Calculating Time Until Voltage Falls in a Circuit with Resistor and Capacitor

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SUMMARY

The discussion focuses on calculating the time until the voltage across a capacitor falls to a specific fraction of its initial value in a circuit containing a resistor and a capacitor. The relevant parameters include a capacitor value of 35 microfarads and a resistor value of 120 ohms. The time constant is calculated using the formula T = cR, and the voltage decay is described by V(Final) = V(initial) e^(-t/cr). Additionally, the behavior of a bulb in a series circuit with an uncharged capacitor is analyzed, confirming that the bulb will initially remain off, then gradually brighten as the capacitor charges.

PREREQUISITES
  • Understanding of RC circuit theory
  • Familiarity with the time constant formula T = cR
  • Knowledge of exponential decay in electrical circuits
  • Basic logarithmic functions for solving equations
NEXT STEPS
  • Study the derivation of the time constant in RC circuits
  • Learn how to apply the exponential decay formula in practical scenarios
  • Explore the behavior of capacitors in series and parallel circuits
  • Investigate the impact of different resistor and capacitor values on circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and hobbyists interested in understanding capacitor charging and discharging behavior in RC circuits.

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If a charged capacitor c=35 micro farads is connected to a resistor R=120 ohms how much time will elapse until the voltage falls to 10^6/D of its original max value?

Ciruit contains a switch, resistor, emf and a capacitor in series.

Relevant equations are Time constant= T=cR and V(Final)=V(initial) e^-t/cr


I think it requires logarithms but i am not sure.

Can you please help me how to answer this.
 
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Another question is:

A circuit containing a switch, Vo, a capacitor and a bulb in series.

The capacior is originally uncharged. Describe the behaviour of the bulb from the instant the switch is on until a long time later.

I think there will be a delay for the bulb to emit light and then it will emit light getting slowly brighter as the capacitor is getting charged up.

Is this right?
 

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