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**Hint:-**The brightness of a bulb depends upon the power dissipated by the bulb. We just have to compare power dissipation of the bulb in initial condition with the final condition after adding another bulb in series with it.

**Complete Step by Step Explanation:**

When we add a resistance in series with an already present resistance in a circuit, then the initial voltage difference across the initial resistance is divided among both resistances after adding them in series according to their resistance values.

This is due to decrease in the current, because according to ohm’s law,

$V = IR$

Where $R$ is the resistance

$I$ is the current passing through $R$

And $V$ is the voltage across $R$ .

Now, if we add another resistance in series with $R$ , equivalent resistance of the circuit will increase, which states that $I$ will decrease, because $V$ will remain constant across the circuit.

Now we know that power $P = \dfrac{{{V^2}}}{R}$

Now, putting value $V = IR$ (by ohm’s law) in above equation, we get,

$P = \dfrac{{{I^2}{R^2}}}{R}$

On solving, we get,

$P = {I^2}R$

Therefore, $P \propto {I^2}$

Now, the total resistance of the circuit is increasing due to addition of another bulb in series with the first bulb of the circuit when there was no second bulb, therefore the current is decreasing due to increase in resistance, so power dissipation will also decrease in the first bulb.

So, the brightness of the bulb will decrease.

**So, the correct answer is option (C).**

**Note:**Since, we added the bulb in series, therefore the current was decreasing and voltage was constant, but if resistance would be added in parallel, then resistance would have been decreased so current would increase, and brightness would increase.

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