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Parallel circuits and other questions.

  1. Dec 24, 2003 #1
    1. will battery last longer in a parallel or series circuit.

    2. I need an easy expirement in which I can find the speed of sound, excluding the use of a tube and a tuning fork in the experiment. If you have any easy hands on experiments to try I will be more than grateful. The more experiments the better.

    Thanks in advance.
  2. jcsd
  3. Dec 24, 2003 #2


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    1. This question is unclear. You need to be more specific.

    2. I assume you mean the speed of sound in air?

    Well, an easy thing to do is to find a source of noise and light -- for example lightning -- and then compare the difference in time for the light and the sound (thunder) to reach you. You can also do this with firecrackers, or a gong or anything similar.

    If you can make a sound detector of some type -- for example a microphone attached to an LED or something similar, then you can put two of them different distances away from your sound source, and an observer on the perpendicular bisector. This will allow you to eliminate the speed of light from your calculations. (The speed of light will probably be a non-issue, that's neither here nor there.) I think that those truly obnoxious dancing flowers/coke cans might work for this.

    If you have a number of sound recorders, then you can measure the speed of sound by setting them all up in some arrangement, and then moving a sound source around. This will give you a differential between the signals that you can measure. This can probably be done with two tape recorders and someone clapping.

    If you record the sounds that an object makes at two different velocities relative to you, then you can calculate the speed of sound from the interfearance provided you know the change in velocity. This one does use a constant frequency sound source, so you might want to nix it.

    If you have an echo off of a known surface, you can calculate the speed of sound from the echo delay.
  4. Dec 24, 2003 #3


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    The life of a battery will be determined by the total resistance of the circuit, not the configuration of the load. 2 4ohm resistors in parallel will be the same load as a single 2 ohm resistor.

    Remember resistance in series is found by

    [tex] R_T = R_1 + R_2 + ... + R_n [/tex]

    and the total resistance of parallel resistors is

    [tex]R_T = \frac {1} {\frac {1} {R_1} + \frac {1} {R_2} + ... + \frac {1} {R_n}} [/tex]
  5. Dec 25, 2003 #4
    Integral answered this question succinctly. A parallel circuit would require at least two components. Given two resistors, the total resistance would be lower in the parallel circuit as shown by the equations in Integral's post. The current would be higher and the rate at which energy is being converted (Power, measured in watts. Joules/Second) is directly proportional to current. P=E*I. This would drain the battery faster than the two resistors in series.
  6. Dec 27, 2003 #5
    Sorry for not being clear on the first question. What I meant was if you had bulbs connected in parallel and ones connected in series, which will run the batter down most rapidly?
  7. Dec 27, 2003 #6
    The bulbs connected in parallel would run the battery down faster because the total resistance of the circuit would be less than if they were connected in series. The reason the total resistance is lower when the bulbs are connected in parallel is simply because there are more paths available for the current to flow. If you keep adding more bulbs (resistance) in parallel, then you are creating more paths in which the current can flow. Since the the total current is higher because of the decreased resistance, the rate at which the energy of the battery is converted to light and heat is higher. Energy/Time equals Power. Power is the rate at which energy is converted. Power is also equivalent to the product of Voltage and Current. P=EI. As you can see here, Power is directly proportional to current and voltage. In this case we are only concerned with the change in current since the voltage of the battery is constant.

    In a series circuit, there is only one path for current and as you add more resistances in series, you are restricting the flow of current through that one path. As you add more bulbs in series the total resistance increases, the current decreases, thus power or the rate of energy conversion decreases. The same amount of energy will be converted in this case but at a slower rate. You'll notice as you add more bulbs in series, all the bulbs will get dimmer. This is because the current is always the same at all points in a series circuit due to there only being one path for current flow. If you increase the resistance by adding more bulbs, the current decreases through each bulb. The rate at which energy is converted by each bulb is less. If you measure the voltage across each bulb in series, you will see that as more bulbs are added, the voltage across each bulb decreases. E=I*R. Since current through each bulb is less, voltage at each bulb is less. Given that each bulb has the same resistance, the total voltage of the source will be divided evenly across each bulb.

    Ideally, as more bulbs are added in parallel the bulbs will not grow dimmer. When resistances are added in parallel, the voltage that is applied across each resistance is equal to the source voltage. Since the voltage is the same across each bulb in parallel, the current through each bulb is the same. The power dissipated is the same. As more bulbs are added, the total current of the circuit will increase as stated before but the current in the individual bulbs stays the same.

    In practice, the bulbs will get a little dimmer when added in parallel. It's not nearly as pronounced as in the series case. Batteries have an internal resistance which affect things:


    That thread should help you understand as well.

    For series circuits, the sum of the voltage drops across each component equal the source voltage. The current is the same through all components. As new components are added, the total resistance increases. The current decreases.

    For parallel circuits, the sum of the current through each individual component equals the total current of the circuit. The voltage at each component is the same. As new components are added, the total resistance decreases and the total current increases.
    Last edited: Dec 27, 2003
  8. Jan 6, 2004 #7
    Thanks guys your participation and clear explanation really helped me straighten things out.:smile:
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