of importance:
| | | |
I must be careful in using the last equation since an alteration in current volition as well alter the resistance whenever the voltage is held abiding. Thus, the first two equations are of greater importance since they represent equations with i independent variable and the other variable held constant.
a. 2 • P (doubling the current will double the power)b. (i/ii) • P (halving the current will double the power)
c. (1/two) • P (doubling the resistance will half the power)
d. 2 • P (halving the resistance will double the power)
e. three • P (tripling the current will triple the power)
f. (1/3) • P (tripling the resistance volition make the power one-third of the original value)
g. (1/100) • P (one-tenth the voltage will brand the power one-hundredth of the original value; observe the square on voltage)
h. 4 • P (two times the voltage will make the power 4 times the original value; observe the square on voltage)
63. An electric apparatus with a electric current of I and a resistance of R is used for t hours during the grade of a month. The cost of operating the appliance at 120-Volts is D dollars. What would be the new cost (in terms of D) if ...
a. ... the usage charge per unit was doubled to 2t?b. ... the usage rate was halved?
c. ... an appliance which drew twice the current (at 120 Volts) were used?
d. ... an appliance with twice the resistance (at 120 Volts) were used?
e. ... an appliance with one-half the resistance (at 120 Volts) were used?
f. ... the usage rate was doubled and an appliance with twice the resistance (at 120 Volts) were used?
g. ... the usage charge per unit was halved and an appliance with twice the current (at 120 Volts) were used?
h. ... the usage charge per unit was quartered and an appliance with twice the current (at 120 Volts) were used?
Answer: See answers beneath.
Similar the previous question, this question tests your understanding of the mathematical relationship between power, electric current, voltage and resistance. But this question also tests your understanding between power, fourth dimension, energy and electricity costs. An electric beak is based upon energy consumption. The energy consumed is measured in terms of kiloWatts•hr and is determined past multiplying the power past the time. Thus, an increase in either fourth dimension or power will lead to an increase in the electricity costs by the same factor. So the central to the question is to use information nearly power and virtually usage rate to decide the free energy consumed and thus the electricity costs. There are two equations of importance in predicting how alterations in current and resistance effect the power:
The first equation shows that the power would increase by the same gene by which the current is increased. The second equation shows that the power would subtract past the same factor that the resistance is increased.
a. The new price would be 2•D.b. The new cost would be (1/ii)•D.
c. The new price would be 2•D.
d. The new toll would be (ane/2)•D.
east. The new cost would be two•D.
f. The new cost would nevertheless exist D.
g. The new cost would nevertheless exist D.
h. The new cost would be (one/2)•D.
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a. If the current at location A is I amperes, then the current at location B is ____ amperes. (Answer in terms of I.)
b. If the electric current at location A is I amperes, so the current at location D is ____ amperes. (Answer in terms of I.)
c. If the current at location A is I amperes, then the electric current at location L is ____ amperes. (Reply in terms of I.)
d. If the voltage of the battery is doubled, then the electric current at location A would exist ____ (2 times, four times, one-half, one-4th, etc.) the original value.
e. If the voltage of the bombardment is doubled, and then the electric current at location B would be ____ (two times, four times, one-half, 1-fourth, etc.) the original value.
f. If the voltage of the battery is doubled, and so the current at location D would be ____ (2 times, four times, one-half, 1-4th, etc.) the original value.
g. Suppose that the resistance of the lite seedling located between points D and Grand is doubled. This would result in the current measured at location Grand to ____ (increase, decrease, not be affected).
h. Suppose that the resistance of the light bulb located between points D and Thousand is doubled. This would result in the electric potential deviation between points D and G to ____ (increase, decrease, not exist affected).
i. Suppose that the resistance of the light seedling located between points D and G is doubled. This would consequence in the current measured at location A to ____ (increase, decrease, not be afflicted).
j. Suppose that the resistance of the light bulb located between points D and G is doubled. This would result in the current measured at location Eastward to ____ (increase, decrease, not be affected).
thou. Suppose that the resistance of the lite seedling located between points D and G is doubled. This would result in the current measured at location K to ____ (increase, decrease, not be affected).
