#1A. Consider the diagram 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, one-half, one-fourth, 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, 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, one-fourth, 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).
#2A. 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.
#3A What is the resistance (in ohms) of a typical 40-Watt light bulb plugged into a 120-Volt outlet in your home?
#4A 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 cost of electricity is $0.13 / kW•hr. (Assume that a month lasts for 30 days.)
Appliance
(with info from labels)
Time
(hours/day)
Power
(Watts)
Energy
Consumed
Cost
($)
Hair Dryer
0.10
(12 Amp)
Coffee Percolator
0.10
(9.0 Amp)
Light Bulb
8
(100 Watt)
Attic Fan
10
(140 Watt)
Microwave Oven
0.25
(8.3 Amps)
#5A. Determine the resistance of a 1500 Watt electric grill connected to a 120-Volt outlet.
#6A Four resistors - 2-Ohms, 5-Ohms, 12-Ohms and 15-Ohms - are placed in series with a 12-Volt battery. Determine the current at and
voltage drop across each resistor.
#7A. Four resistors - 2-Ohms, 5-Ohms, 12-Ohms and 15-Ohms - are placed in parallel with a 12-Volt battery. Determine the current at and
voltage drop across each resistor.
1. A combination circuit is shown in the diagram at the right.
Use the diagram to answer the following questions.
a. The current at location A is _____ (greater than, equal to, less
than) the current at location B.
b. The current at location B is _____ (greater than, equal to, less
than) the current at location E.
c. The current at location G is _____ (greater than, equal to, less
than) the current at location F.
d. The current at location E is _____ (greater than, equal to, less
than) the current at location G.
e. The current at location B is _____ (greater than, equal to, less
than) the current at location F.
f. The current at location A is _____ (greater than, equal to, less than) the current at location L.
f. The current at location H is _____ (greater than, equal to, less than) the current at location I.
2. Consider the combination circuit in the diagram at the right.
Use the diagram to answer the following questions. (Assume
that the voltage drops in the wires themselves in negligibly
small.)
a. The electric potential difference (voltage drop) between
points B and C is _____ (greater than, equal to, less than) the
electric potential difference (voltage drop) between points J and
K.
b. The electric potential difference (voltage drop) between
points B and K is _____ (greater than, equal to, less than) the
electric potential difference (voltage drop) between points D
and I.
c. The electric potential difference (voltage drop) between
points E and F is _____ (greater than, equal to, less than) the
electric potential difference (voltage drop) between points G
and H.
d. The electric potential difference (voltage drop) between points E and F is _____ (greater than, equal to, less than) the electric potential
difference (voltage drop) between points D and I.
e. The electric potential difference (voltage drop) between points J and K is _____ (greater than, equal to, less than) the electric potential
difference (voltage drop) between points D and I.
f. The electric potential difference between points L and A is _____ (greater than, equal to, less than) the electric potential difference (voltage
drop) between points B and K.
3. Use the concept of equivalent resistance to determine the unknown resistance of the identified resistor that would make the circuits
equivalent.
4. Analyze the following circuit and determine the values of the total resistance, total current, and the current at and voltage drops across
each individual resistor.
1. Answer: The current outside the branches of a combination circuit is everywhere the same. The current inside of the
branches is always less than that outside of the branches. When comparing the current of two parallel-connected resistors, the
resistor with the least resistance will have the greatest current. The current within a single branch will be the
sameabove and below the resistor.
a. The current at location A is equal to the current at
location B.
b. The current at location B is greater than the current at
location E.
c. The current at location G is less than the current at
location F.
d. The current at location E is greater than the current at
location G.
e. The current at location B is greater than the current at
location F.
f. The current at location A is equal to the current at
location L.
g. The current at location H is less than the current at location I.
2. Answer. The voltage drop across a resistor is dependent upon the current in the resistor and the resistance of the
resistor. In situations in which the current is the same for both resistors (such as for series-connected resistors), the resistor
with the greatest resistance will have the greatest voltage
drop.
a. The electric potential difference (voltage drop)
between points B and C is greater than the electric
potential difference (voltage drop) between points J and
K.
b. The electric potential difference (voltage drop)
between points B and K is greater than the electric
potential difference (voltage drop) between points D and
I.
c. The electric potential difference (voltage drop)
between points E and F is equal to the electric potential
difference (voltage drop) between points G and H.
d. The electric potential difference (voltage drop)
between points E and F is equal to the electric potential
difference (voltage drop) between points D and I.
e. The electric potential difference (voltage drop) between points J and K is greater than the electric potential difference
(voltage drop) between points D and I.
f. The electric potential difference between points L and A is equal to the electric potential difference (voltage drop) between
points B and K.
For parallel-connected resistors:
1/Req = 1/R1 + 1/R2 = 1 / (6
) + 1 / (6
Req = 3
For series-connected resistors:
Req = R1 + R2 + R3 = 3
Req = 11
+3
+5
) = 2 / (6
)
For parallel-connected resistors:
1/Req = 1/R1 + 1/R2 = 1 / (12
) + 1 / (6
) = 3 / (12
)
) = 3 / (12
)
Req = 4
For series-connected resistors:
Req = R1 + R2 + R3 = 9
+4
+5
Req = 18
For parallel-connected resistors:
1/Req = 1/R1 + 1/R2 = 1 / (12
) + 1 / (6
Req = 4
For series-connected resistors:
Req = R1 + R2 + R3 + R4 = 3
Req = 18
+6
4
+5
#4
The first step is to simplify the circuit by replacing the two parallel resistors with a single resistor with an equivalent
resistance. The equivalent resistance of a 4 and 6 resistor placed in parallel can be determined using the usual formula
for equivalent resistance of parallel branches:
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 ...
