If I connect the +ve terminal of one battery to a terminal of a bulb

[varadgautam]

If I connect the +ve terminal of one battery to a terminal of a bulb and the other terminal of the bulb to the -ve terminal of an other battery, why doesn't it glow? There is still potential difference across the bulb's terminals and so the charge should flow.

 

[Lsos]

I'm thinking for the same reason water doesn't flow out of closed faucet/ hose. There's pressure pushing it, but if the path is blocked/ cut, then it's just not going to flow.

 

[Vanadium 50]

How does the current flow?

 

[Philip Wood]

There will be excess electrons on the negative terminal of one battery, and a deficit of electrons on the positive terminal of the other battery. So when you connect the bulb across these terminals there will be a p.d. across the bulb. But this p.d. will disappear in a very short time as electrons flow through the bulb neutralising the charges on the batteries' joined terminals (those connected to the bulb). [If the batteries are identical a symmetry argument shows that there can be no net charge remaining on the joined terminals]. And that's it; there won't now be a p.d.across the bulb until the circuit is completed.

There will, of course, still be a p.d. across the terminals of one battery (i.e. between its connected terminal and its free terminal), and the same for the other battery. But it isn't either of these p.d.s which is being connected across the bulb.

 

[varadgautam]

There will be excess electrons on the negative terminal of one battery, and a deficit of electrons on the positive terminal of the other battery. So when you connect the bulb across these terminals there will be a p.d. across the bulb. But this p.d. will disappear in a very short time as electrons flow through the bulb neutralising the charges on the batteries' joined terminals (those connected to the bulb). [If the batteries are identical a symmetry argument shows that there can be no net charge remaining on the joined terminals]. And that's it; there won't now be a p.d.across the bulb until the circuit is completed.

The same thing should happen if I connect the bulb to a single battery, since there too will be excess electrons on the -ve terminal and these will flow through the bulb to the +ve terminal.

There will, of course, still be a p.d. across the terminals of one battery (i.e. between its connected terminal and its free terminal), and the same for the other battery. But it isn't either of these p.d.s which is being connected across the bulb.

Let's say the potential of the positive terminal of a battery is V and the potential of the negative terminal is zero. So, the potential difference is V. If I take another battery, similar to this battery, and connect the bulb to the +ve of one of the batteries and - of the other. The potential difference across the bulb is still V and therefore it should glow.

 

[Vanadium 50]

Calculate how much charge has moved when you've had the bulb lit for, I don't know, say a minute. Then calculate the force on the two batteries for the change imbalance. You'll quickly see why this doesn't happen.

 

[sophiecentaur]

Conversely, you could say that, at the instant of connecting things up, you may have had a slight potential difference between the tow battery terminals. Only a few picoCoulombs would have needed to flow for this PD to have reduced to zero. Not enough to make the bulb flash even for an instant.

As Vanadium says - the Force between a couple of 1Coulomb Charges held 1m apart is pretty fearsome!

 

[Philip Wood]

varadgautam: What is the basis of your claim that there is a p.d. across the bulb (even when there is not a complete circuit)? [Clearly I didn't manage to convince you that there wasn't (except for a very short time when the bulb is first connected)].

 

[RedX]

Here's the way I like to see it. Imagine that you connect two parallel plates to the unconnected ends of your batteries. Then charge will flow through the bulb until the potential across the two plates equals the sum of the potentials of the batteries. Then flow will cease, because that's what capacitors (in series) do to DC current. When flow ceases the light bulb will not shine.

Now imagine shrinking the plates to zero size. The capacitance will decrease, so that it won't take much charge to get the capacitor to the battery voltages.

So Kirchoff's rule says the the voltage of the batteries added to voltage drop across the capacitor is zero, so no current will flow.

The charges at the unconnected terminals of the battery act like a capacitor that stops flow.

 

[Philip Wood]

RedX. Love it.

 

[varadgautam]

Got it, thanks for the help.