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60. A resistor with a resistance of R is connected to a battery 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 onethird the resistance is used?
g. ... onefifth the voltage and onefourth the resistance is used?
Answer: See answers below. This question tests your understanding of the currentvoltageresistance relationship. The current is directly proportional to the voltage and inversely proportional to the resistance. Any alteration in the voltage will result in the same alteration of the current. So doubling or tripling the voltage will cause the current to be doubled or tripled. On the other hand, any alteration in the resistance will result in the opposite or inverse alteration of the current. So doubling or tripling the resistance will cause the current to be onehalf or onethird the original value. a. The new current will be 0.5 • I. 

Ohm's Law 
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61. A wire of length L and crosssectional 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 crosssectional area of the wire is doubled?
c. ... the length of the wire is doubled and the crosssectional area of the wire is doubled?
d. ... the length of the wire is tripled and the crosssectional area of the wire is doubled?
e. ... the length of the wire is halved and the crosssectional area of the wire is tripled?
f. ... the length of the wire is tripled and the crosssectional 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?
Answer: See answers below. This question tests your understanding of the variables which effect 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 material, L is length of wire, and A is crosssectional area of the wire). The resistance is directly proportional to the resistivity, directly proportional to the wire length, and inversely proportional to the crosssectional area. Any alteration in the resistivity or the length will result in the same alteration in the resistance of the wire. And any alteration in the crosssectional area of the wire will result in the opposite or inverse alteration in the resistance of the wire. a. The new resistance will be 2•R. 

Resistance 
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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 connected to a 120Volt outlet. What would be the new power rating (in terms of P) if ...
a. ... the current is doubled (and the same 120Volt outlet is used)?b. ... the current is halved (and the same 120Volt outlet is used)?
c. ... the resistance is doubled (and the same 120Volt outlet is used)?
d. ... the resistance is halved (and the same 120Volt outlet is used)?
e. ... the current is tripled (and the same 120Volt outlet is used)?
f. ... the resistance is tripled (and the same 120Volt outlet is used)?
g. ... the same appliance is powered by a 12Volt supply?
h. ... the same appliance is powered by a 240Volt supply?
Answer: See answers below. This question tests your understanding of the mathematical relationship between power, current, voltage and resistance. There are three equations of importance:
One must be careful in using the last equation since an alteration in current will also alter the resistance whenever the voltage is held constant. Thus, the first two equations are of greater importance since they represent equations with one independent variable and the other variable held constant. a. 2 • P (doubling the current will double the power) 

Power Revisited 
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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 appliance at 120Volts 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 onehalf 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 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 below. Like the previous question, this question tests your understanding of the mathematical relationship between power, current, voltage and resistance. But this question also tests your understanding between power, time, energy and electricity costs. An electric bill is based upon energy consumption. The energy consumed is measured in terms of kiloWatts•hour and is determined by multiplying the power by the time. Thus, an increase in either time or power will lead to an increase in the electricity costs by the same factor. So the key to the question is to use information about power and about 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 first equation shows that the power would increase by the same factor by which the current is increased. The second equation shows that the power would decrease by the same factor that the resistance is increased. a. The new cost would be 2•D. 

Power: Putting Charges to Work  Power Revisited 
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64. Consider the diagram at the right of a parallel circuit. Each light bulb has an identical resistance of R and the battery voltage is V. Use the labeled points on the diagram to answer the following questions.
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, 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 current at location A would be ____ (two times, four times, onehalf, onefourth, etc.) the original value.
e. If the voltage of the battery is doubled, then the current at location B would be ____ (two times, four times, onehalf, onefourth, 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, onehalf, onefourth, etc.) the original value.
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 G to ____ (increase, decrease, not be affected).
h. Suppose that the resistance of the light bulb 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 be affected).
i. 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 A to ____ (increase, decrease, not be affected).
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 E to ____ (increase, decrease, not be affected).
k. 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 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 total circuit current is measured. This current will ultimately divide into three pathways, with each pathway carrying the same current (since each pathway has the same resistance). Location D is a branch location; onethird of the charge passes through this branch. Location B represents a location after a point at which onethird of the charge has already branched off to the light bulb between points D and G. So at location B, there is twothirds of the current remaining. And location L is a location in the last branch; so onethird of the charge passes through location L. d.  f. The current at every branch location and in the total circuit is simply equal to the voltage drop across the branch (or across the total circuit) divided by the resistance of the branch (or of the total circuit). As such, the current is directly proportional to the voltage. So a doubling of the voltage will 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 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 branch) is simply equal to the voltage gained by the charge in passing through the battery. For a parallel circuit, the only means of altering a branch voltage drop is to alter the battery voltage. i.  k. Altering the resistance of a light bulb in a specific branch can alter the current in that branch and the current in the overall circuit. The current in a branch is inversely proportional to the resistance of the branch. So 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). However, 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. 

Parallel Circuits 
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65. If the current at a given point in a circuit is 2.5 Amps, then how many electrons pass that point on the circuit in a time period of 1 minute.
Answer: 9.375 x 10^{20} electrons The current (I) is the rate at which charge passes a point 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 minute is equivalent to 60 seconds can lead to the amount of Coulombs moving pass the point. Q = I•t = (2.5 C/s)•(60 s) = 150 Coulombs The charge of a single electron is equal to 1.6 x 10^{19} C. So 150 Coulombs must be a lot of electrons. The actual number can be computed as shown: 

Electric Current 
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66. What is the resistance (in ohms) of a typical 40Watt light bulb plugged into a 120Volt outlet in your home?
Answer: 360 Ohms The power dissipated in a circuit is given by the equation P = I•V. Substituting in V/R for the current can lead to an equation relating the resistance (R) to the voltage (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. 

