Calculating Voltage Decay in a Resistor-Capacitor Circuit

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SUMMARY

The discussion focuses on calculating voltage decay in a resistor-capacitor (RC) circuit, specifically with a capacitor of 35 microfarads and a resistor of 120 ohms. The time constant (T) is determined using the formula T = cR, leading to a time calculation for the voltage to drop to 10^6/D of its initial value. Additionally, the behavior of a bulb in a circuit with an uncharged capacitor is analyzed, confirming that the bulb will initially remain off, then gradually brighten as the capacitor charges over time.

PREREQUISITES
  • Understanding of RC circuits and their components
  • Familiarity with the time constant formula T = cR
  • Knowledge of exponential decay in voltage (V(Final) = V(initial) e^-t/cr)
  • Basic logarithmic functions for solving equations
NEXT STEPS
  • Study the implications of time constants in RC circuits
  • Learn how to derive voltage decay equations for different capacitor and resistor values
  • Explore the behavior of bulbs in RC circuits during charging and discharging phases
  • Investigate the effects of varying resistor values on charging time and voltage decay
USEFUL FOR

Electrical engineering students, hobbyists working with RC circuits, and anyone interested in understanding capacitor charging behavior in practical applications.

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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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