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 consists of the battery and light bulb 1. The more current flows through a light bulb, the brighter it shines.When two resistors a wired in series, their equivalent resistance is Req=R1+R2. By contrast, when two resistors are wired in parallel, their net resistance is given by 1/Req=1/R1 + 1/R2. A.) When bulb 1 I screwed in, but bulb 2 in unscrewed, the power generated in bulb 1 is:4 watts12 watts36 watts48 wattsB.) Which statement is false?Some of the energy produced by the light bulb takes the form of heatThe batter is the source of all the electrons flowing in the circuitThe current entering the light bulb equals the current leaving the light bulbThe potential in the wire to the left of the light bulb differs from the potential in the wire to the right of that bulb.C.) Bulb 2 is now screwed in. As a result, bulb 1:Turns offBecomes DimmerStays about the same brightnessBecomes brighterD. With both light bulbs screwed in, the current through the battery is1.2 amperes2 amperes4 amperes6 amperesE.) With only bulb 1 screwed in, a NeverQuit 12-V battery goes dead in 24 days. With both light bulbs screwed in, a NeverQuit 12-V battery goes dead in:12 days14 days16 days18 days

Answer 1

Given data:

* The resistance of bulb 1 is,

`R_1=3\text{ ohm}`

* The resistance of bulb 2 is,

`R_2=6\text{ ohm}`

* The voltage across the battery is V = 12 volts.

Solution:

(A). As bulb 2 is unscrewed, thus, the only resistance of the circuit is due to bulb 2.

Thus, the power generated in bulb 1 is,

`P=(V^2)/(R_1)`

Substituting the known values,

`\begin{gathered} P=(12^2)/(3) \\ P=48\text{ watts} \end{gathered}`

Thus, the power generated in bulb 1 is 48 watts.

Hence, the fourth option is the correct answer.