Question

What happens to an the resistance/brightness of a bulb in a simple circuit when you swap the wires in the circuit for ideal wires?

If this circuit has one bulb in series with a parallel pair of two bulbs, how does the second bulb added in parallel's brightness change based when the wire is ideal?

Answer 1

Everything depends on the power dissipated by an element. Greater the power dissipated, greater will be the brightness. Provided

If we consider a bulb with wires swapped for ideal wires, this reduces the resistances of the overall circuit increasing the current flowing in the circuit .

With increase in current ,the power dissipated by the bulb also increases hence it will glow brighter.

Let us consider a circuit with practical wires in it having some resistance. There would be some voltage drop accross the wires. Thus the voltage accross the parallel combination of bulbs is relatively lesser than we expect it to be when we use ideal wires, i.e. the wires with no resistance. When we use ideal wires, the voltage accross the parallel combination of bulbs is greater. Thus the current through the bulbs increases which in turn increases the brightness.

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