Physics _ Chap 13
Three bulbs have power ratings for use on a 120 volt line: 60 watt, 100 watt and 150 watt. The one with the largest value of resistance has power rating
60 W.
A 9 ohm resistor and a 16 ohm resistor are connected in series in a circuit with a 5.0 volt battery. Assuming negligible internal resistance in the battery, the current in the 16 ohm resistor will be
0.20 A.
The potential difference across the terminals of a storage battery not connected in any circuit is observed to be 12 V. When it is connected to an external resistance the potential difference across the terminals is observed to be 11 V while the current in the external resistance is 3 A. What is the internal resistance of the battery?
0.33 ohm
One ampere is equivalent to
1 C/s.
The voltage drop across a resistor is 4.0 V for a current of 1.0 A in the resistor. What is the current that will produce a voltage drop of 8.0 V across the resistor?
2 A
A student finds two ancient meters with numbered scales but no unit markings. He thinks that one is an ammeter and one is a voltmeter but doesn't know which is which. He measures the resistance of each, finding that meter A has a resistance of 5000 ohms and meter B has a resistance of 0.10 ohm. He can conclude that
A is the voltmeter and B is the ammeter.
Two arrangements of a battery, bulb and wire are shown in the diagram. Which of the arrangements will light the bulb?
A only
An ohm is equal to which of these?
A volt per ampere
Two circuit diagrams are shown. Which one, if either, will cause the lightbulb to light?
B
A 1.5 V flashlight battery, a 10 ohm resistor, and a flashlight lamp are available. How should the lamp be connected in a circuit so that it would glow the brightest?
Connect it directly to the battery; don't use the resistor
You have exactly 4 resistors: one 3 Ω, one 4 Ω, one 5 Ω, and one 6 Ω. How can you combine these to make a 2 Ω resistor? (The symbol Ω stands for "ohm.")
Connect the 3 Ω resistor in parallel with the 6 Ω resistor
Power supply : Series connection: Parallel connection: What is the best way to ensure that a 40 watt bulb and a 60 watt bulb have the same voltage applied to them?
Connect the bulbs in parallel with the power supply
What is the best way to ensure that a 40 watt bulb and a 60 watt bulb have the same current within them?
Connect the bulbs in series with the power supply
Resistors of 10 ohms, 20 ohms, and 30 ohms are connected in series with a battery. The 10 ohm resistor is closest to the positive terminal and the 30 ohm resistor is closest to the negative terminal. What happens as the current flows through this circuit?
None of the resistors consume current
Two resistors are connected in series with a battery shown in the diagram. R1 is less than R2. Which of the two resistors, if either, has the greatest voltage difference across it?
R2
In the circuit diagram shown, R1, R2, and R3 are three resistors of different values. R3 is greater than R2, and R2 is greater than R1. ε is the electromotive force of the battery whose internal resistance is negligible. Which of the three resistors has the greatest current flowing through it?
R3
A person discovers that the 5 ampere fuse in his car's brake light circuit is blown. Not having a spare 5 ampere fuse available, he replaces it with a new 10 ampere fuse. What is likely to happen as he uses the brakes?
The circuit overheats and the brake lights do not work
Item27 0/0.25 points awarded Item Scored eBook Print References Item 27 Item 27 0 of 0.25 points awarded Item Scored A person discovers that the 5 ampere fuse in his car's brake light circuit is blown. Not having a spare 5 ampere fuse available, he replaces it with a new 10 ampere fuse. What is likely to happen as he uses the brakes?
The circuit overheats and the brake lights do not work
Three resistors - R1 = 24 Ω, R2 = 38 Ω, and R3 = 34 Ω are connected to a 12-V battery as shown in the figure. The internal resistance of the battery is negligible. What is the current coming from the battery?
The current from the battery is calculated using Ohm's law as follows I=VR where V is the voltage of the battery, and R is the total resistance. The total resistance in the circuit is calculated as follows Rseries=24 Ω+38 Ω+34 Ω Rseries=96 Ω Substituting values in the equation, we get I=12 V96 Ω=0.125 A
Three identical resistors, where R1 = 32 Ω, are connected in parallel with one another as shown. Then the parallel combination is connected to a 12-V battery whose internal resistance is negligible. How much current flows through each resistor in the combination?
