# Discharging Capacitor: Differential Equation in terms of Vs, C, R

• Tekneek
In summary, the voltage across the capacitor is 10V when t<0 and the switch is closed at t=0. The differential equation in terms of Vs, C, and R can be found by balancing the current at node Vc(t). Assuming Vs=0 at t<0, the current through the resistor is Vc/R. The polarity of the voltage across the capacitor is not explicitly stated, so it is assumed to stay at +10V for all time.
Tekneek

## Homework Statement

The voltage across the capacitor is 10V when t<0
The switch is closed at t=0

I have to find the diff eq. in terms of Vs, C, R by balancing the current at node Vc(t). But i am not sure whether the Vs is 0 or 10V at t<0

## The Attempt at a Solution

I am assuming Vs = 0 at t<0 since the voltage across capacitor is 10V. Also, I am not sure what current balance at the node means...I mean when the switch is closed the current through the capacitor is the same as current through the resistor. So,

iR = iC
(Vc/R) = C (dVc/dt)

How do i write this in term of Vs? Isn't the current through the resistor Vc/R since Vs is 0 at the beginning?

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I'm not sure what you mean when you say "I am assuming Vs=0...". I think you mean that you are assuming that the voltage at the negative end of Vs is zero - which would be a fine assumption. It doesn't really matter what point you take as your "ground" or zero voltage level.

If you take Vs- as zero, then obviously Vs+ will be "Vs" and Vc(0) will be at (Vs-10v). Or, perhaps, (Vs+10V), since the polarity of the voltage across the capacitor is not explicitly specified.

The voltage across the capacitor is 10V when t<0
Are you sure this is what it says--"across the capacitor"? Can you quote precisely the wording of the question?

You have to specify the polarity of the 10V across C before proceeding with the problem. It could be +10V or -10V.

Since the source is not mentioned in the problem statement you should asume it stays at +10V for all time -infinity to +infinity.

The differential equation for a discharging capacitor can be written as:

Vc(t) = Vs * e^(-t/RC)

Where Vc(t) is the voltage across the capacitor at time t, Vs is the initial voltage across the capacitor, R is the resistance in the circuit, and C is the capacitance of the capacitor.

In this specific scenario, since the voltage across the capacitor is 10V at t<0 and the switch is closed at t=0, we can assume that Vs = 10V. This means that the initial voltage across the capacitor is 10V and it will start to discharge once the switch is closed.

The current balance at the node refers to the Kirchhoff's Current Law, which states that the sum of currents entering a node must equal the sum of currents leaving the node. In this case, the current through the resistor and the current through the capacitor must be equal.

Therefore, we can write the differential equation as:

(Vs - Vc(t))/R = C * dVc/dt

Where Vs is the initial voltage across the capacitor and Vc(t) is the voltage across the capacitor at time t.

I hope this helps clarify your understanding of the differential equation for a discharging capacitor.

## 1. What is a discharging capacitor?

A discharging capacitor is a circuit component that stores electrical energy in the form of an electric charge. When a capacitor is connected to a circuit, it charges up and stores energy. However, when the circuit is disconnected, the capacitor will slowly release this stored energy, known as discharging.

## 2. What is the differential equation for discharging capacitor?

The differential equation for a discharging capacitor is given by dV/dt = -V/RC, where V is the voltage across the capacitor, R is the resistance in the circuit, and C is the capacitance of the capacitor.

## 3. How does the discharging of a capacitor affect voltage and current?

As a capacitor discharges, the voltage across it decreases exponentially, while the current through it decreases linearly. This means that the rate of change of voltage is much faster than the rate of change of current.

## 4. How does the resistance and capacitance of a circuit affect the discharging of a capacitor?

A higher resistance in the circuit will lead to a slower discharging of the capacitor, as there is more opposition to the flow of current. On the other hand, a larger capacitance will result in a slower rate of voltage decrease during discharging, as the capacitor can store more charge.

## 5. What is the time constant of a discharging capacitor?

The time constant of a discharging capacitor is the time it takes for the voltage across the capacitor to decrease to 36.8% of its initial voltage. It is given by the equation RC, where R is the resistance and C is the capacitance of the circuit.

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