If The Diameter Is Doubled And The Length Is Tripled, By What Factor Is The Resistance Changed?
Electric Circuits Review
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Part D: Qualitative Relationships Between Variables
sixty. A resistor with a resistance of R is continued to a bombardment with a voltage of V to produce a 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 ane-third the resistance is used?
thou. ... one-fifth the voltage and one-fourth the resistance is used?
Reply: 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. Whatever amending in the voltage volition result in the aforementioned alteration of the current. And then doubling or tripling the voltage will cause the current to be doubled or tripled. On the other mitt, any alteration in the resistance will result in the contrary or changed amending of the current. And so doubling or tripling the resistance will crusade the electric current to be half or one-tertiary the original value.
a. The new current will be 0.five • I.b. The new current will be 2 • I.
c. The new current will be one.v • I.
d. The new current will be four • I.
e. The new electric current will exist 0.25 • I.
f. The new current will be xv • I.
g. The new electric current will be 0.8 • I.
61. A wire of length 50 and cross-exclusive 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-sectional surface area of the wire is doubled?
c. ... the length of the wire is doubled and the cross-exclusive area of the wire is doubled?
d. ... the length of the wire is tripled and the cross-sectional area of the wire is doubled?
east. ... the length of the wire is halved and the cross-sectional area of the wire is tripled?
f. ... the length of the wire is tripled and the cross-sectional 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 by the equation R = Rho • L / 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, straight proportional to the wire length, and inversely proportional to the cross-sectional area. Any amending in the resistivity or the length will result in the same alteration in the resistance of the wire. And any alteration in the cross-sectional area of the wire volition result in the opposite or inverse alteration in the resistance of the wire.
a. The new resistance will be ii•R.b. The new resistance will exist 0.five•R.
c. The new resistance will nonetheless be R.
d. The new resistance volition be 1.5•R.
e. The new resistance will be (one/six)•R.
f. The new resistance will be 6•R.
thou. The new resistance volition be 12•R. (Halving the diameter will make the area 1-fourth the size since expanse is directly proportional to the foursquare of the diameter.)
h. The new resistance will be (3/4)•R. (Doubling the diameter will brand the area iv times the size since surface area is directly proportional to the square of the bore.)
62. An electric appliance 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 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 aforementioned 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?
Answer: See answers below.
This question tests your understanding of the mathematical relationship betwixt power, current, voltage and resistance. There are three equations of importance:
| | | |
One must be careful in using the last equation since an alteration in electric current will also change the resistance whenever the voltage is held constant. 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 electric current will double the ability)b. (ane/2) • P (halving the electric current will double the power)
c. (one/2) • P (doubling the resistance will half the power)
d. 2 • P (halving the resistance volition double the ability)
e. 3 • P (tripling the current will triple the power)
f. (i/3) • P (tripling the resistance will make the power i-third of the original value)
g. (1/100) • P (one-tenth the voltage will brand the power i-hundredth of the original value; discover the square on voltage)
h. four • P (2 times the voltage will make the power four times the original value; observe 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 class 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 rate 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?
one thousand. ... the usage rate was halved and an appliance with twice the current (at 120 Volts) were used?
h. ... the usage rate was quartered and an appliance with twice the current (at 120 Volts) were used?
Answer: See answers beneath.
Like the previous question, this question tests your understanding of the mathematical relationship between ability, current, voltage and resistance. But this question also tests your understanding between ability, time, energy and electricity costs. An electrical bill is based upon free energy consumption. The energy consumed is measured in terms of kiloWatts•hour and is determined by multiplying the ability past the time. Thus, an increase in either time or ability volition lead to an increment in the electricity costs by the same factor. So the fundamental to the question is to use information about power and most usage rate to determine the 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 beginning equation shows that the power would increase by the same factor by which the current is increased. The 2nd equation shows that the ability would decrease by the same factor that the resistance is increased.
a. The new toll would be two•D.b. The new cost would be (1/2)•D.
c. The new price would be two•D.
d. The new cost would be (1/2)•D.
e. The new price would be 2•D.
f. The new cost would still exist D.
g. The new price would still be D.
h. The new cost would exist (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 current at location D is ____ amperes. (Answer in terms of I.)
c. If the current at location A is I amperes, then the current at location L is ____ amperes. (Answer in terms of I.)
d. If the voltage of the battery is doubled, then the electric current at location A would be ____ (two times, 4 times, one-half, one-fourth, etc.) the original value.
due east. If the voltage of the battery is doubled, so the current at location B would exist ____ (two times, 4 times, one-half, i-fourth, etc.) the original value.
f. If the voltage of the battery is doubled, then the electric current at location D would exist ____ (two times, four times, one-half, i-fourth, etc.) the original value.
