*Notes to a video lecture on http://www.unizor.com*

__Direct Current - Ohm's Law - Problems 4__

*Problem A*

Given a circuit presented on a picture below.

Initially, a red switch is in position

**A**to fully charge a

*capacitor*of capacity

**C**from a battery producing a direct current with voltage

**V**.

When a capacitor is fully charged, a switch is moved to position

**B**, disconnecting a capacitor from a battery and forming a new circuit that includes only a fully charged capacitor and a

*resistor*of resistance

**R**.

When a switch is in position

**B**, a

capacitor starts discharging its charge through a resistor. It's charge

will gradually diminish to zero, when all excess electrons on its one

plate will flow through a resistor to a plate with deficiency of

electrons.

During this process of discharge the electric current in a circuit that

contains a capacitor and a resistor will change from some maximum value

in the beginning of this process to zero, when the discharge is

completed.

Find the charge on a capacitor

*and an electric current flowing trough a resistor*

**Q(t)***as functions of time*

**I(t)***.*

**t***Solution*

Assume, our switch is in position

**A**, and we are at the charging stage, when the battery of voltage

*is charging a capacitor of capacity*

**V***.*

**C**The capacity of a capacitor is defined as the constant ratio of a charge

accumulated by a capacitor to a voltage applied to its plate (see

"Capacitors" lecture in the "Electromagnetism - Electric Field"

chapter):

**C = Q/V**Therefore, the full charge of a capacitor at the end of the first stage of charging is

*.*

**V·C**Then we flip a switch into position

**B**, starting the second stage - discharging of a capacitor through a resistor.

At the beginning of this second stage a capacitor is fully charged. So, at time

*its charge is*

**t=0**

**Q(0) = V·C**The charge on a capacitor at any time produces a voltage between its plates

**V(t) = Q(t)/C**This voltage produces a current flowing through a resistor

*that, according to the Ohm's Law, should be equal to*

**I(t)**

**I(t) = V(t)/R**From the two equations above we conclude

**Q(t)/C = I(t)·R**This is our first equation that connects two time-dependent (that is,

functions of time) variables - an electric current in a circuit

*and a charge on a capacitor*

**I(t)***.*

**Q(t)**The second functional equation is, basically, a definition of an

electric current as the rate of electric charge flowing in a circuit

(that is,

*amperage*is how much electricity in

*coulombs*flows through a circuit per unit of time - a

*second*).

Mathematically speaking, an electric current is the first derivative of

an electric charge by time, taken with a sign that depends on the

direction of the change of the charge (

*plus*if the charge is increasing and

*minus*if decreasing):

**I(t) = −**d**Q(t)/**d**t**Considering the charge

*is decreasing and, therefore,*

**Q(t)**its derivative is negative, while we would like the electric current to a

be a positive number, we have to use a minus sign in this equation.

This is our second functional equation (that happens to be differential)

connecting two functions - an electric current in a circuit

*and a charge on a capacitor*

**I(t)***.*

**Q(t)**Now we have two functional equations, one of them is differential, and an initial condition:

**Q(t)/C = I(t)·R**

**I(t) = −**d**Q(t)/**d**t**

**Q(0) = V·C**It's up to our mathematical skills to solve this system of equations.

First, we substitute

*from the second equation into the first, getting a differential equation for*

**I(t)**

**Q(t)**

**Q(t)/C = −R·**d**Q(t)/**d**t**This can be converted into

*d*

**Q(t)/Q(t) = −**d**t/(R·C)**or

*d*[

*]*

**ln(Q(t))***[*

**=**d*]*

**−t/(R·C)**If differentials of two functions are equal, the functions themselves

are just separated by a constant that can be determined using the

initial condition. Let denote that constant as

*.*

**K**

**ln(Q(t)) = −t/(R·C) + K**or, applying an exponent to both sides of this equation,

**Q(t) = e**^{K}·e^{−t/(R·C)}It's time to use the initial condition

*to determine the multiplier*

**Q(0)=V·C***.*

**e**^{K}For

*the right side of an expression for*

**t=0***equals to*

**Q(t)***. Therefore, this multiplier equals to*

**e**^{K}*.*

**V·C**Therefore, the final expression for a charge on a capacitor as a function of time

*is*

**Q(t)**

**Q(t) = V·C·e**^{−t/(R·C)}So, a charge on a capacitor is exponentially diminishing.

From the expression of

*we can find the expression on an electric current going through a resistor, using the equation*

**Q(t)**

**I(t) = −**d**Q(t)/**d**t**from which follows

*[*

**I(t) = −**d*]*

**V·C·e**^{−t/(R·C)}*[*

**/**d**t =**

= −V·C·d= −V·C·

*]*

**e**^{−t/(R·C)}

**/**d**t =**

= −(V·C)·(−1/(R·C))·e

= (V/R)·e= −(V·C)·(−1/(R·C))·e

^{−t/(R·C)}== (V/R)·e

^{−t/(R·C)}*Answer*

**Q(t) = (V·C)·e**^{−t/(R·C)}The multiplier

*is the initial full charge of a capacitor.*

**V·C**

**I(t) = (V/R)·e**^{−t/(R·C)}The multiplier

*is the current that would flow through a resistor, if there were no capacitor. This follows from the Ohm's Law.*

**V/R**