Explain too where it asks The circuit at right contains a battery, a bulb, a swi

Question

Explain too where it asks

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

Explanation / Answer

1a.

At the start, there’s no charge on the capacitor. When you make the connection, current flows to give one plate of the capacitor a negative charge and the other a positive charge, until the potential difference across the capacitor is the same as the potential difference across the battery. In order to flow, the current has to go through the resistor – if it were a bulb, we’d see it light for a little bit and then get dimmer and go out.

it was obvious that the capacitor reached the same voltage as the battery by our "conductors touching have equal voltage" rule. It is not as obvious with the bulb. The bottom of the capacitor will obviously have the same

  voltage as the bottom of the battery, and the top must too, because the resistor must have the same voltage on both sides OR CURRENT WOULD FLOW THROUGH IT. Thus we know that when the current stops flowing the voltage across the capacitor is equal to the voltage of the bulb. We could also have gotten this by doing a Kirchoff's loop. +(Battery voltage difference) -IR - (Capacitor voltage difference)=0 implies the battery and capacitor voltage differences must be the same when the current finally goes to zero

1b.

Opening switch 1 disconnects the battery; closing switch 2 connects another wire that would allow current to flow from one plate of the capacitor to the other, again through the resistor. That happens until the capacitor plates return to neutral.

How long would the bulb light during discharge compared to during charging? Well, we know that it's how big the resistor is and how big the capacitor is that matters. If there were no resistor during charging, the charging would happen immediately. But the resistor slows the process down with a time constant RC. It's the same time constant for both charging and discharging, so if you were just looking at the bulb, you'd have no way of distinguishing charging from discharging

c)

1)

When the switch is first closed the current through the bulbs is half the current through bulb A in part a. This may seem obvious - twice the resistance, half the current. But if you are having trouble thinking about this with the capacitor there, go back to Kirchoff's loop. +(Battery voltage difference) -IR -IR- (Capacitor voltage difference)=0. When the switch is just closed, there is no charge on the capacitor and thus no voltage drop, so I=(battery voltage)/2R, just the same as if there was no capacitor there. But this current immediately starts to drop as the capacitor charges up, although the current lasts longer than that in bulb A because there is less current = slower charging. So, the bulbs are dimmer but light longer and in the end when there is no current the capacitor voltage difference equals the battery difference just like in a.

2)

Again, the bulbs look exactly the same during discharge as they do in charging, dimmer than A but lasting longer. (Note, if wire on the right connecting the bulbs through switch 2 had a resistor or another capacitor or battery, the discharge would be different! Discharges aren't always the same as charges; they just are in this case since R and C are the same….) At the end, the capacitor gets completely discharged and has zero voltage difference.

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