1 / Req = 1 / (3
) + 1 / (6
1 / Req = 0.500
-1
Req = 1 / (0.500
-1)
)
Req = 2.00
Based on this calculation, it can be said that the two branch resistors (R 2 and R3) can be replaced by a single resistor with a
resistance of 2 . This 2 resistor is in series with R1 and R4. Thus, the total resistance is
Rtot = R1 + 2
+ R4 = 2
+2
+4
Rtot = 8
Now the Ohm's law equation ( V = I • R) can be used to determine the total current in the circuit. In doing so, the total
resistance and the total voltage (or battery voltage) will have to be used.
Itot =
Vtot / Rtot = (24 V) / (8
)
Itot = 3.0 Amp
The 3.0 Amp current calculation represents the current at the battery location. Yet, resistors R 1 and R4 are in series and the
current in series-connected resistors is everywhere the same. Thus,
Itot = I1 = I4 = 3.0 Amp
For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I 2 +
I3 must equal 3.0 Amp. There are an infinite possibilities of I2 and I3 values which satisfy this equation. Determining the
amount of current in either branch will demand that we use the Ohm's law equation. But to use it, the voltage drop across the
branches must first be known. To determine the voltage drop across the parallel branches, the voltage drop across the two
series-connected resistors (R1 and R4) must first be determined. The Ohm's law equation ( V = I • R) can be used to determine
the voltage drop across each resistor. These calculations are shown below.
V1 = I1 • R1 = (3.0 Amp) • (2
) = 6.0 V
V4 = I4 • R4 = (3.0 Amp) • (4
) = 12 V
This circuit is powered by a 24-volt source. Thus, the cumulative voltage drop of a charge traversing a loop about the circuit is
24 volts. There will be a 18.0 V drop (6.0 V + 12.0 V) resulting from passage through the two series-connected resistors
(R1 and R4). The voltage drop across the branches must be 6.0 volts to make up the difference between the 24 volt total and the
18.0 volt drop across R1 and R4. Knowing the voltage drop across the parallel-connected resistors (R1 and R4) allows one to use
the Ohm's law equation ( V = I • R) to determine the current in the two branches.
I2 =
V2 / R2 = (6.0 V) / (3
) = 2.0 A
I3 =
V3 / R3 = (6.0 V) / (6
) = 1.0 A
#1A Answer
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.
# 2A 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:
# electrons = 150 C • (1 electron / 1.6 x 10-19 C) = 9.375 x 1020 electrons
#3A Answer: 360 Ohms
The power dissipated in a circuit is given by the equation P = I•
relating the resistance (R) to the voltage drop (
V. Substituting in
V/R for the current can lead to an equation
V) and the power (P).
P = I•
V=(
V/R)•
V=
V2 / R
Rearrangement of the equation and substitution of known values of power (40 Watts) and voltage (120 V) leads to the following solution.
R=
V2 / P = (120 V)2 / (40 Watts) = 360 Ohm
# 4A 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 cost of electricity is $0.13 / kW•hr. (Assume that a month lasts for 30 days.)
Appliance
Time
Power
Energy
Cost
(with info from labels)
(hours/day)
(Watts)
Consumed
($)
0.10
1440 W
4.32 kW•h
$0.56
0.10
1080 W
3.24 kW•h
$0.42
8
100 W
24 kW•h
$3.12
10
140 W
42 kW•h
$5.46
0.25
996 W
7.47 kW•h
$0.97
Hair Dryer
(12 Amp)
Coffee Percolator
(9.0 Amp)
Light Bulb
(100 Watt)
Attic Fan
(140 Watt)
Microwave Oven
(8.3 Amps)
Total
$10.53
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.
# 5A. 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•
resistance (R) to the voltage drop (
V. Substituting in V/R for the current can lead to an equation relating the
V) and the power (P).
P = I•
V=(
V/R)•
V=
V2 / R
Rearrangement of the equation and substitution of known values of power (1500 Watts) and voltage (120 V) leads to the following solution.
R=
V2 / P = (120 V)2 / (1500 Watts) = 9.6 Ohms
#6A. Four resistors - 2-Ohms, 5-Ohms, 12-Ohms and 15-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 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 individual resistances. That is
RTot = R1 + R2 + R3 + R4
RTot = 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 (VTot/RTot).
ITot =
VTot/RTot = (12 Volt) / (34 Ohm) = 0.35294 Amps
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.
V1 = I1 • R1 = (0.35294 Amps) • (2 Ohms) = 0.71 V
V2 = I2 • R2 = (0.35294 Amps) • (5 Ohms) = 1.76 V
V3 = I3 • R3 = (0.35294 Amps) • (12 Ohms) = 4.24 V
V4 = I4 • R4 = (0.35294 Amps) • (15 Ohms) = 5.29 V
#7A. Four resistors - 2-Ohms, 5-Ohms, 12-Ohms and 15-Ohms - are placed in parallel with a 12-Volt 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 / RTot) is simply the sum of the reciprocals of individual resistances. That is
1 / RTot = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4
1 / RTot = 1 / 2 ½ + 1 / 5 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 total resistance value can be used to
determine the total current through the circuit.
ITot =
VTot/RTot = (12 Volt) / (1.17647 Ohm) = 10.2 Amps
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.
I1 =
V1 / R1 = (12 Volts) / (2 Ohms) = 6.00 Amp
I2=
V2 / R2 = (12 Volts) / (5 Ohms) = 2.40 Amp
I3=
V3 / R3 = (12 Volts) / (12 Ohms) = 1.00 Amp
I4=
V4 / R4 = (12 Volts) / (15 Ohms) = 0.80 Amp