a. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point B in 2 seconds.
b. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point B in 1 second.
c. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point B in 4 seconds.
d. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point B in 4 seconds.
e. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point C in 4 seconds.
f. Suppose that the resistance of the light bulb located between points A and B is increased. This would cause the current through the other light bulb to ____ (increase, decrease, remain the same).
56. Consider the diagram at the right of a series circuit. Each light bulb in the circuit has an identical resistance. Use the labeled points on the diagram to answer the following questions. Each question may have one, less than one, or more than one answer.
a. The electric potential at point A is the same as the electric potential at point(s) B. Include all that apply, if any apply.
b. The electric potential at point C is the same as the electric potential at point(s) D and E. Include all that apply, if any apply.
c. The electric potential at point F is the same as the electric potential at point(s) G. Include all that apply, if any apply.
d. The electric potential at point I is the same as the electric potential at point(s) H. Include all that apply, if any apply.
e. The electric potential difference between points A and B is the same as the electric potential difference between points C and D (and C and E; and D and E; and F and G; and H and I). Include all that apply, if any apply.
f. The electric potential difference between points A and C is the same as the electric potential difference between points E and F (and C and F; and D and F; and E and G; and C and G; and D and G; and G and H; and F and H; and G and I; and F and I). Include all that apply, if any apply.
g. The electric potential difference between points A and F is the same as the electric potential difference between points E and H (and between any other two sets of points that are on the opposite side of two adjacent bulbs ... such as points C and H or points D and H or points E and I, etc.). Include all that apply, if any apply.
h. The electric potential difference between points D and H is the same as the electric potential difference between points A and F (and between any other two sets of points that are on the opposite side of two adjacent bulbs ... such as points C and H or points D and I or points E and I, etc.). Include all that apply, if any apply.
i. The current at point A is the same as the current at point(s) C (and at every other point on the circuit). Include all that apply, if any apply.
j. The current at point E is the same as the current at point(s) F (and at every other point on the circuit). Include all that apply, if any apply.
k. The current at point G is the same as the current at point(s) G (and at every other point on the circuit). Include all that apply, if any apply.
Answer: See answers above.
In an electric circuit, the electric potential for a moving charge is gained in the battery and lost in a light bulb (or some resistor found in the external circuit). So the electric potential of a charge is the same for any two points which are not separated by a battery or by a light bulb. (a through d)
In this circuit, the light bulbs have the same resistance; thus, each light bulb causes the same drop in potential (electric potential difference). So the electric potential difference will be the same between any two points which are distanced by a light bulb, or by two light bulbs. (e through h)
Finally, in a series circuit the current is the same at every point along the circuit. Since charge is conserved and since there is no location in a circuit where the charge is accumulating, there must be the same charge flow rate at all locations. (i through k)

[ #52  #53  #54  #55  #56  #57  #58  #59 ]
57. Consider the diagram at the right of a parallel circuit. Each light bulb in the circuit has an identical resistance. Use the labeled points on the diagram to answer the following questions. Each question may have one, less than one, or more than one answer.
a. The electric potential at point A is the same as the electric potential at point(s) B (and at any other point before the light bulbs ... such as at points C, D, E and F). Include all that apply, if any apply.
b. The electric potential at point D is the same as the electric potential at point(s) B (and at any other point before the light bulbs ... such as at points A, C, E and F). Include all that apply, if any apply.
c. The electric potential at point J is the same as the electric potential at point(s) K (and at any other point after the light bulbs ... such as at points G, H, I, and L. Include all that apply, if any apply.
d. The electric potential difference between points A and J is the same as the electric potential difference between points B and K (and any other combination of two points located on opposite sides of the light bulbs ... such as points A and K, or points A and L, or points F and G, or points D and G, etc., etc.). Include all that apply, if any apply.
e. The electric potential difference between points D and G is the same as the electric potential difference between points A and J (and any other combination of two points located on opposite sides of the light bulbs ... such as points A and K, or points A and L, or points F and G, or points E and H, etc., etc.). Include all that apply, if any apply.
f. The current at point A is the same as the current at point(s) J. Include all that apply, if any apply.
g. The current at point B is the same as the current at point(s) K. Include all that apply, if any apply.
h. The current at point C is the same as the current at point(s) L (and points F and I and E and H and D and G). Include all that apply, if any apply.
i. The current at point D is the same as the current at point(s) G (and at points F and I and C and E and H). Include all that apply, if any apply.
j. If the light bulb located between points D and G were to be replaced by a bulb of greater resistance, then the current at point(s) A, D, G, and J would be decreased. Include all that apply, if any apply.
k. If the light bulb located between points D and G were to be replaced by a bulb of greater resistance, then the electric potential difference between points  and  would be increased. Include all that apply, if any apply. (None apply.)
l. If the light bulb located between points D and G were to go out, then the current would decrease at point(s) A, D, G and J. Include all that apply, if any apply.
Answer: See answers above.
In an electric circuit, the electric potential for a moving charge is gained in the battery and lost in a light bulb (or some resistor found in the external circuit). So the electric potential of a charge is the same for any two points which are not separated by a battery or by a light bulb. Even if the circuit is a parallel circuit, any point between the positive terminal of a battery and light bulb will have the same electric potential; and any point located between the  terminal of the battery and a location after passage through the resistor of a branch have the same electric potential. (a through e)
In this circuit, the light bulbs have the same resistance. Thus, when charge reaches the branching location an equal amount of charge will chose the middle branch as does the left branch and the right branch. So with the same resistance, the current in every branch is the same. And the current prior to the branching location and after the branches come together is the same. Finally, locations B and K are at locations where current for two of the three branches will be passing; these points will have the same current. (e through i)
In j, removing the bulb in the first branch will not affect the current in the other branches. Such a modification will only reduce the overall circuit current. Less branches would result in more overall resistance and less overall current. Yet the current through the second branch is still the voltage drop across the second branch (which is the battery voltage) divided by the resistance of the second branch. Since removing the bulb in the first branch does not alter either quantity, the current in the middle branch is not altered.
In k, the electric potential across a branch is simply equal to the voltage of the battery. Removing a light bulb will not alter the voltage of the battery. Thus, there is no effect.
In l, if the light bulb in the first branch burns out, the same effect will occur as occurred in part j.

