Question - Change in brightness

Question

I have a circuit with two identical lamps, one battery and one switch (see image). I'm pretty sure that if the switch is closed the lamp marked A would get brighter (due to the power being able to take a path with less resistance than going over the other lamp).

What I'm wondering about is how I should properly explain this result. Which laws are involved with it? I need to be able to make a rather formal explanation that brings up exactly why lamp A gets brighter when the switch closes.

Anyone can point me in the right direction?

PS: I know this is probably a rather basic question, but it's part of a large number of very mixed questions that deal with my ability to explain stuff, so I don't actually know much about electrical circuits overall...

Answer 1

So you have selected two bulbs with identical resistance. When the switch is open, these two bulbs form a voltage divider. In your case, the voltage is divided equally due to same resistance. Also, same current is flowing through both of those bulbs. So the circuit consists of two resistors in series. And the current flows accordingly.

Now, you close the switch. Notice how the ends of the "not A" bulb are shorted. This implies there is no any voltage across this bulb so it should not glow ideally. But due to certain resistance of the wires and the switch itself, there is small potential difference across this bulb giving slight glow.

The cicuit after closing the switch now has only one resistance in its path.

The bulb A now has full battery voltage across its terminals. Since the circuit now has only one resistacne, more current flows through the circuit giving you the brighter glow of the bulb.

^{simulate this circuit – Schematic created using CircuitLab}

In the first case you have the current of 5mA. In the second case you have the current of 10 mA. Twice the current !!. Do calculate the power now yourself and see the result.

The resistances are only for demonstrating the problem. The actual resistance of the bulbs is different from those I have used.

Answer 2

Only Ohm's law is enough I guess.

Assuming both bulbs are ideal and of same resistance.

In the first case when switch is at open position, both the bulbs are in series. Equal amount of current will pass through both bulbs, which will be: $$I_1=\frac {V_{battery}}{2R}$$ In the second case when switch is closed, the second bulb is shorted across and hence the current drawn is only due to the first bulb, which will be: $$I_2=\frac {V_{battery}}{R}$$ Power developed across the bulb A in both cases is: $$P = I^2R$$ Clearly \$ I_2 > I_1 \$ , So bulb A will light brighter in the second case.