What i need is just short but decent answer/explanation. Thanks for all the help

Question

What i need is just short but decent answer/explanation. Thanks for all the help.

The circuit at right contains a battery, a bulb, a switch, and a capacitor. The capacitor is initially uncharged. Bulb A Describe the behavior of the bulb in the two situations below 1. Battery Capacitor The switch is first moved to position I. Describe the behavior of the bulb from just after the switch is closed until a long time later. Explain. a. b. The switch is now moved to position 2. Describe the behavior of the bulb from just afier the switch is closed until a long time later. Explain your reasoning. A second identical bulb is now added to the circuit as shown. The capacitor is discharged. c. Bulb B Bulb C Capacitor i. The switch is now moved to position 1 Battery Describe the behavior of bulbs B and C from just afier the switch is closed until a long time later. Explain. How does the initial brightness of bulb C compare to the initial brightness of bulb A in question 1? Ex plain your reasoning.

Explanation / Answer

1)

a- a moment after the switch one is conected to the batery the bul is going to be off because the capacitor is off, and the batery will charge the capacitor, and when it happens the bulb will turn on.

b- when you move the switch to the position 2, it will be turned on until the carge at the capacitor goes away through the bulb.

c- it happens the same as part a, the bulbs are going to be on when the capacitor start charging. you have to remember that when the capacitor is discharged it have no current, so it is open. so the bulbs have no currents either..

- the brightness of the bulb A its going to be more than C because in C it shares voltage with bulb B and the capacitor. but A just shares voltage with tha capacitor.

- when the capacitor is pass a log time, the capacitor is fully charged with the energy of the batery, and the will be in parallel, so they will have the same voltage..

- a moment after whe move the switch, the capacitor acts like a batery just like the part 1- but it will discharge faster because it has two bulbs-

Bulb C might be less than bulb A because of bulb B.

in position 2. the voltage in the capacitor will be less than the batery because the charge in the capacitor went through the bulbs until it discharged.

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