Problem: Two batteries (in series), a bulb, an uncharged capacitor, and a switch are connected together in a circuit with the switch open. (Note, the positive end of the battery is connected to the positive terminal of the capcitor) 1. what happens to the current in this circuit and the voltage across the bulb after the switch is closed? (Do they increase or decrease) How do your answers to question relate to the presence of the capacitor in the circuit?2. What happens to the voltage across the capacitor over time when starting with a partially-charged capacitor? Why? 3. If you re-ran the lightbulb-timing experiment, except you started with a partially-charged capacitor, what would be different and why?

FREE Expert Solution

(1)

i=i0e-t/RC=VbatteryRe-t/RC

From the charging current expression, the current in the circuit decreases with time. 

The voltage across the bulb is given by ohm's law; V = iR.

The resistance of the bulb is constant. When i decreases, the voltage across the bulb decreases.

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Problem Details

Two batteries (in series), a bulb, an uncharged capacitor, and a switch are connected together in a circuit with the switch open. (Note, the positive end of the battery is connected to the positive terminal of the capcitor) 

1. what happens to the current in this circuit and the voltage across the bulb after the switch is closed? (Do they increase or decrease) How do your answers to question relate to the presence of the capacitor in the circuit?

2. What happens to the voltage across the capacitor over time when starting with a partially-charged capacitor? Why? 

3. If you re-ran the lightbulb-timing experiment, except you started with a partially-charged capacitor, what would be different and why?

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