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So I have two questions, but they're kind of linked so I'm putting them in one.

1) I have a simple circuit with a battery connected to three lamps in series. If I then add a second battery but connect it in parallel to the first, why does the potential difference across one of the lamps and the current through all of the lamps stay the same as before? I would have thought that an extra battery should mean that more work is being done on the charges, meaning that the current should increase.

2) Why, in a parallel circuit, is the potential difference across each resistor equal to the potential difference across the battery and why do the p.d.s add to give the p.d. of the battery in a series circuit?

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    $\begingroup$ I answered your question as first posed, but I don't want to chase down your evolving edits. If you want to ask a new or different question, there's no problem with asking questions in succession, but it's difficult when you substantively change the question into something else once it's been answered. $\endgroup$ – tom10 Apr 2 '16 at 15:37
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Adding another battery in parallel could allow the supply of more current, but if the current you need (for your circuit and battery voltage) is already sufficiently supplied by the first battery then you won't see any difference.

Note also, that in generally you shouldn't connect batteries in parallel, and it can even be damaging or unsafe to do so. For example, one battery can try to charge the other which can lead to excessive heating.

If you want to increase the voltage you can connect the batteries in series.

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Hint for your second question :

Apply kirchoff's loop law in series circuit.

And in parallel circuits P.D. remains same as there is no resistance for voltage drop[in ideal wired circuit]

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Are you aware of Kirchoff's Laws? A good analogy is to think of a battery applying an electric pressure difference across the ends of the circuit. For components in parallel, one side of the arrangement is at one potential supplied by the battery, and the other side of the arrangement is at the other potential so that the potential difference across all the components must be the same.

For components in series, what must be the same is the current through the components must be the same, there are no "electric leaks" and charge that enters one side of the circuit is the same as the charge that leaves the components. If you replace "charge" by "rate of flow of charge", which is current, then you can see that this means the same current flows through components in series. An analogy with water pipes is that the pressure drops across several pipes connected in series will be the same as the pressure drop across all of the pipes and this is the pressure applied by the source. In this case replace the term "pressure difference" by "electric pressure difference" by "potential difference.

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  1. When you connect similar batteries in parallel, the effective voltage they offer is equal to one. So there will be no change in $V$ so no change in $I$.

  2. Charges will flow from one place to another until potential at both places are equal. In a parallel connection, the ends of the resistance and the batteries are connected together so the potential at all right ends will be same and similarly at the left ends.

This is just the beginning of the thought processes, just think about what charges like and where they would want to go like and you will get it.

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