Answer: See answers above.
a. - c. Location A is outside or before the branching locations; it represents a location where the full excursion electric current is measured. This current will ultimately split into three pathways, with each pathway carrying the same current (since each pathway has the aforementioned resistance). Location D is a branch location; i-3rd of the charge passes through this branch. Location B represents a location later on a point at which one-third of the accuse has already branched off to the low-cal bulb betwixt points D and One thousand. So at location B, at that place is 2-thirds of the electric current remaining. And location L is a location in the last branch; so one-third of the charge passes through location L.
d. - f. The current at every branch location and in the total excursion is simply equal to the voltage drop across the branch (or across the full circuit) divided by the resistance of the branch (or of the total excursion). As such, the current is directly proportional to the voltage. Then a doubling of the voltage volition double the current at every location.
g. The current at a branch location is simply the voltage across the branch divided by the resistance of the branch. So the electric current at location G is inversely proportional to the resistance of the branch. Doubling the resistance will cause the current to be decreased by a factor of 2.
h. The voltage drop across the first branch (or any co-operative) is simply equal to the voltage gained by the charge in passing through the battery. For a parallel excursion, the simply means of altering a branch voltage drib is to alter the battery voltage.
i. - yard. Altering the resistance of a lite bulb in a specific co-operative tin alter the current in that branch and the electric current in the overall circuit. The current in a branch is inversely proportional to the resistance of the co-operative. And so increasing the resistance of a co-operative will decrease the current of that branch and the overall current in the circuit (as measured at location A). Nevertheless, the current in the other branches are dependent solely upon the voltage drops of those branches and the resistance of those branches. So while altering the resistance of a single branch alters the current at that branch location, the other branch currents remain unaffected.
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Part E: Trouble-Solving and Circuit Assay
65. If the current at a given point in a circuit is ii.five Amps, then how many electrons pass that signal on the circuit in a time flow of 1 minute.
Answer: ix.375 x 1020 electrons
The current (I) is the rate at which charge passes a signal on the circuit in a unit of time. So I = Q/t. Rearranging this equation leads to Q = I•t. Recognizing that a current of 2.5 Amps is equivalent to 2.5 Coulombs per second and that 1 infinitesimal is equivalent to 60 seconds tin lead to the corporeality of Coulombs moving pass the betoken.
Q = I•t = (2.5 C/due south)•(threescore s) = 150 Coulombs
The charge of a unmarried electron is equal to 1.6 ten x-19 C. Then 150 Coulombs must be a lot of electrons. The actual number can be computed equally shown:
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66. What is the resistance (in ohms) of a typical 40-Watt light bulb plugged into a 120-Volt outlet in your abode?
Reply: 360 Ohms
The power dissipated in a circuit is given past the equation P = I• Δ 5. Substituting in Δ V/R for the current can lead to an equation relating the resistance (R) to the voltage driblet ( Δ Five) and the power (P).
Rearrangement of the equation and substitution of known values of power (40 Watts) and voltage (120 V) leads to the following solution.
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67. Determine the length of nichrome wire (resistivity value = 150 x 10-8 ohm•g) required to produce a i.00 mAmp current if a voltage of 1.5 Volts is impressed across it. The diameter of the wire is 1/sixteen-thursday of an inch. (GIVEN: 2.54 cm = 1 inch)
Answer: two.0 x x3 m
This is clearly an do in unit conversion (or at least unit awareness). The resistance (R) of a wire is related to the resistivity (Rho), the length (L) and the cross-sectional area (A) by the equation
This can be rearranged to solve for length
The diameter is given; cross-sectional area is simply given by PI•R2. The radius of the wire is 1-one-half the diameter - 1/32 inch.
Since the unit of length for the Rho value is meters, the cross-exclusive area volition be adamant in k2 earlier substitution into the equation. The fact that the conversion involves squared units makes this problem even more trickier. The conversion factors will accept to be squared to accomplish the conversion successfully.
The final quantity which must be determined earlier computing the wire length is the bodily resistance of the wire. Resistance is related to voltage ( Δ V) and current (I) past the equation R = Δ V / I. Standard units are Ohms, Volts and Amps. Here, the current is given in milliAmps (mAmps); substitution to Amps must exist performed before substitution.
Now R, A and Rho can be substituted into the length equation to determine the length of the wire in meters.
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68. Determine the full monthly cost of using the following appliances/household wires for the given amount of time if each is plugged into a 120-Volt household outlet. The cost of electricity is $0.13 / kW•hr. (Assume that a month lasts for 30 days.)
| (with info from labels) | (hours/day) | (Watts) | Consumed | ($) |
| Hair Dryer (12 Amp) | | | | |
| Coffee Percolator (9.0 Amp) | | | | |
| Light Bulb (100 Watt) | | | | |
| Attic Fan (140 Watt) | | | | |
| Microwave Oven (8.3 Amps) | | | | |
Total | | |||
Answer: Meet table in a higher place.
