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52. Consider the diagram at the right to answer the following questions.

a. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point B in 2 seconds.

a. less than 4

b. 4

c. more than 4

d. impossible to make such a prediction without knowledge of the resistances.

b. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point B in 1 second.

a. less than 4

b. 4

c. more than 4

d. impossible to make such a prediction without knowledge of the resistances.

c. If 4 Coulombs of charge flow past point A in 2 seconds, then ___ Coulombs of charge flow past point B in 4 seconds.

a. less than 4

b. 4

c. more than 4

d. impossible to make such a prediction without knowledge of the resistances.

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).

This question tests your understanding of current as the rate at which charge (expressed here in Coulombs) flows past a point on a circuit. Current is found as the charge/time ratio. For a series circuit such as this one, the current is everywhere the same. a. b. c. d. e. f. |

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53. The diagram at the right shows three identical light bulbs wired in series. Several points along the circuit are labeled with letters. Compare the electric potential and the electric potential energy of the various points. For each comparison, use a greater than (>), less than (<), or approximately equal to (=) symbol.

Electric Potential

ComparisonPotential Energy

ComparisonV _{A}=V_{B}PE _{A}=PE_{B}V _{B}>V_{C}PE _{B}>PE_{C}V _{C}=V_{D}PE _{C}=PE_{D}V _{D}>V_{E}PE _{D}>PE_{E}V _{E}=V_{F}PE _{E}=PE_{F}V _{F}>V_{G}PE _{F}>PE_{G}V _{G}=V_{H}PE _{G}=PE_{H}V _{H}<V_{A}PE _{H}<PE_{A}

Point A corresponds
to the positive terminal of the battery. At point A, charge possesses
the greatest amount of electric potential energy. It has a high voltage
at this location. In the process of moving through the circuit to the -
terminal, the charge will have its electric potential energy converted
to light energy (and heat energy) in the light bulbs. It will |

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54. Use proper schematic symbols to construct a diagram of a circuit powered by a 6-Volt battery that consists of two 3-ohm resistors connected in series. Place ammeters in series at a location such that the current through each resistor can be measured and in a location such that the overall current in the circuit can be measured. On the schematic diagram, use an unbroken arrow to indicate the direction of conventional current. Finally, indicate the ammeter readings on the diagram.

Since the two resistors are in series, the total or equivalent resistance is simply the sum of the individual resistances. The total resistance is 6 ohms. The total current in the circuit can be found by the ratio of battery voltage to total resistance: _{TOT}
= V_{TOT} / R_{TOT} = (6 V) /
(6 ohms) = 1 AmpSince it is a series circuit, the current through the battery is the same as the current through each of the resistors. |

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55. Use proper schematic symbols to construct a diagram of a circuit powered by a 6-Volt battery that consists of two 3-ohm resistors connected in parallel. Place an ammeter in series with each of the individual resistors in a manner that the current through each resistor can be measured. Place a third ammeter in a location such that the overall current in the circuit can be measured. On the schematic diagram, use an unbroken arrow to indicate the direction of conventional current. Finally, indicate the ammeter readings on the diagram.

The voltage across each branch is equivalent to the voltage of the battery. The current through a branch is simply the V/R ratio where V = 6 Volts and R = 3 ohms. This calculation leads to the conclusion that 2 Amps is the current in each branch. The current outside the branches is simply the sum of the current in the branches. So there will be a current of 4 Amps outside the branches. As an alternative to
this solution, the equivalent resistance could first be determined
using the 1/R |

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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) ____. Include all that apply, if any apply.

b. The electric potential at point C is the same as the electric potential at point(s) ____. Include all that apply, if any apply.

c. The electric potential at point F is the same as the electric potential at point(s) ____. Include all that apply, if any apply.

d. The electric potential at point I is the same as the electric potential at point(s) ____. 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 ___ and ____. 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 ___ and ____. 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 ___ and ____. 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 ___ and ____. Include all that apply, if any apply.

i. The current at point A is the same as the current at point(s) ___. Include all that apply, if any apply.

j. The current at point E is the same as the current at point(s) ___. Include all that apply, if any apply.

k. The current at point G is the same as the current at point(s) ___. Include all that apply, if any apply.

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) |

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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) ____. Include all that apply, if any apply.

b. The electric potential at point D is the same as the electric potential at point(s) ____. Include all that apply, if any apply.

c. The electric potential at point J is the same as the electric potential at point(s) ____. 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 ___ and ____. 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 ___ and ____. Include all that apply, if any apply.

f. The current at point A is the same as the current at point(s) ___. Include all that apply, if any apply.

g. The current at point B is the same as the current at point(s) ___. Include all that apply, if any apply.

h. The current at point C is the same as the current at point(s) ___. Include all that apply, if any apply.

i. The current at point D is the same as the current at point(s) ___. 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) ___ 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.

l. If the light bulb located between points D and G were to *go
out*, then the current would decrease at point(s) ___. Include
all
that apply, if any apply.

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 In j, removing the bulb in the first branch will not effect 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 j. |

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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.

The total resistance (or equivalent resistance) can first be determined using the equation for series circuits. _{Tot}
= R_{1} + R_{2} + R_{3}
= 5 Ohms + 10 Ohms + 15 Ohms = 30 OhmsOnce known, the R _{Tot}
= (V_{Tot}) / (R_{Tot}) = (120
V) / (30 Ohms) = 4 AmpsFor a series circuit,
the current through each resistor is the same as the total circuit
current. Thus, I 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 (V _{1}
= I_{1} • R_{1} = (4 Amps) • (5
Ohms) = 20 Volts
V V |

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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.

The total resistance (or equivalent resistance) can first be determined using the equation for series circuits. _{Tot}
= 1 / R_{1} + 1 / R_{2} + 1 / R_{3}
= 1 / (5 Ohms + 1 / (10 Ohms) + 1 / (15 Ohms)
R Once known, the R _{Tot}
= (V_{Tot}) / (R_{Tot}) = (120
V) / (2.727 Ohms) = 44.0 AmpsThe 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 (V In this circuit, the branch currents can be computed by using the V = I•R equation for each resistor. This is shown below. _{1}
= V_{1} / R_{1} = (120 Volts) /
(5 Ohms) = 24 Amps
I I |

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