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Electrical Circuits Review
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Part D: Qualitative Relationships Between Variables
lx. A resistor with a resistance of R is connected to a battery with a voltage of V to produce a electric current of I. What would exist 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?
eastward. ... the voltage is halved and the resistance is doubled?
f. ... five times the voltage and one-third the resistance is used?
g. ... one-5th the voltage and 1-fourth the resistance is used?
Respond: Run into answers below.
This question tests your agreement of the current-voltage-resistance relationship. The electric current is directly proportional to the voltage and inversely proportional to the resistance. Whatsoever alteration in the voltage volition result in the same alteration of the current. Then doubling or tripling the voltage will cause the electric current to exist doubled or tripled. On the other hand, any amending in the resistance volition issue in the contrary or inverse alteration of the current. So doubling or tripling the resistance volition cause the current to be one-half or i-third the original value.
a. The new current will be 0.5 • I.b. The new current volition be 2 • I.
c. The new current will be 1.5 • I.
d. The new electric current will be 4 • I.
e. The new current will exist 0.25 • I.
f. The new current will be 15 • I.
g. The new current volition be 0.eight • I.
61. A wire of length L and cross-exclusive expanse 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-sectional area of the wire is doubled?
c. ... the length of the wire is doubled and the cantankerous-sectional surface area of the wire is doubled?
d. ... the length of the wire is tripled and the cantankerous-sectional area of the wire is doubled?
eastward. ... the length of the wire is halved and the cantankerous-sectional area of the wire is tripled?
f. ... the length of the wire is tripled and the cross-sectional surface 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 bore of the wire is doubled?
Respond: See answers below.
This question tests your understanding of the variables which outcome the resistance of a wire. The resistance of a wire expressed by the equation R = Rho • 50 / A (where Rho is the resistivity of the fabric, L is length of wire, and A is cantankerous-sectional area of the wire). The resistance is directly proportional to the resistivity, directly proportional to the wire length, and inversely proportional to the cantankerous-exclusive area. Any amending in the resistivity or the length will consequence in the aforementioned alteration in the resistance of the wire. And whatsoever amending in the cantankerous-sectional expanse of the wire will result in the opposite or inverse amending in the resistance of the wire.
a. The new resistance will be 2•R.b. The new resistance volition be 0.5•R.
c. The new resistance will all the same exist R.
d. The new resistance will be ane.five•R.
e. The new resistance will exist (1/vi)•R.
f. The new resistance volition be 6•R.
1000. The new resistance will be 12•R. (Halving the diameter will make the area one-fourth the size since area is directly proportional to the square of the diameter.)
h. The new resistance will be (iii/4)•R. (Doubling the diameter will brand the area iv times the size since expanse is directly proportional to the square of the bore.)
62. An electric appliance with a current of I and a resistance of R converts free energy to other forms at a rate of P when connected 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 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 aforementioned 120-Volt outlet is used)?
g. ... the same appliance is powered by a 12-Volt supply?
h. ... the same appliance is powered by a 240-Volt supply?
Respond: See answers below.
This question tests your understanding of the mathematical human relationship betwixt power, current, voltage and resistance. There are 3 equations of importance:
| | | |
One must be conscientious in using the last equation since an alteration in current will also modify the resistance whenever the voltage is held abiding. Thus, the first 2 equations are of greater importance since they represent equations with one contained variable and the other variable held abiding.
a. 2 • P (doubling the current will double the power)b. (ane/2) • P (halving the current volition double the power)
c. (1/two) • P (doubling the resistance will half the power)
d. 2 • P (halving the resistance volition double the power)
east. three • P (tripling the current volition triple the power)
f. (ane/3) • P (tripling the resistance will make the power one-tertiary of the original value)
yard. (1/100) • P (one-tenth the voltage will make the power one-hundredth of the original value; observe the square on voltage)
h. 4 • P (two times the voltage will make the ability iv times the original value; discover the square on voltage)
63. An electric appliance with a current of I and a resistance of R is used for t hours during the course of a month. The cost of operating the apparatus at 120-Volts is D dollars. What would be the new toll (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 electric current (at 120 Volts) were used?
d. ... an appliance with twice the resistance (at 120 Volts) were used?
due east. ... an appliance with 1-half the resistance (at 120 Volts) were used?
f. ... the usage charge per unit 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 apparatus with twice the electric current (at 120 Volts) were used?