The current through each resistor is given by I=V/R1=12 V/32 Ω I=0.375 A
Three identical resistors, where R1 = 32 Ω, are connected in parallel with one another as shown. Then the parallel combination is connected to a 12-V battery whose internal resistance is negligible. What is the total current through the combination?
The current through the combination is determined using the following formula I=VR where V is the voltage and R is the equivalent resistance of the circuit. The equivalent resistance of this parallel combination is determined as follows 1Rparallel=1R1+1R2+1R3=132 Ω+132 Ω+132 Ω 1Rparallel=10.0938 Ω Rparallel=10.6667 Ω Substituting the values in the equation I=VR , we get I=VR=12 V10.6667 Ω I=1.125 A
Two resistors, each having a resistance of 10 Ω, are connected in parallel. What is the equivalent resistance of this combination?
The equivalent resistance is calculated as follows: 1/Rparallel=1/R1+1/R2=1/10 Ω+1/10 Ω=1/5 Ω Rparallel=5 Ω
Three resistors of 15 Ω, 6 Ω, and 10 Ω are connected in parallel with one another. What is the equivalent resistance of this combination?
The equivalent resistance of this combination is 1Rparallel=1R1+1R2+1R3=115 Ω+16 Ω+110 Ω 1Rparallel=10.3333 Ω Rparallel=3 Ω
Three identical resistors, where R1 = 32 Ω, are connected in parallel with one another as shown. Then the parallel combination is connected to a 12-V battery whose internal resistance is negligible. What is the equivalent resistance of this parallel combination?
The equivalent resistance of this parallel combination is determined as follows 1/Rparallel=1/R1+1/R2+1/R3=1/32 Ω+1/32 Ω+1/32 Ω 1/Rparallel=1/0.0938 Ω Rparallel=10.6667 Ω
Three resistors - R1 = 24 Ω, R2 = 38 Ω, and R3 = 34 Ω are connected to a 12-V battery as shown in the figure. The internal resistance of the battery is negligible. What is the total resistance in the circuit?
The equivalent series resistance of a circuit is calculated as follows Rseries=R1+R2+R3+......Rn The total resistance in the circuit is calculated as follows Rseries=24 Ω+38 Ω+34 Ω Rseries=96 Ω
Bulbs A, B, C, and D are identical and connected to a battery as shown. The bulb with the least current passing through it is
both bulbs A and B.
A current of 6.5 A flows through a battery for 3 minutes. How much charge passes through the battery in that time?
The quantity of charge passing through the battery is determined using the following formula: q=It where I is the current flowing through the battery (A), q is the charge (C), and t is the time in seconds. Substituting the values in the equation, we get q=(6.5 A)×(180 s)= 1170 C
A current of 1.122 A flows through a resistor with a voltage difference of 115 V across it. Determine the resistance of this resistor.
The resistance is calculated using Ohm's law. We know that V=IR . From this, we get R=VI Substituting the values in the equation, we get R=115 V/1.122 A=102.4955 Ω where I is the current, R is the resistance, and V is the voltage.
The voltage across each resistor is determined using Ohm's law. V=IR The total series resistance is determined using the following formula: Rseries= R1+ R2+ R3 +R4=4×28 Ω Rseries= 112 Ω The current flowing through each resistor is determined using Ohm's law. V= IR, I= VRseries Substituting the values in the equation, we get I=18 V 112 Ω= 0.1607 A where I is the current, R is the resistance, and V is the voltage. V= IR=(0.1607 A)×(28 Ω)=4.5 V where I is the current, R is the resistance, and V is the voltage.
The total resistance is calculated using the formula: Rseries=R1+R2=50 Ω + 28 Ω=78 Ω The current flowing through each resistor is determined using Ohm's law. V= IR, I= VRseries Substituting the values in the equation, we get I=12 V78 Ω=0.1538 A where I is the current, R is the resistance, and V is the voltage.
Three resistors - R1 = 24 Ω, R2 = 38 Ω, and R3 = 34 Ω are connected to a 12-V battery as shown in the figure. The internal resistance of the battery is negligible. What is the voltage difference across the 38-Ω resistor?
