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 are wired in series, their equivalent resistance is . By contrast, when two resistors are wired in parallel, their net resistance is given by .

Figure 1

  
When bulb 1 is screwed in, but bulb 2 is unscrewed, the power generated in bulb 1 is:

bulb 2 bulb 1 12V

A. 4 watts

B. 12 watts

C. 36 watts

D. 48 watts

  
Which of the following is FALSE?

A. Some of the energy produced by the light bulb takes the form of heat.

B. The battery is the source of all the electrons flowing around the circuit.

C. The current entering the light bulb equals the current leaving the light bulb.

D. The potential in the wire to the left of the light bulb differs from the potential in the wire to the right of that bulb.

  
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

  
With both light bulbs screwed in, the current through the battery is:

A. 1.2 amperes.

B. 2 amperes

C. 4 amperes

D. 6 amperes

  
With only light bulb 1 screwed in, a NeverQuit 12-volt battery goes dead in 24 days. With both light bulbs screwed in, a NeverQuit 12-volt battery goes dead in:

A. 12 days

B. 14 days

C. 16 days

D. 18 days

Answer 1

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