Answer: See answers below.
Like the previous question, this question tests your understanding of the mathematical human relationship between power, current, voltage and resistance. But this question also tests your understanding between ability, time, energy and electricity costs. An electrical bill is based upon energy consumption. The energy consumed is measured in terms of kiloWatts•60 minutes and is determined by multiplying the power past the fourth dimension. Thus, an increase in either time or power will pb to an increase in the electricity costs past the aforementioned factor. So the key to the question is to apply information about power and nigh usage charge per unit to determine the energy consumed and thus the electricity costs. At that place are two equations of importance in predicting how alterations in current and resistance event the power:
The outset equation shows that the power would increase by the aforementioned gene by which the electric current is increased. The 2nd equation shows that the power would decrease past the same factor that the resistance is increased.
a. The new cost would exist 2•D.b. The new cost would be (1/two)•D.
c. The new cost would be 2•D.
d. The new cost would be (one/2)•D.
e. The new price would be 2•D.
f. The new cost would notwithstanding exist D.
g. The new cost would still be D.
h. The new cost would be (1/2)•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 current at location A is I amperes, and then the electric current at location D is ____ amperes. (Answer in terms of I.)
c. If the current at location A is I amperes, and then the current at location L is ____ amperes. (Answer in terms of I.)
d. If the voltage of the battery is doubled, then the current at location A would exist ____ (two times, four times, one-half, one-fourth, etc.) the original value.
e. If the voltage of the battery is doubled, then the current at location B would exist ____ (two times, four times, one-half, one-fourth, etc.) the original value.
f. If the voltage of the battery is doubled, then the current at location D would be ____ (two times, four times, one-half, i-fourth, etc.) the original value.
grand. 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 G to ____ (increment, decrease, not be afflicted).
h. Suppose that the resistance of the calorie-free seedling located between points D and G is doubled. This would result in the electric potential difference between points D and G to ____ (increase, decrease, not exist afflicted).
i. Suppose that the resistance of the light bulb located between points D and Yard is doubled. This would result in the current measured at location A to ____ (increase, subtract, not exist affected).
j. Suppose that the resistance of the low-cal bulb located between points D and M is doubled. This would upshot in the electric current measured at location E to ____ (increase, decrease, not exist affected).
k. Suppose that the resistance of the calorie-free seedling located between points D and G is doubled. This would result in the current measured at location Thou to ____ (increment, decrease, not exist afflicted).
Answer: Come across answers higher up.
a. - c. Location A is outside or before the branching locations; it represents a location where the total circuit current is measured. This current volition ultimately divide into 3 pathways, with each pathway carrying the same current (since each pathway has the same resistance). Location D is a branch location; one-third of the charge passes through this co-operative. Location B represents a location after a point at which one-3rd of the charge has already branched off to the light seedling between points D and K. So at location B, at that place is two-thirds of the current remaining. And location Fifty is a location in the last branch; so one-third of the accuse passes through location L.
d. - f. The current at every branch location and in the total circuit is simply equal to the voltage drop beyond the branch (or beyond the total excursion) divided by the resistance of the branch (or of the total excursion). Equally such, the current is directly proportional to the voltage. And so a doubling of the voltage will double the current at every location.
g. The current at a co-operative location is simply the voltage across the branch divided by the resistance of the co-operative. So the current at location M is inversely proportional to the resistance of the branch. Doubling the resistance volition cause the current to be decreased by a factor of 2.
