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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 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 one-half or one-third the original value.

a. The new current will be 0.5 • I.

b. The new current will be 2 • I.

c. The new current will be 1.5 • I.

d. The new current will be 4 • I.

e. The new current will be 0.25 • I.

f. The new current will be 15 • I.

g. The new current will be 0.8 • I.

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 cross-sectional area of the wire). The resistance is directly proportional to the resistivity, directly proportional to the wire length, and inversely proportional to the cross-sectional 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 cross-sectional 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.

b. The new resistance will be 0.5•R.

c. The new resistance will still be R.

d. The new resistance will be 1.5•R.

e. The new resistance will be (1/6)•R.

f. The new resistance will be 6•R.

g. 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 (3/4)•R. (Doubling the diameter will make the area four times the size since area is directly proportional to the square of the diameter.)

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)

b. (1/2) • P (halving the current will double the power)

c. (1/2) • P (doubling the resistance will half the power)

d. 2 • P (halving the resistance will double the power)

e. 3 • P (tripling the current will triple the power)

f. (1/3) • P (tripling the resistance will make the power one-third of the original value)

g. (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 power four times the original value; observe the square on voltage)

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.

b. The new cost would be (1/2)•D.

c. The new cost would be 2•D.

d. The new cost would be (1/2)•D.

e. The new cost would be 2•D.

f. The new cost would still be D.

g. The new cost would still be D.

h. The new cost would be (1/2)•D.

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; one-third of the charge passes through this branch. Location B represents a location after a point at which one-third of the charge has already branched off to the light bulb between points D and G. So at location B, there is two-thirds of the 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 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.

Answer: 9.375 x 10²⁰ 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:

Answer: 360 Ohms

The power dissipated in a circuit is given by the equation P = I•ΔV. Substituting in ΔV/R for the current can lead to an equation relating the resistance (R) to the voltage drop (ΔV) 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.

Answer: 2.0 x 10³ 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 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 one-half the diameter - 1/32 inch.

Since the unit 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 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 (ΔV) and current (I) by the equation R = ΔV / 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.

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•ΔV. 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.

Answer: 0.59 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 cross-sectional 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•R². 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:

The area is in m² 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•R²) can be used to determine the radius.

The radius is simple twice the diameter. So d = 0.59014 cm.

Answer: 9.6 Ohms

The power dissipated in a circuit is given by the equation P = I•ΔV. Substituting in V/R for the current can lead to an equation relating the resistance (R) to the voltage drop (ΔV) 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.

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

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.

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

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.