Your lab partner builds a circuit with a single battery, three identical bulbs, and a single...

Your lab partner builds a circuit with a single battery, three identical bulbs, and a single switch. You can see the bulbs but not the wiring of the circuit. With the switch closed, bulbs A and B are equally bright, and bulb C is not lit at all. When the switch is opened, the brightness of bulb A does not change, the brightness of bulb B decreases, and bulb C "turns on". Now (with the switch open), bulb A is the brightest, and bulbs B and C are equally bright (but dimmer than A)

Construct a circuit diagram showing bulbs A-C, the switch, and the battery. Explain your reasoning.

Expert Solution

according to the statement of the problem, the brightness of the bulb A remains unaffected of whether the switch is open or closed. so that means Bulb A is in parallel with the battery.

When switch is closed, the bulb C does not lit and when it is open, Bulb C lits. this suggests that switch is connected in parallel with bulb A

when switch is open bulb C is lit and brightness of B and C becomes same but less than A , that means B and C are in series.

genius_generous answered 2 years ago

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