h. The voltage drop across the outset branch (or whatever branch) is just equal to the voltage gained by the charge in passing through the bombardment. For a parallel circuit, the merely means of altering a branch voltage drop is to alter the battery voltage.
i. - chiliad. Altering the resistance of a light bulb in a specific branch tin alter the current in that branch and the electric current in the overall circuit. The electric current in a branch is inversely proportional to the resistance of the co-operative. So increasing the resistance of a branch will decrease the current of that branch and the overall current in the excursion (as measured at location A). However, the current in the other branches are dependent solely upon the voltage drops of those branches and the resistance of those branches. And so while altering the resistance of a single branch alters the current at that co-operative location, the other co-operative currents remain unaffected.
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Part Eastward: Problem-Solving and Circuit Analysis
65. If the current at a given point in a circuit is 2.5 Amps, then how many electrons pass that signal on the circuit in a fourth dimension period of ane minute.
Answer: 9.375 x x20 electrons
The current (I) is the rate at which accuse passes a point on the circuit in a unit of fourth dimension. 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 one infinitesimal is equivalent to sixty seconds can atomic number 82 to the amount of Coulombs moving laissez passer the point.
Q = I•t = (2.five C/southward)•(60 due south) = 150 Coulombs
The charge of a unmarried electron is equal to 1.6 x 10-19 C. And so 150 Coulombs must be a lot of electrons. The actual number can be computed as shown:
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66. What is the resistance (in ohms) of a typical xl-Watt light bulb plugged into a 120-Volt outlet in your home?
Answer: 360 Ohms
The power prodigal in a excursion is given by the equation P = I• Δ V. Substituting in Δ V/R for the electric 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 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 ten-eight ohm•m) required to produce a 1.00 mAmp current if a voltage of ane.5 Volts is impressed across information technology. The diameter of the wire is 1/sixteen-th of an inch. (GIVEN: 2.54 cm = 1 inch)
Answer: two.0 x ten3 m
This is conspicuously an exercise in unit conversion (or at least unit of measurement awareness). The resistance (R) of a wire is related to the resistivity (Rho), the length (L) and the cross-exclusive area (A) past the equation
This tin be rearranged to solve for length
The diameter is given; cantankerous-sectional area is simply given by PI•R2. The radius of the wire is ane-half the diameter - ane/32 inch.
Since the unit of measurement of length for the Rho value is meters, the cross-sectional area will be determined in m2 before substitution into the equation. The fact that the conversion involves squared units makes this trouble even more trickier. The conversion factors will have to be squared to attain the conversion successfully.
The final quantity which must be adamant before computing the wire length is the actual resistance of the wire. Resistance is related to voltage ( Δ V) and current (I) by the equation R = Δ Five / I. Standard units are Ohms, Volts and Amps. Here, the current is given in milliAmps (mAmps); substitution to Amps must exist performed before commutation.
At present 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. Decide the total monthly price of using the following appliances/household wires for the given corporeality of time if each is plugged into a 120-Volt household outlet. The cost of electricity is $0.13 / kW•hr. (Assume that a month lasts for thirty days.)
| (with info from labels) | (hours/day) | (Watts) | Consumed | ($) |
| Hair Dryer (12 Amp) | | | | |
| Coffee Percolator (9.0 Amp) | | | | |
| Light Seedling (100 Watt) | | | | |
| Attic Fan (140 Watt) | | | | |
| Microwave Oven (eight.3 Amps) | | | | |
Total | | |||
Answer: See table in a higher place.
The power is either explicitly stated (as in the case of the light seedling) or calculated using P = I• Δ 5. In this instance, the voltage is 120 Volts. The free energy consumed is the Power•fourth dimension. It is useful to express this quantity in the same units for which one is charged for it - kiloWatt • hour. The adding involves converting power in Watts to kiloWatts by dividing by 1000 and then multiplying by the time in hours/month and then multiplying by 30 days/month. The cost in dollars is simply the kiloWatt•hours of energy used multiplied past the cost of $0.13/kW•hr.