The total resistance of the circuit is R=R1+R2+R3=24 Ω+38 Ω+34 Ω R=96 Ω The current through the 38-Ω resistor is I=VR=12 V96 Ω I=0.125 A The voltage difference across the 38-Ω resistor is ΔV=IR=0.125 A×38 Ω ΔV=4.75 V
In the circuit shown, consider R1 = 3.5 Ω as the internal resistance, Rbt, of the battery. As shown, it can be considered in series with the battery and R2 = 19 Ω as the load resistance Rbulb. What is the current flowing through the 19-Ω resistor?
The total resistance of the circuit is R=Rbt+Rbulb=3.5 Ω + 19 Ω R=22.5 Ω The current through the 19-Ω resistor is I=VR=6 V22.5 Ω I=0.2667 A
In the circuit shown, consider R1 = 3.5 Ω as the internal resistance, Rbt, of the battery. As shown, it can be considered in series with the battery and R2 = 19 Ω as the load resistance Rbulb. What is the voltage difference ΔVbulb across the 19-Ω resistor?
The total resistance of the circuit is R=Rbt+Rbulb=3.5 Ω + 19 Ω R=22.5 Ω The current through the 19-Ω resistor is I=VR=6 V22.5 Ω I=0.2667 A The voltage difference ΔVbulb across the 19-Ω resistor is calculated as follows: ΔV=IR=0.2667 A×19 Ω ΔV=5.0667 V
Four 28-Ω resistors are connected in series to a 18-V battery of negligible internal resistance. Calculate the current flowing through each resistor.
The total series resistance is determined using the following formula: Rseries= R1+ R2+ R3 +R4=4×28 Ω Rseries= 112 Ω The current flowing through each resistor is determined using Ohm's law. V= IR, I= VRseries Substituting the values in the equation, we get I=18 V 112 Ω= 0.1607 A where I is the current, R is the resistance, and V is the voltage.
A 50-Ω resistor and a 28-Ω resistor are connected in series to a 12-V battery of negligible resistance. What is the voltage difference across each resistor?
The voltage across each resistor is determined using Ohm's law. The total resistance is calculated using the formula: Rseries=R1+R2=50 Ω + 28 Ω=78 Ω The current flowing through each resistor is determined using Ohm's law. V= IR, I= VRseries Substituting the values, in the equation we get I=12 V78 Ω=0.1538 A where I is the current, R is the resistance, and V is the voltage. v1=IR1=0.1538 A×50 Ω v1=7.6923 V v2=IR2=0.1538 A×28 Ω v2= 4.3077 V where I is the current, R is the resistance, and V is the voltage.
Four 28-Ω resistors are connected in series to a 18-V battery of negligible internal resistance. Determine the voltage difference across each resistor.
The voltage across each resistor is determined using Ohm's law. V=IR The total series resistance is determined using the following formula: Rseries= R1+ R2+ R3 +R4=4×28 Ω Rseries= 112 Ω The current flowing through each resistor is determined using Ohm's law. V= IR, I= VRseries Substituting the values in the equation, we get I=18 V 112 Ω= 0.1607 A where I is the current, R is the resistance, and V is the voltage. V= IR=(0.1607 A)×(28 Ω)=4.5 V where I is the current, R is the resistance, and V is the voltage.
A current of 1.5 A is flowing through a resistance of 34 Ω. What is the voltage difference across this resistance?
The voltage across the resistance is determined using Ohm's law. Ohm's law states that V=IR where V is the voltage, I is the current, and R is the resistance. Substituting the values in the equation, we get V=1.5 A×34 Ω =51 V where I is the current, R is the resistance, and V is the voltage.
The unit for electromotive force (emf) is
V.
What energy source increases the potential energy of the water behind the dam of a hydroelectric power plant?
a pump
Which normally have the larger resistance?
a voltmeter
Two resistors are connected in series with a battery shown in the diagram. R1 is less than R2. Which of the two resistors, if either, has the greatest current flowing through it?
both have the same current flowing through them
Which of these appliances is most likely to cause an overload problem when connected to a circuit that already has other appliances drawing current from it? A. an electric shaver B. a coffee maker C. a television set
coffee maker
The current through a circuit consisting of a single battery and single resistor is measured to be 0.25 ampere. Another identical battery and another identical resistor are added to the circuit. The current passing through the circuit is
dependent on how the battery and resistor are added to the circuit.