Power Revisited 
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67. Determine the length of nichrome wire (resistivity value = 150 x 10^{8} ohm•m) required to produce a 1.00 mAmp current if a voltage of 1.5 Volts is impressed across it. The diameter of the wire is 1/16th of an inch. (GIVEN: 2.54 cm = 1 inch)
Answer: 1.98 x 10^{3} m This is clearly an exercise 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 crosssectional area (A) by the equation This can be rearranged to solve for length The diameter is given; crosssectional area is simply given by PI•R^{2}. The radius of the wire is onehalf the diameter  1/32 inch. Since the unit of length for the Rho value is meters, the crosssectional area will be determined in m^{2} 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 final quantity which must be determined before computing the wire length is the actual resistance of the wire. Resistance is related to voltage (V) and current (I) by the equation R = V / I. Standard units are Ohms, Volts and Amps. Here, the current is given in milliAmps (mAmps); substitution to Amps must be performed before substitution. Now R, A and Rho can be substituted into the length equation to determine the length of the wire in meters. 

Resistance  Ohm's Law 
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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 120Volt 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: See table above. The power is either explicitly stated (as in the case of the light bulb) or calculated using P = I•V. In this case, the voltage is 120 Volts. The energy consumed is the Power•time. It is useful to express this quantity in the same units for which one is charged for it  kiloWatt • hour. The calculation 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 by the cost of $0.13/kW•hr. 

Power: Putting Charges to Work  Power Revisited 
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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, then what is the diameter of the wire. (Given: 1609 m = 1 mile. Resistivity of Cu = 1.7 x 10^{8} ohm•m )
Answer: 0.590 cm Like Question #67, this is another exercise in unit conversion and unit awareness. The resistance (R) of a wire is related to the resistivity (Rho), the length (L) and the crosssectional area (A) by the equation This can be rearranged to solve for crosssectional area The area is related to the radius by the equation A = PI•R^{2}. The plan will involve determining the Area, then the radius, then the diameter of the wire. First, note that the given information is: Rho = 1.7 x 10^{8} ohm•m; L = 1 mi = 1609 m; R = 10 Ohm. By substitution, the Area can be determined: A = 2.7353 x 10^{6} m^{2} The area is in m^{2} 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 will have to be squared to accomplish the conversion successfully. Now the area equation (A = PI•R^{2}) can be used to determine the radius. R = 0.29507 cm The radius is simple twice the diameter. So d = 0.59014 cm. 

Resistance  Ohm's Law 
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70. Determine the resistance of a 1500 Watt electric grill connected to a 120Volt outlet.
Answer: 9.6 Ohms The power dissipated in a circuit is given by the equation P = I•V. Substituting in V/R for the current can lead to an equation relating the resistance (R) to the voltage (V) and the power (P). Rearrangement of the equation and substitution of known values of power (1500 Watts) and voltage (120 V) leads to the following solution. 

Power Revisited 
71. Four resistors  2Ohms, 5Ohms, 12Ohms and 15Ohms  are placed in series with a 12Volt battery. Determine the current at and voltage drop across each resistor.
Answer: See diagram below. The diagram below 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 (R_{Tot}) is simply the sum of the individual resistances. That is R_{Tot} = 2 ½ + 5 ½ + 12 ½ + 15 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 current for the circuit and can be determined by finding the ratio of battery voltage to overall resistance (V_{Tot}/R_{Tot}). 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 below. V_{2} = I_{2} • R_{2} = (0.35294 Amps) • (5 Ohms) = 1.76 V V_{3} = I_{3} • R_{3} = (0.35294 Amps) • (12 Ohms) = 4.24 V V_{4} = I_{4} • R_{4} = (0.35294 Amps) • (15 Ohms) = 5.29 V 

Series Circuits 
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72. Four resistors  2Ohms, 5Ohms, 12Ohms and 15Ohms  are placed in parallel with a 12Volt battery. Determine the current at and voltage drop across each resistor.
Answer: See diagram below. The diagram below depicts the parallel circuit using schematic symbols. Note that there is a branching, consistent with the notion of a parallel circuit. For a parallel circuit, the reciprocal of overall resistance (1 / R_{Tot}) is simply the sum of the reciprocals of individual resistances. That is 1 / R_{Tot} = 1 / 2 ½ + 1 / 5 ½ + 1 / 12 ½ + 1 / 15 ½ = 0.850 / ½ R_{Tot} = 1.17647 ½ The series of three resistors supplies an overall, total or equivalent resistance of ~1.18 Ohms. This total resistance value can be 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 by the resistance of each resistor. For series circuits, the voltage drop across each resistor is the same as the voltage gained by the charge in the battery (12 Volts in this case). The branch current calculations are shown below. I _{2} = V_{2} / R_{2} = (12 Volts) / (5 Ohms) = 2.40 Amp I _{3} = V_{3} / R_{3} = (12 Volts) / (12 Ohms) = 1.00 Amp I _{4} = V_{4} / R_{4} = (12 Volts) / (15 Ohms) = 0.80 Amp 

Parallel Circuits 
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