Power - Physics
Card 1 of 12
A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be 
, what should the resistance of the heating element be if the radiated power is to total 
?
A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be
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We are asked to calculate the power dissipated through a resistor connected to a battery. We know that 
. Substituting the values given above, 
. Solving for R yields 
.
We are asked to calculate the power dissipated through a resistor connected to a battery. We know that
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What is the RMS voltage of a 
peak-to-peak AC current?
What is the RMS voltage of a
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This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is 
peak-to-peak. Thus, the peak voltage is half of this, or 
. By definition, to convert peak voltage to RMS voltage, we must divide the peak voltage by 
:
This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is
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A 
battery is connected to two 
light bulbs that are arrange in parallel. What is the total power dissipated by the light bulbs?
A
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The power dissipated in each light bulb is 
. Since there are two light bulbs, the total power dissipated is 
The power dissipated in each light bulb is
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A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be , what should the resistance of the heating element be if the radiated power is to total ?
A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be
Tap to reveal answer
We are asked to calculate the power dissipated through a resistor connected to a battery. We know that . Substituting the values given above, . Solving for R yields .
We are asked to calculate the power dissipated through a resistor connected to a battery. We know that
← Didn't Know|Knew It →
What is the RMS voltage of a peak-to-peak AC current?
What is the RMS voltage of a
Tap to reveal answer
This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is peak-to-peak. Thus, the peak voltage is half of this, or . By definition, to convert peak voltage to RMS voltage, we must divide the peak voltage by :
This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is
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A battery is connected to two light bulbs that are arrange in parallel. What is the total power dissipated by the light bulbs?
A
Tap to reveal answer
The power dissipated in each light bulb is . Since there are two light bulbs, the total power dissipated is
The power dissipated in each light bulb is
← Didn't Know|Knew It →
A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be , what should the resistance of the heating element be if the radiated power is to total ?
A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be
Tap to reveal answer
We are asked to calculate the power dissipated through a resistor connected to a battery. We know that . Substituting the values given above, . Solving for R yields .
We are asked to calculate the power dissipated through a resistor connected to a battery. We know that
← Didn't Know|Knew It →
What is the RMS voltage of a peak-to-peak AC current?
What is the RMS voltage of a
Tap to reveal answer
This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is peak-to-peak. Thus, the peak voltage is half of this, or . By definition, to convert peak voltage to RMS voltage, we must divide the peak voltage by :
This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is
← Didn't Know|Knew It →
A battery is connected to two light bulbs that are arrange in parallel. What is the total power dissipated by the light bulbs?
A
Tap to reveal answer
The power dissipated in each light bulb is . Since there are two light bulbs, the total power dissipated is
The power dissipated in each light bulb is
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A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be , what should the resistance of the heating element be if the radiated power is to total ?
A resistive heating element can be modeled as a resistor connected across the terminals of a battery.
If the battery is selected to be
Tap to reveal answer
We are asked to calculate the power dissipated through a resistor connected to a battery. We know that . Substituting the values given above, . Solving for R yields .
We are asked to calculate the power dissipated through a resistor connected to a battery. We know that
← Didn't Know|Knew It →
What is the RMS voltage of a peak-to-peak AC current?
What is the RMS voltage of a
Tap to reveal answer
This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is peak-to-peak. Thus, the peak voltage is half of this, or . By definition, to convert peak voltage to RMS voltage, we must divide the peak voltage by :
This question requires two steps: the first is to calculate the peak voltage of the AC current, and the second requires conversion of the peak voltage to an RMS voltage. We are told that the current is
← Didn't Know|Knew It →
A battery is connected to two light bulbs that are arrange in parallel. What is the total power dissipated by the light bulbs?
A
Tap to reveal answer
The power dissipated in each light bulb is . Since there are two light bulbs, the total power dissipated is
The power dissipated in each light bulb is
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