If the voltage consumed by the lamps in the figure below is higher than the battery supply, what would happen?

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You don't consume voltage. You apply voltage.

What you consume is the current.

Now coming to your question. Here we assume that filament of the 3V lamps have same resistance and double in case of 6V lamp

The applied voltage will get divided across each lamps in their proportional resistance

ratio = source voltage/ (lamp design voltages)

voltage ratio = 9V/(3V+3V+6V) = 9/12 = ¾

voltage across 3V lamp = 3V(¾) = (9/4)V = 2.25V

voltage across 6V lamp = 6V(¾) = (9/4)V = 4.5V

So voltage across both 3V lamps will be 2.25V each and across 6V lamp will be 4.5V

That means all the bulbs will glow at about ¾ or 75% of their rated intensity

impossible to answer as you do not know the resistance of each bulb. The voltage is just the designed operating voltage, nothing to do with resistance.

Whoever wrote this question is an idiot.

A guess at what the intended question: if (a very big IF) each bulb is designed for equal brightness, then

n each is equal. That means the 6 volt bulb woul

the current at it's rated voltage, compared to the others.

Also assuming the bulbs are linear (which they are not). Linear means resistance does not change with voltage, and it does on incandescent bulbs, by a factor of about 10.

Th

e resistance of the 6 volt bulb would be:

equal power, twice the voltage, means half the current, and resistance, R = E/I, is 4 times the other bulbs.

Call the resistance of the 3 volt bulbs R, the the 6 volt one is 4R. Total R is 4R + R+R = 6R

I = E/R = 9/(6R) = 3/(2R)

voltage across each of the 3 volt bulbs is E = IR = R•3/(2R) = 1.5 volts

voltage across the 6 volt bulbs is E = IR = 4R•3/(2R) = 6 volts

but as I said above, this it totally bogus. The assumptions I made above that are just guesses.