Calculating Voltage Decay in a Resistor-Capacitor Circuit

In summary, the conversation discusses the behavior of a circuit with a charged capacitor and a resistor in series, and the relevant equations for calculating the time constant and final voltage. It also addresses the behavior of a circuit with an uncharged capacitor and a bulb in series, concluding that there will be a delay before the bulb emits light and it will gradually become brighter as the capacitor charges.
  • #1
SMUDGY
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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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  • #2
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?
 
  • #3


To calculate the voltage decay in a resistor-capacitor circuit, we can use the equation V(t) = V(0)e^(-t/RC), where V(t) is the voltage at time t, V(0) is the initial voltage, R is the resistance, and C is the capacitance.

In this case, we have V(0) = 35 microfarads, R = 120 ohms, and we are looking for the time t until the voltage falls to 10^6/D of its original maximum value.

To find t, we can rearrange the equation as t = -RCln(V(t)/V(0)). Substituting in the values, we get t = - (120 ohms)(35 microfarads) ln(10^6/D), where D is the desired fraction of the original maximum voltage.

To solve for t, we will need to use logarithms. We can use the natural logarithm (ln) on both sides to get rid of the e and then use the logarithm rules to simplify the equation.

For example, if D = 2, then the equation becomes t = - (120 ohms)(35 microfarads) ln(10^6/2) = - (120 ohms)(35 microfarads) ln(500,000) ≈ 1.13 seconds.

If you are not familiar with logarithms, I recommend consulting a math or physics textbook or reaching out to a tutor for assistance.
 

1. How do you calculate the voltage decay in a resistor-capacitor circuit?

The voltage decay in a resistor-capacitor circuit can be calculated using the formula V = V0 * e^(-t/RC), where V is the voltage at time t, V0 is the initial voltage, R is the resistance, and C is the capacitance.

2. Why is it important to calculate voltage decay in a resistor-capacitor circuit?

Calculating voltage decay in a resistor-capacitor circuit is important because it helps us understand how the circuit behaves over time. This information is crucial in designing and analyzing electronic circuits.

3. What factors affect the voltage decay in a resistor-capacitor circuit?

The voltage decay in a resistor-capacitor circuit is affected by the values of the resistance and capacitance, as well as the initial voltage. Additionally, the time constant (RC) of the circuit also plays a significant role in determining the rate of voltage decay.

4. How does the voltage decay in a resistor-capacitor circuit change with time?

The voltage decay in a resistor-capacitor circuit follows an exponential decay curve. This means that initially, the voltage decreases rapidly, but as time goes on, the rate of decay slows down until it reaches a steady-state voltage.

5. Can voltage decay be reversed in a resistor-capacitor circuit?

No, voltage decay cannot be reversed in a resistor-capacitor circuit. However, the voltage can be increased by applying a higher initial voltage or by decreasing the resistance or capacitance values in the circuit.

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