The power is either explicitly stated (every bit in the case of the light bulb) or calculated using P = I• Δ V. In this example, the voltage is 120 Volts. The energy consumed is the Ability•time. It is useful to express this quantity in the same units for which one is charged for information technology - kiloWatt • hour. The calculation involves converting power in Watts to kiloWatts by dividing by thousand and and so multiplying by the time in hours/calendar month and and then multiplying by 30 days/month. The price in dollars is simply the kiloWatt•hours of free energy used multiplied by the toll of $0.13/kW•hr.
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69. If the copper wire used to bear telegraph signals has a resistance of 10 ohms for every mile of wire, then what is the bore of the wire. (Given: 1609 one thousand = one mile. Resistivity of Cu = 1.7 ten 10-8 ohm•m )
Respond: 0.59 cm
Like Question #67, this is some other exercise in unit conversion and unit sensation. The resistance (R) of a wire is related to the resistivity (Rho), the length (L) and the cross-exclusive area (A) by the equation
This can be rearranged to solve for cross-sectional area
The area is related to the radius by the equation A = PI•Rii. The plan volition involve determining the Area, so the radius, so the diameter of the wire.
First, annotation that the given information is: Rho = 1.7 x x-8 ohm•m; L = 1 mi = 1609 m; R = 10 Ohm. By substitution, the Area tin be determined:
A = 2.7353 x 10-half-dozen yard2
The expanse is in chiliad2 units. Since the diameter of wires is typically expressed in centimeters or millimeters, a conversion will be performed. The fact that the conversion involves squared units makes this conversion even more trickier. The conversion factors volition have to exist squared to attain the conversion successfully.
Now the area equation (A = PI•R2) can be used to make up one's mind the radius.
R = 0.29507 cm
The radius is simple twice the diameter. So d = 0.59014 cm.
70. Determine the resistance of a 1500 Watt electrical grill connected to a 120-Volt outlet.
Answer: ix.6 Ohms
The power dissipated in a circuit is given by the equation P = I• Δ V. Substituting in Five/R for the current can lead to an equation relating the resistance (R) to the voltage drop ( Δ V) and the power (P).
Rearrangement of the equation and exchange of known values of power (1500 Watts) and voltage (120 V) leads to the following solution.
71. Four resistors - two-Ohms, five-Ohms, 12-Ohms and 15-Ohms - are placed in series with a 12-Volt bombardment. Determine the electric current at and voltage drop beyond each resistor.
Respond: See diagram below.
The diagram below depicts the series circuit using schematic symbols. Note that at that place is no branching, consistent with the notion of a series excursion.
For a series circuit, the overall resistance (RTot) is just the sum of the individual resistances. That is
RTot = 2 ½ + 5 ½ + 12 ½ + 15 ½ = 34 ½
The serial of iii resistors supplies an overall, total or equivalent resistance of 34 Ohms. Since there is no branching, the current is the aforementioned through each resistor. This electric current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to overall resistance (5Tot/RTot).
The electric current through the battery and through each of the resistors is ~0.353 Amps. The voltage drib across each resistor is equal to the I•R product for each resistor. These calculations are shown below.
Δ 52 = Iii • Rtwo = (0.35294 Amps) • (5 Ohms) = 1.76 V
Δ V3 = Ithree • R3 = (0.35294 Amps) • (12 Ohms) = 4.24 V
Δ 5iv = I4 • R4 = (0.35294 Amps) • (15 Ohms) = five.29 V
72. Four resistors - 2-Ohms, 5-Ohms, 12-Ohms and 15-Ohms - are placed in parallel with a 12-Volt bombardment. Determine the electric current at and voltage drib beyond each resistor.
Respond: Come across diagram below.
The diagram beneath depicts the parallel circuit using schematic symbols. Note that in that location is a branching, consequent with the notion of a parallel circuit.
For a parallel excursion, the reciprocal of overall resistance (1 / RTot) is just the sum of the reciprocals of individual resistances. That is
1 / RTot = 1 / 2 ½ + 1 / 5 ½ + 1 / 12 ½ + 1 / xv ½ = 0.850 / ½
RTot = i.17647 ½
The serial of iii resistors supplies an overall, total or equivalent resistance of ~1.18 Ohms. This total resistance value tin exist used to determine the total current through the circuit.
Since there is branching, the total current will be equal to the sum of the currents at each resistor. The current at each resistor is the voltage drop across each resistor divided past the resistance of each resistor. For series circuits, the voltage drop across each resistor is the same as the voltage gained by the accuse in the battery (12 Volts in this instance). The branch current calculations are shown beneath.
I 2 = Δ Fiveii / R2 = (12 Volts) / (5 Ohms) = two.forty Amp
I 3 = Δ 53 / Rthree = (12 Volts) / (12 Ohms) = 1.00 Amp
I 4 = Δ V4 / R4 = (12 Volts) / (15 Ohms) = 0.80 Amp
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