g. Suppose that the resistance of the light seedling located between points D and G is doubled. This would event in the current measured at location 1000 to ____ (increment, decrease, not be affected).
h. Suppose that the resistance of the lite bulb located between points D and Thou is doubled. This would result in the electrical potential divergence between points D and G to ____ (increase, decrease, not be affected).
i. Suppose that the resistance of the light bulb located between points D and Grand is doubled. This would result in the current measured at location A to ____ (increase, subtract, not be afflicted).
j. Suppose that the resistance of the light bulb located betwixt points D and G is doubled. This would effect in the electric current measured at location Due east to ____ (increment, decrease, not be affected).
g. 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 M to ____ (increment, decrease, not be affected).
Reply: See answers above.
a. - c. Location A is exterior or earlier the branching locations; information technology represents a location where the total excursion current is measured. This current will ultimately divide into three pathways, with each pathway carrying the same electric current (since each pathway has the same resistance). Location D is a branch location; one-third of the charge passes through this branch. Location B represents a location later on a point at which i-tertiary of the charge has already branched off to the light bulb between points D and One thousand. So at location B, there is ii-thirds of the current remaining. And location L is a location in the last co-operative; so one-tertiary of the charge passes through location L.
d. - f. The current at every branch location and in the full circuit is just equal to the voltage drop across the branch (or beyond the total excursion) divided by the resistance of the branch (or of the full circuit). Equally such, the current is directly proportional to the voltage. Then a doubling of the voltage will double the current at every location.
thou. The current at a co-operative location is only the voltage across the branch divided past the resistance of the branch. So the current at location G is inversely proportional to the resistance of the branch. Doubling the resistance will cause the current to exist decreased past a factor of 2.
h. The voltage drib across the start co-operative (or whatsoever branch) is simply equal to the voltage gained by the charge in passing through the battery. For a parallel excursion, the merely means of altering a branch voltage drop is to alter the battery voltage.
i. - k. Altering the resistance of a light seedling in a specific co-operative can alter the electric current in that branch and the current in the overall excursion. The electric current in a branch is inversely proportional to the resistance of the co-operative. Then increasing the resistance of a branch will decrease the current of that branch and the overall current in the circuit (as measured at location A). All the same, the electric 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 unmarried co-operative alters the current at that branch location, the other branch currents remain unaffected.
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Part E: Trouble-Solving and Circuit Analysis
65. If the current at a given signal in a circuit is ii.5 Amps, then how many electrons laissez passer that point on the circuit in a time period of 1 infinitesimal.
Answer: nine.375 ten 10xx electrons
The electric current (I) is the rate at which accuse passes a bespeak on the excursion in a unit of time. So I = Q/t. Rearranging this equation leads to Q = I•t. Recognizing that a electric current of 2.5 Amps is equivalent to two.5 Coulombs per second and that 1 infinitesimal is equivalent to 60 seconds can atomic number 82 to the corporeality of Coulombs moving laissez passer the point.
Q = I•t = (two.5 C/due south)•(60 s) = 150 Coulombs
The charge of a unmarried electron is equal to 1.6 x 10-19 C. So 150 Coulombs must be a lot of electrons. The actual number tin can be computed every bit 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 home?
Respond: 360 Ohms
The power dissipated in a circuit is given past the equation P = I• Δ V. Substituting in Δ V/R for the current can lead to an equation relating the resistance (R) to the voltage drop ( Δ 5) and the power (P).
Rearrangement of the equation and exchange of known values of power (xl Watts) and voltage (120 V) leads to the following solution.
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67. Decide the length of nichrome wire (resistivity value = 150 x ten-8 ohm•chiliad) required to produce a 1.00 mAmp electric current if a voltage of i.5 Volts is impressed across it. The diameter of the wire is 1/xvi-th of an inch. (GIVEN: ii.54 cm = one inch)
Answer: two.0 x 103 m
This is conspicuously an exercise in unit conversion (or at least unit sensation). The resistance (R) of a wire is related to the resistivity (Rho), the length (L) and the cross-sectional expanse (A) by the equation
This tin be rearranged to solve for length
The diameter is given; cross-sectional area is simply given by PI•Rii. The radius of the wire is one-half the diameter - i/32 inch.
Since the unit of measurement of length for the Rho value is meters, the cross-sectional area will be adamant in m2 before substitution into the equation. The fact that the conversion involves squared units makes this problem even more trickier. The conversion factors will have to be squared to accomplish the conversion successfully.
