## Tuesday, September 29, 2020

### UNIZOR.COM - Physics4Teens - Electromagnetism - Ohm's Law - Problems 4

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 Q(t) and an electric current flowing trough a resistor I(t) as functions of time t.

Solution

Assume, our switch is in position A, and we are at the charging stage, when the battery of voltage V is charging a capacitor of capacity 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 t=0 its charge is

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 I(t) that, according to the Ohm's Law, should be equal to

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 I(t) and a charge on a capacitor 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) = −dQ(t)/dt

Considering the charge Q(t) is decreasing and, therefore,
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 I(t) and a charge on a capacitor 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) = −dQ(t)/dt

Q(0) = V·C

It's up to our mathematical skills to solve this system of equations.

First, we substitute I(t) from the second equation into the first, getting a differential equation for Q(t)

Q(t)/C = −R·dQ(t)/dt

This can be converted into

dQ(t)/Q(t) = −dt/(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) = eK·e−t/(R·C)

It's time to use the initial condition Q(0)=V·C to determine the multiplier eK.

For t=0 the right side of an expression for Q(t) equals to eK. Therefore, this multiplier equals to V·C.

Therefore, the final expression for a charge on a capacitor as a function of time Q(t) is

Q(t) = V·C·e−t/(R·C)

So, a charge on a capacitor is exponentially diminishing.

From the expression of Q(t) we can find the expression on an electric current going through a resistor, using the equation

I(t) = −dQ(t)/dt

from which follows

I(t) = −d[V·C·e−t/(R·C)]/dt =

= −V·C·
d
[e−t/(R·C)]/dt =

= −(V·C)·(−1/(R·C))·e−t/(R·C) =

= (V/R)·e−t/(R·C)

Q(t) = (V·C)·e−t/(R·C)

The multiplier V·C is the initial full charge of a capacitor.

I(t) = (V/R)·e−t/(R·C)

The multiplier V/R is the current that would flow through a resistor, if there were no capacitor. This follows from the Ohm's Law.

## Tuesday, September 8, 2020

### 3-Phase AC Problem: UNIZOR.COM - Physics4Teens - Electromagnetism - Alte...

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

Problems on AC Induction

Problem A

Three-phase generator has four wires coming out from it connected to a "star" with three phase wires carrying sinusoidal EMF shifted by 120°=2π/3 from each other and one neutral wire.

The angular speed of a generator's rotor is ω.

Assume that the peak difference in electric potential between each phase wire and a neutral one is E.

Describe the difference in electric potential between each pair of phase wires as a function of time.

Solution

The difference in electric potential between phase wires and a neutral one can be described as

Phase 1: E1(t)=E·sin(ωt)

Phase 2: E2(t)=E·sin(ωt−2π/3)

Phase 3: E3(t)=E·sin(ωt+2π/3)

The difference in electric potential between phase 1 wire and phase 2 wire can be represented as

E1,2(t) = E1(t) − E2(t)

Similarly, the difference in electric potential between two other pairs of phase wires is

E2,3(t) = E2(t) − E3(t)

E3,1(t) = E3(t) − E1(t)

Let's calculate all these voltages.

E1,2(t) = E·sin(ωt) − E·sin(ωt−2π/3) =

= E·sin((ωt−π/3)+π/3) − E·sin((ωt−π/3)−π/3) =

[substitute φ=ωt−π/3]

= E·sin(φ+π/3)−E·sin(φ−π/3) =

= E·
[sin(φ+π/3)−sin(φ−π/3)]

Let's simplify the trigonometric expression.

sin(φ+π/3) − sin(φ−π/3) =

= sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) −

− sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) =

= 2cos(φ)·sin(π/3) =

= √3cos(φ) =

= √3cos(ωt−π/3)

Therefore,

E1,2(t) = E√3cos(ωt−π/3)

As we see, the electric potential between phase 1 and phase 2 wires is also sinusoidal (since cos(x)=sin(x+π/2), cos(ωt−π/3) equals to sin(ωt+π/6)),
but shifted in time, and its peak voltage is greater than the peak
voltage between a phase wire and a neutral one by a factor of √3.

Similar factor difference of √3 is between effective voltages of these pairs of wires.

Analogous calculations for the other pairs of phase wires produce the following.

E2,3(t) = E·sin(ωt−2π/3) − E·sin(ωt+2π/3) =

[substitute φ=ωt]

= sin(φ)·cos(2π/3) −

− cos(φ)·sin(2π/3) −

− sin(φ)·cos(2π/3) −

− cos(φ)·sin(2π/3) =

= −2cos(φ)·sin(2π/3) =

= −√3cos(φ) =

= −√3cos(ωt)

Therefore,

E2,3(t) = −E√3cos(ωt)

Also the same factor difference of √3 relative to phase/neutral voltage.

Finally, the third phase/phase voltage calculations produce the following.

E3,1(t) = E·sin(ωt+2π/3) − E·sin(ωt) =

= E·sin((ωt+π/3)+π/3) − E·sin((ωt+π/3)−π/3) =

[substitute φ=ωt+π/3]

= E·sin(φ+π/3)−E·sin(φ−π/3) =

= E·
[sin(φ+π/3)−sin(φ−π/3)]

Let's simplify the trigonometric expression.

sin(φ+π/3) − sin(φ−π/3) =

= sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) −

− sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) =

= 2cos(φ)·sin(π/3) =

= √3cos(φ) =

= √3cos(ωt+π/3)

Therefore,

E3,1(t) = E√3cos(ωt+π/3)

Again, the same factor difference of √3 relative to phase/neutral voltage.

CONCLUSION

The voltage between any two phase wires is in magnitude greater than phase to neutral voltage by √3. Both are sinusoidal, but one is shifted in time relatively to another.

The phase to phase effective voltage (which is by √2 less then peak voltage) also is greater than phase to neutral by the same √3.

EXAMPLES

If phase to neutral effective voltage is 127V, the phase to phase effective voltage is 220V.

If phase to neutral effective voltage is 220V, the phase to phase effective voltage is 380V.