[ #52  #53  #54  #55  #56  #57  #58  #59 ]
58. Consider the diagram below of a series circuit. For each resistor, use arrows to indicate the two locations where one would have to tap with the leads of a voltmeter in order to measure the voltage drop across the individual resistor. Finally, indicate the ammeter readings and the voltage readings.
Answer: See diagram above.
The total resistance (or equivalent resistance) can first be determined using the equation for series circuits.
R_{Tot} = R_{1} + R_{2} + R_{3} = 5 Ohms + 10 Ohms + 15 Ohms = 30 Ohms
Once known, the R_{Tot} value can be used with the battery voltage (ΔV_{Tot}) to determine the total current in the circuit.
I_{Tot} = (ΔV_{Tot}) / (R_{Tot}) = (120 V) / (30 Ohms) = 4 Amps
For a series circuit, the current through each resistor is the same as the total circuit current. Thus, I_{1} = I_{2} = I_{3} = 4 Amps.
The voltage drop across a resistor can be determined with a voltmeter by tapping with the leads on the metal wires on the opposite sides of the resistor. By so doing, the voltmeter determines the difference in voltage (i.e., voltage drop or electric potential difference) between the two locations where the leads were tapped. In this circuit, the expected voltage drops (ΔV_{1}, ΔV_{2}, and ΔV_{3 }) can be computed by determining the IR product for each resistor. This is shown below.
ΔV_{1} = I_{1} • R_{1} = (4 Amps) • (5 Ohms) = 20 Volts
ΔV_{2} = I_{2} • R_{2} = (4 Amps) • (10 Ohms) = 40 Volts
ΔV_{3} = I_{3} • R_{3} = (4 Amps) • (15 Ohms) = 60 Volts

[ #52  #53  #54  #55  #56  #57  #58  #59 ]
59. Consider the diagram below of a parallel circuit. For each resistor, use arrows to indicate the two locations where one would have to tap with the leads of a voltmeter in order to measure the voltage drop across the individual resistor. Finally, indicate the ammeter readings and the voltage readings.
Answer: See diagram above.
The total resistance (or equivalent resistance) can first be determined using the equation for series circuits.
1 / R_{Tot} = 1 / R_{1} + 1 / R_{2} + 1 / R_{3} = 1 / (5 Ohms + 1 / (10 Ohms) + 1 / (15 Ohms)
R_{Tot} = 2.727 Ohms
Once known, the R_{Tot} value can be used with the battery voltage (ΔV_{Tot}) to determine the total current in the circuit.
I_{Tot} = (ΔV_{Tot}) / (R_{Tot}) = (120 V) / (2.727 Ohms) = 44.0 Amps
The voltage drop across a resistor can be determined with a voltmeter by tapping with the leads on the metal wires on the opposite sides of the resistor. By so doing, the voltmeter determines the difference in voltage (i.e., voltage drop or electric potential difference) between the two locations where the leads were tapped. For a series circuit, the expected voltage drop across each resistor (ΔV_{1}, ΔV_{2}, and ΔV_{3 }) is the same as the total voltage drop. Thus, ΔV_{1} = ΔV_{2} = ΔV_{3} = 120 Volts.
In this circuit, the branch currents can be computed by using the ΔV = I•R equation for each resistor. This is shown below.
R_{1} = ΔV_{1} / R_{1} = (120 Volts) / (5 Ohms) = 24 Amps
I_{2} = ΔV_{2} / R_{2} = (120 Volts) / (10 Ohms) = 12 Amps
I_{3} = ΔV_{3} / R_{3} = (120 Volts) / (15 Ohms) = 8 Amps

[ #52  #53  #54  #55  #56  #57  #58  #59 ]