The last quantity which must exist adamant before calculating the wire length is the bodily resistance of the wire. Resistance is related to voltage ( Δ V) and current (I) by the equation R = Δ V / I. Standard units are Ohms, Volts and Amps. Here, the current is given in milliAmps (mAmps); exchange to Amps must exist performed earlier substitution.
Now R, A and Rho tin be substituted into the length equation to determine the length of the wire in meters.
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68. Determine the total 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 price of electricity is $0.13 / kW•hr. (Presume that a month lasts for 30 days.)
| (with info from labels) | (hours/twenty-four hour period) | (Watts) | Consumed | ($) |
| Hair Dryer (12 Amp) | | | | |
| Java Percolator (ix.0 Amp) | | | | |
| Light Bulb (100 Watt) | | | | |
| Cranium Fan (140 Watt) | | | | |
| Microwave Oven (8.3 Amps) | | | | |
Total | | |||
Answer: Run into tabular array above.
The power is either explicitly stated (equally in the case of the light bulb) or calculated using P = I• Δ V. In this case, 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 information technology - kiloWatt • hour. The calculation involves converting power in Watts to kiloWatts past dividing by chiliad and then multiplying by the time in hours/month and then multiplying past 30 days/month. The cost in dollars is simply the kiloWatt•hours of free energy used multiplied by 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 x ohms for every mile of wire, then what is the bore of the wire. (Given: 1609 m = ane mile. Resistivity of Cu = 1.7 x x-eight ohm•m )
Answer: 0.59 cm
Like Question #67, this is another exercise in unit conversion and unit of measurement sensation. The resistance (R) of a wire is related to the resistivity (Rho), the length (Fifty) and the cross-sectional area (A) by the equation
This can exist rearranged to solve for cross-sectional area
The area is related to the radius by the equation A = PI•R2. The plan volition involve determining the Area, and then the radius, then the diameter of the wire.
Showtime, note that the given information is: Rho = i.7 x ten-8 ohm•m; Fifty = 1 mi = 1609 m; R = 10 Ohm. By exchange, the Area can be determined:
A = 2.7353 x ten-6 mii
The area is in k2 units. Since the bore 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 be squared to accomplish the conversion successfully.
At present the surface area equation (A = PI•Rii) can be used to make up one's mind the radius.
R = 0.29507 cm
The radius is elementary twice the diameter. So d = 0.59014 cm.
lxx. Determine the resistance of a 1500 Watt electric grill connected to a 120-Volt outlet.
Answer: 9.6 Ohms
The power dissipated in a circuit is given by the equation P = I• Δ 5. Substituting in V/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 ability (1500 Watts) and voltage (120 V) leads to the following solution.
71. Four resistors - 2-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: Come across diagram below.
The diagram beneath depicts the series circuit using schematic symbols. Note that there is no branching, consistent with the notion of a series circuit.
For a series circuit, the overall resistance (RTot) is simply the sum of the private resistances. That is
RTot = 2 ½ + five ½ + 12 ½ + xv ½ = 34 ½
The series of three resistors supplies an overall, total or equivalent resistance of 34 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall electric current for the circuit and can be determined by finding the ratio of bombardment voltage to overall resistance (VTot/RTot).
The current through the bombardment 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 Fiveii = Itwo • R2 = (0.35294 Amps) • (5 Ohms) = one.76 Five
Δ Vthree = Ithree • R3 = (0.35294 Amps) • (12 Ohms) = 4.24 5
Δ 54 = I4 • Riv = (0.35294 Amps) • (15 Ohms) = 5.29 V
72. Four resistors - 2-Ohms, 5-Ohms, 12-Ohms and xv-Ohms - are placed in parallel with a 12-Volt battery. Determine the current at and voltage drop across each resistor.
Answer: See diagram beneath.
The diagram beneath depicts the parallel circuit using schematic symbols. Note that there is a branching, consistent with the notion of a parallel excursion.
For a parallel circuit, the reciprocal of overall resistance (ane / RTot) is only the sum of the reciprocals of individual resistances. That is
1 / RTot = 1 / 2 ½ + ane / five ½ + ane / 12 ½ + 1 / 15 ½ = 0.850 / ½
RTot = 1.17647 ½
The series of three resistors supplies an overall, full or equivalent resistance of ~i.18 Ohms. This total resistance value can be used to determine the total current through the excursion.
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 drop across each resistor is the same every bit the voltage gained by the charge in the battery (12 Volts in this case). The branch current calculations are shown below.
I 2 = Δ Vtwo / Rtwo = (12 Volts) / (5 Ohms) = 2.xl Amp
I three = Δ 5three / R3 = (12 Volts) / (12 Ohms) = i.00 Amp
I 4 = Δ V4 / R4 = (12 Volts) / (15 Ohms) = 0.80 Amp
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