Lecture 20

Readings

http://umdberg.pbworks.com/w/page/105379521/Example%3A%20Batteries%20in%20series%20and%20parallel

http://umdberg.pbworks.com/w/page/106206591/Example%3A%20A%20complex%20network

http://icircuitapp.com

V o l t a g e (V) = C u r r e n t (I) * R e s i s t a n c e (R)

Batteries in Series

C u r r e n t (I) = \frac{# O f B a t t e r i e s (n) * E l e c t r i c F i e l d o f B a t t e r y (E)}{E x t e r n a l R e s i s t a n c e (R) + (# O f B a t t e r i e s (n) * I n t e r n a l R e s i s t a n c e (r))}

Batteries in Parallel

C u r r e n t (I) = \frac{# O f B a t t e r i e s (n) * E l e c t r i c F i e l d o f B a t t e r y (E)}{I n t e r a n l R e s i s t a n c e (r) + (# O f B a t t e r i e s (n) * E x t e r n a l R e s i s t a n c e (R))}

If the external resistence is higher than batteries in parallel will live long and if the internal resistence is higher then the batteries in series will be live long.
Usually, the internal resistence of batteries are very low so connecting in parallel will make the batteries live long.
But if the external resistance is high in this case the batteries will live long but you will not get much power from it.

C o n d u c t a n c e (g) = C o n d u c t i v i t y o f M a t e r i a l (σ) * \frac{a r e a}{l e n g t h}

As the length increases, conductance decreases
As the diameter decreases, conductance decreases
Each conductance is associated with its own "battery"
Each conductance has a current that flows through it
The total current through the membrane is the sum of the three currents: Sodium , Potassium, Leak

Batteries in a Circuit

Basic:

V o l t a g e (V) = C u r r e n t (I) * R e s i s t a n c e (R)

C u r r e n t (I) = \frac{V o l t a g e (V)}{R e s i s t a n c e (R)}

P o w e r = \frac{J o u l e s}{1 S e c o n d} = \frac{E n e r g y}{T i m e} = W a t t s (w)

"Power":
- = "brightness" of a light bulb
- $Power\ (P) = Current\ (I)^2 * Resistance\ (R)$
  - $V = I * R$
- $Power\ (P) = \frac{Voltage\ (V)^2}{Resistance\ (R)}$
  - $I = \frac{V}{R}$
- Use whichever of the three forms above is most convenient

If batteries are in series:

$V_0$ potential.
Choice of "zero" is arbitrary, but the zero can only be at 1 spot (1 wire)

If batteries are in parallel:

Doesn't really change the circuit.
$V_0$ across resistor
Any benefit at all?
- 1 battery only has so much "capacity" , in the sense of the number of charges it can put out.
- Batteries "run down"
- Two batteries in parallel can keep light on longer

Complex Circuit - Kirchhoff's Rules

https://opentextbc.ca/universityphysicsv2openstax/chapter/kirchhoffs-rules/

https://cnx.org/contents/[email protected]:[email protected]/Problems

https://www.slader.com/textbook/9781938168161-university-physics-volume-2-1st-edition/488/problems/37/#

V o l t a g e (V) = C u r r e n t (I) * R e s i s t a n c e (R)

I_{i n} = I_{o u t}

I_{1} = I_{2} + I_{3}

I_{3} = I_{1} - I_{2}

$I_{in} = I_{out} \ce{->} I_1 = I_2 + \left(I_1 + I_2\right)$
Loop 1:
- $V_1 - (I_1 * R_1) - V_2 - ((I_1 - I_2)*R_3) = 0$
Loop 2:
- $V_2 - (I2*R2) - ((I_2-I_1)*R_3) = 0$
$I_1$ $I_2$
So 2 equations and 2 unknowns
$I_2$

V_{2} + (I_{1} * R_{3}) = I_{2} * (R_{2} + R_{3})

I_{2} = \frac{V_{2} + (I_{1} * R_{3})}{R_{2} + R_{3}}

Plug into 1st equation:

V_{1} - (I_{1} * R_{1}) - V_{2} - ((I_{1} - \frac{V_{2} + (I_{1} * R_{3})}{R_{2} + R_{3}}) * R_{3}) = 0

$I_1$ :

I_{1} = \frac{V 1 - V_{2} + \frac{V_{2} * R_{3}}{R_{2} + R_{3}}}{R_{1} + \frac{R_{2} * R_{3}}{R_{2} + R_{3}}}

Algebra can be intense:
- https://www.symbolab.com/
- https://www.mathway.com/Algebra
Point is, set up loops where net change in voltage for entire loop is zero
Then use guassian elimination to solve system