The resistance of a 50 watt bulb is ______ that of a 100 watt bulb, if both run on the same voltage.
double
What is the difference, if any, of electric current and electric charge?
electric current is moving electric charges
Three identical resistors can be connected in any combination. It's not necessary to include all three in any given combination. The number of distinct values of resistance that can be made is
five.
Three resistors - R1 = 24 Ω, R2 = 38 Ω, and R3 = 34 Ω are connected to a 12-V battery as shown in the figure. The internal resistance of the battery is negligible. What is the current through the 24-Ω resistor?
he total resistance of the circuit is R=R1+R2+R3=24 Ω+38 Ω+34 Ω R=96 Ω The current through the 38-Ω resistor is I=VR=12 V96 Ω I=0.125 A
In which situation, if either, does the electrical signal travel fastest?
in an electrical wire
In which setup would it make the most sense to connect a fuse or circuit breaker with other elements in a circuit?
in series
A flashlight bulb typically has a small metallic button at one end, and this button is surrounded by ceramic material. The chief purpose of the ceramic material is to
insulate the metal button from the other contact point.
If the current through a certain resistance is doubled, how does the power dissipated in that resistor change?
it quadrupoles
When current passes through a series combination of resistors how does the current change, if at all, as it goes through each successive resistor in the combination?
it remains the same each time
When an axon is stimulated a voltage spike is created. What happens to the voltage spike after it is created?
it travels down the axon
When a battery is being used in a circuit, how will the voltage across its terminals be different from when no current is being drawn from the battery?
it will be larger when it is being used
If we decrease the potential difference across a resistance in a circuit, how will the current flowing through that resistance change
it will decrease
In the circuit diagram shown, R1, R2 and R3 are three resistors of different values. R3 is greater than R2, and R2 is greater than R1. ε is the electromotive force of the battery whose internal resistance is negligible. If we disconnect R2 how will the current through R3 change?
it will decrease
f we decrease the potential difference across a resistance in a circuit, how will the current flowing through that resistance change?
it will decrease
Consider the circuit shown. What would happen to the brightness of the bulb if we connected a wire between points A and B?
it would decrease
Suppose that we use an uncoated metal clamp to hold the wires in place in the battery-and-bulb circuit shown. How bright will the bulb be?
much dimmer than without the clamp
Which, if either, is larger in a simple battery-and-bulb circuit?
neither; the current is the same at both places
How does a bimetallic strip break a circuit when things heat up?
one metal expands faster than the other, bending and breaking contact with the rest of the circuit
A voltmeter is always connected in
parallel.
In many ways, an electric circuit is like the plumbing system in your home. The voltage on an electric circuit corresponds to the ________ in a plumbing system.
pressure
Two identical lightbulbs are installed in two sockets connected in parallel and power is then applied to the combination so that both bulbs light. If one of the bulbs is then removed from its socket, the other one will
remain equally bright.
Suppose that the appliances connected to a household circuit were connected in series rather than parallel. What disadvantage would there be to this arrangement?
should one appliance fail, the whole system would be down
Consider the two signs shown in the diagram, located in different physics labs. Which of the two, if either, would be reason for greater caution?
the 10,000 V
The current (measured in amperes) in a circuit is
the amount of charge (measured in Coulombs) that passes a point (in the circuit) in 1 second.
Electrical current is
the charge per unit time.
A 12 ohm resistor and a 24 ohm resistor are connected in series with a 6.0 V battery. The correct statement is
the current in each resistor is the same.
In the circuit shown, the circle with an A in it represents an ammeter. Which of these statements is correct?
the meter will draw a significant current from the battery
Power supplied by a battery is calculated as
the potential difference times the current.
The resistance of a circuit element is a measure of
the ratio of the voltage difference between element ends to the current in the element.
What does the power being delivered by a battery depend on?
the resistance of the circuit
Resistances connected in series all have
the same current.
Resistances connected in parallel all have
the same potential difference.
Power dissipated by a resistance may be calculated as
the square of the current times the resistance.
If the resistance in a circuit using an ideal (no internal resistance) battery is increased, then
the voltage difference maintained by the battery does not change.
An unmarked resistor is being used in a circuit. To determine its resistance by making voltage and current measurements, one should connect
the voltmeter in parallel with the resistor, and the ammeter in series with the resistor.
In the circuit shown, the circle with a V in it represents a voltmeter. Which of the following statements is correct?
the voltmeter is in the correct position for measuring the voltage difference across R.h
