II Charging and discharging capacitors A Suppose an uncharge

II. Charging and discharging capacitors A. Suppose an uncharged capacitor is connected in series with a battery and bulb as shown Predict the behavior of the bulb when the switch is closed. Explain. I. Set up the circuit and check your prediction. If your prediction is in conflict with your observation, how can you account for your observation? Without using a voltmeter, determine the potential difference across the capacitor at the following times: 2. just after the switch is closed. Explain how you can tell. (Hint: Compare the brightness of the bulb to the brightness of a bulb connected to a battery in a single-bullb circuit without a capacitor.) a long time after the switch is closed. Explain how you can tell Use a voltmeter to check your predictions. (Hint: Be sure to discharge the capacitor completely after each observation.)

Solution

Based on the Chegg guidelines since you have asked 5 independant questions, I am bound to answer 4 subparts of only one chronologically appearing. Please upload all the questions individually if you need the answer. Also specificy the doubt.
A1. The bulb will tend to light up initially when the circuit is closed and will gradually diminish in intensity and will shut down after some time. The rule of thumb is when a capacitor is connected in a direct circuit always remember that it works like a copper wire and the bulb gets the voltage required, but as and when time passes slowly it works like a break point in a circuit and the DC behaves like an open circuit.
A2. Voltage = EMF of Cell at time t = inf
Voltage = 0 at time t = 0 . Explanation based on a rule suggested above
It can also be seen from the decaying voltage expression of capacitor = E*exp(-t/tao)