L07

Chapter 27 DC Circuits

DC Circuits

Combining Resistors and

Capacitors

Time Dependent Circuits

Work done by a battery on charge





Here εB  εA

W

ε  [volts!]

q

dW

ε  [volts!]

dq

Real Battery and Single Loop circuits… What’s the current ?









Conservation of energy:

Kirchoff’s first Law: Sum of voltages in a closed loop is zero.



Va    ir  iR  Va

Real Circuit with ammeter   ir  iR  0

and voltmeter 

i

rR

PR  i 2 R

 2R



(r  R) 2

Equivalent Resistance

Resistors in Series









Series requirements V  V1  V2  ...  Vn

– Conservation of energy Apply Ohm’s Law to each resistor

– Potential differences add

– Current is constant IReq  IR1  IR2  ...  IRn



Req  R1  R2  ...  Rn

Resistors in parallel

Parallel requirements

– Charge conservation

– Currents must add

– Potential difference is

same across each resistor





i  i1  i2  i3

1 1 1 1

Apply Ohm’s Law to each resistor    ... 

Req R1 R2 Rn

V V V V

   ... 

Req R1 R2 Rn

Example 27-2

What is current through

battery?







What is current

through i2 and i3 ?

Kirchhoff’s Rules

1 The algebraic sum of the currents Signs for Rule 2

entering a junction is zero. The direction of travel when traversing

(Conservation of Charge) the loop is from a to b.

2 The algebraic sum of the changes in

electric potential difference around

any closed circuit loop is zero.

(Conservation of Energy)

Problem 27-3



Find the currents in each of the

three legs of the circuit,





i1 , i2 , i3



Three unknowns, need three equations.

Also, since batteries are in the loops, cannot reduce the resistances

since none in parallel or series

Example or Applying Kirchhoff’s

Rules

Apply Kirchhoff’s second

rule to the closed path in red,

traversing it clockwise



5.0V  3.0I1  5.0I 2  0

Apply Kirchhoff’s second rule

Apply Kirchhoff’s first rule to

to the closed path in green,

the three wire junction at

traversing it clockwise

the bottom of the diagram

Note the sign changes for

I 2  I1  I 3  0 some of the elements





5.0V  3.0I1  10.0V  7.0I 3  0

Another, example: applying Kirchhoff’s Rules



I 2  I1  I 3  0



5.0V  3.0I1  5.0I 2  0



 5.0V  3.0I1  7.0I 3  0



Solve the equations

simultaneously for the values I1  0.141A

if I. If I is negative the I 2  0.915 A

current is in the opposite

direction I 3  0.774 A

RC Circuits and Time dependence

Time dependence

Recall Lab 7! Resistor

slows down the charging

of the capacitor



Time dependent behavior

(transient) 2 cases: switch at

“a” or at “b”



a) Charging

b) discharging

a) Charging the Capacitor









Note:   RC

is called the time constant

What are the units?

In position “a” Charging the

Capacitor

  Vc t   VR t   0



time dependent!

Use Kirchhoff’s Voltage

Loop rule

q t 

-  IR  0

First order Differential Equation. C

Solution, integrate once. Did this in dq t  q t  

 

lab-6. dt RC R

Can check that these are the 

q (t)  C 1  e t RC 

solutions by differentiation 

Vc (t)   1  e t RC



What’s VR across resistor?

Find the current and multiply by R







q (t)  C 1  e t RC 

dq (t)

I (t)  t ake derivat ive and get

dt



I (t)  e t RC ....or ,

R

dq (t)

VR  R

dt

  e t RC

Discharging Position b)







Kirchhoff's Voltage Law



Vc  VR  0

dq (t) q (t)



dt RC

q (t)  q0 e t RC  CV0 e t RC

Vc (t)  V0 e t RC

Example: Time Constant



R=10

C=1F



t 1 / 2  RC ln 2

How long does it take the  6.93 sec

capacitor to reach ½ its final

charge, if the capacitor is

uncharged at t = 0?