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69. If the copper wire used to carry telegraph signals has a resistance of 10 ohms for every mile of wire, and then what is the diameter of the wire. (Given: 1609 m = ane mile. Resistivity of Cu = 1.7 x 10-viii ohm•thousand )
Reply: 0.59 cm
Similar Question #67, this is another do in unit conversion and 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 exist rearranged to solve for cross-sectional expanse
The area is related to the radius by the equation A = PI•R2. The program will involve determining the Area, then the radius, then the bore of the wire.
First, note that the given information is: Rho = one.7 x x-eight ohm•yard; L = 1 mi = 1609 m; R = 10 Ohm. By substitution, the Area can be determined:
A = 2.7353 10 x-6 mii
The area 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 than trickier. The conversion factors will have to be squared to attain the conversion successfully.
Now the expanse equation (A = PI•R2) tin can be used to determine the radius.
R = 0.29507 cm
The radius is unproblematic twice the diameter. And so d = 0.59014 cm.
70. Determine the resistance of a 1500 Watt electrical grill connected to a 120-Volt outlet.
Reply: 9.6 Ohms
The power dissipated in a circuit is given by the equation P = I• Δ V. Substituting in V/R for the current can lead to an equation relating the resistance (R) to the voltage drop ( Δ Five) and the ability (P).
Rearrangement of the equation and substitution of known values of ability (1500 Watts) and voltage (120 5) leads to the following solution.
71. 4 resistors - ii-Ohms, 5-Ohms, 12-Ohms and fifteen-Ohms - are placed in series with a 12-Volt battery. Determine the current at and voltage drop across each resistor.
Answer: See diagram below.
The diagram below depicts the serial circuit using schematic symbols. Notation that there is no branching, consistent with the notion of a series circuit.
For a series circuit, the overall resistance (RTot) is merely the sum of the individual resistances. That is
RTot = ii ½ + five ½ + 12 ½ + 15 ½ = 34 ½
The serial of 3 resistors supplies an overall, full or equivalent resistance of 34 Ohms. Since at that place is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of bombardment voltage to overall resistance (VTot/RTot).
The current through the battery and through each of the resistors is ~0.353 Amps. The voltage drop across each resistor is equal to the I•R product for each resistor. These calculations are shown beneath.
Δ V2 = I2 • Rii = (0.35294 Amps) • (5 Ohms) = 1.76 5
Δ Viii = Ithree • R3 = (0.35294 Amps) • (12 Ohms) = iv.24 5
Δ 54 = I4 • R4 = (0.35294 Amps) • (fifteen Ohms) = 5.29 5
72. 4 resistors - 2-Ohms, 5-Ohms, 12-Ohms and fifteen-Ohms - are placed in parallel with a 12-Volt bombardment. Determine the current at and voltage drop across each resistor.
Answer: Encounter diagram beneath.
The diagram below depicts the parallel circuit using schematic symbols. Notation that there is a branching, consequent with the notion of a parallel excursion.
For a parallel circuit, the reciprocal of overall resistance (1 / RTot) is simply the sum of the reciprocals of private resistances. That is
1 / RTot = ane / 2 ½ + ane / five ½ + 1 / 12 ½ + 1 / 15 ½ = 0.850 / ½
RTot = 1.17647 ½
The series of three resistors supplies an overall, total or equivalent resistance of ~1.18 Ohms. This full resistance value can be used to make up one's mind the total current through the circuit.
Since there is branching, the total electric 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 by the resistance of each resistor. For serial circuits, the voltage driblet across each resistor is the same equally the voltage gained past the charge in the bombardment (12 Volts in this case). The branch current calculations are shown below.
I 2 = Δ Vtwo / R2 = (12 Volts) / (5 Ohms) = two.40 Amp
I 3 = Δ V3 / R3 = (12 Volts) / (12 Ohms) = 1.00 Amp
I 4 = Δ 54 / R4 = (12 Volts) / (xv Ohms) = 0.80 Amp
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