Question

A 12-volt battery is connected to two light bulbs, as drawn in figure 1 light bulb 1 has resistance 3 ohms, while light bulb 2 has resistance 6 ohms. The battery has essentially no internal resistance, and all the wires are essentially resistanceless, too. When a light bulb is unscrewed, no current flows through that branch of the circuit. For instance, if light bulb 2 is unscrewed, current flows only around the lower loop of the circuit, which consist of the battery and light bulb 1. The more current flows through a light bulb, their equivalent resistance is $R_{eq}=R_1+R_2$. By contrast, when two reisistors are wired in parallel, their net resisance is given by
$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$
Bulb 2 is now screwed in as a result, bulb 1 :

• A. turns off
• B. becomes dimmer
• C. stays about the same brightness
• D. becomes brighter

Answer 1

The correct option is C stays about the same brightness

When only bulb 1 is connected, voltage across it is = 12V

Now bulb 2 is connected in parallel with bulb 1, so both bulbs will have 12V across them.

As voltage across bulb 1 is not changing so current through it will not change so brightness will not change.