Transcript Multi-loop and RC Circuits

Physics 121: Electricity &
Magnetism – Lecture 8
DC Circuits
Dale E. Gary
Wenda Cao
NJIT Physics Department
emf and emf devices
V+
W
V-





The term emf comes from the outdated phrase electromovitive force.
emf devices include battery, electric generator, solar cell, fuel cell,……
emf devices are sources of charge, but also sources of voltage (potential difference).
emf devices must do work to pump charges from lower to higher terminals.
Source of emf devices: chemical, solar, mechanical, thermal-electric energy.
October 24, 2007
Emf





We need a symbol
for emf,
and we will use a script  to
represent emf.  is the potential
difference between terminals of
an emf device.
The SI unit for emf is Volt (V).
We earlier saw that there is a
relationship between energy,
charge, and voltage dqV  dW
So,
dW

dq
Power dW  dq  idt  Pdt
P  i




Real emf device:
Ideal emf device:
V=
(open loop)
V=
(close loop)
(open or close loop) V <



October 24, 2007
Work, Energy, and EMF
1.
The following circuit contains two ideal rechargeable batteries A
and B, a resistance R, and an electric motor M that can lift an
object by using energy it obtains from charge carriers in the circuit.
Which statement is correct?
A.
Battery B lost chemical energy
Battery B charge Battery A
Battery B provides energy to Motor M
Battery B provides energy to heat R
All of the above are true
B.
C.
D.
E.
October 24, 2007
Kirchhoff’s Rules

Junction Rule: At any junction,
the sum of the currents must
equal zero:
i  0
junction


Loop Rule: The sum of the potential
differences across all elements
around any closed circuit loop must
be zero:
 V  0
closedloop

For a move through a
resistance in the
direction of current, the
change in potential is –iR;
in the opposite direction
it is +iR.
For a move through an
ideal emf device in the
direction of the emf
arrow, the change in
potential is +; in the
opposite direction it is - .
October 24, 2007
A Single-Loop Circuit

Travel clockwise from a:
Vba  Vb - Va  
Vda  Vd - Va  0
Vcb  Vc - Vb  0
b
Vdc  Vd - Vc  -iR
 - iR  0

 V  + 0 - iR + 0  0 i 

c
Vbc  Vb - Vc  0
Vab  Va - Vb  -
 V 0 + iR + 0 -   0
d

Vad  Va - Vd  0
closedloop
i
R
closedloop
Vcd  Vc - Vd  iR
R
Travel counterclockwise from a:
Vda  Vd - Va  0
b
Vcd  Vc - Vd  iR
Vbc  Vb - Vc  0
Vab  Va - Vb  -
 V 0 + iR + 0 -   0
closedloop
c
iR -   0
i

d
R
October 24, 2007
Resistances in Series

Junction Rule: When a potential difference V is
applied across resistances connected in series, the
resistances have identical currents i:
i  i1  i2  i3

Loop Rule: The sum of the potential differences
across resistances is equal to the applied potential
difference V:

i

(a)  - iR1 - iR2 - iR3  0
R1 + R2 + R3
(b)  - iR  0

eq
i
Req  R1 + R2 + R3
Req

The equivalent resistance of a series combination of
resistors is the numerical sum of the individual
resistances and is always greater than any individual
resistance.
n
Req   Ri
i 1
October 24, 2007
Resistances in Parallel

When a potential difference V is applied across
resistances connected in parallel, the resistances all
have that same potential difference V.
V  V1  V2  V3
(a) Junction Rule:
i1 
V
V
V
1
1
1 
, i2 
, i3 
i  i1 + i2 + i3  V  +
+ 
R1
R2
R3
 R1 R2 R3 
(b) Loop Rule:
V
i

V - iReq  0
Req

1
1
1
1

+
+
Req
R1 R2 R3
The inverse of the equivalent resistance of two or
more resistors in a parallel combination is the sum
of the inverse of the individual resistances.
n
Furthermore, the equivalent resistance is always 1
1

less than the smallest resistance in the group.
Req
i 1 Ri
October 24, 2007
Resistors in series and parallel
2. Four resistors are connected as shown in
figure. Find the equivalent resistance
between points a and c.
c
A.
B.
C.
D.
E.
4 R.
3 R.
2.5 R.
0.4 R.
Cannot determine
from information given.
R

R
R
a
R
October 24, 2007
Capacitors in series and parallel
3. Four capacitors are connected as shown
in figure. Find the equivalent capacitance
between points a and c.
c
A.
B.
C.
D.
E.
4 C.
3 C.
2.5 C.
0.4 C.
Cannot determine
from information given.
C

C
C
C
a
October 24, 2007
Example: Real Battery

Real battery has internal resistance to the internal movement of charge.
 - ir - iR  0

Current:

Potential difference:
clockwise:

Power
i

R+r
Va +  - ir  Vb Vb - Va   - ir   -

R+r
r

R+r
P  iV  i( - ir )  i - i 2 r
October 24, 2007
R
Example: Multiple Batteries
What is the potential difference and power
between the terminals of battery 1 and 2?
 Current i in this single-loop: (counterclockwise)

- 1 + ir1 + iR + ir2 +  2  0
i

1 -  2
r1 + R + r2
 0.2396 A
Potential difference:
1  4.4V ,  2  2.1V , r1  2.3, r2  1.8, R  5.5
(clockwise)
Vb - ir1 + 1  Va
Va - Vb  -ir1 + 1  +3.84V
Va -  2 - ir2  Vc
Va - Vc   2 + ir2  +2.53V

Power:
i 
P 1  iVab  0.92W
P 2  iVac  0.60W
PR  i 2 R  0.32W
P 1  P 2 + PR
A battery (EMF) absorbs power (charges
up) when i is opposite to .
October 24, 2007
Potential difference and battery
4. According to the circuit shown, which
statement is correct?
A.
B.
C.
D.
E.
Vb = + 8.0 V
Va = - 8.0 V
Vb-Va = - 8.0 V
Va = - 12.0 V
Vb = + 12.0 V
October 24, 2007
Multiloop Circuits
Determine junctions, branches and loops.
 Label arbitrarily the currents for each
branch. Assign same current to all
element in series in a branch.
 The directions of the currents are
assumed arbitrarily; negative current
result means opposite direction.
 Junction rule:

i1 + i3  i2
i1 + i3  i2

You can use the junction rule as often as
you need. In general, the number of
times you can use the junction rule is
one fewer than the number of junction
points in the circuit.
October 24, 2007
Multiloop Circuits




Determine loop and choose moving
direction arbitrarily.
When following the assumed current
direction, ir is negative and voltage drops;
Reverse when going against the assumed
current; Emf is positive when traversed
from – to +, negative otherwise.
Loop rule:
badb: left-hand loop in counterclockwise
1 - i1 R1 + i3 R3  0
bdcb: right-hand loop in counterclockwise
- i3 R3 - i2 R2 -  2  0
You can apply the loop rule as often as
needed as long as a new circuit element or
a new current appears in each new
equation.
i1 + i3  i2

Equivalent loop and wise
badcb: big loop in counterclockwise
1 - i1R1 - i2 R2 -  2  0
bcdb: right-hand loop in clockwise
 2 + i2 R2 + i3 R3  0
October 24, 2007
Multiloop Circuits

In general, to solve a particular circuit
problem, the number of independent
equations you need to obtain from the two
rules equals the number of unknown
currents.
i1 + i3  i2
i1 + i3  i2
1 - i1 R1 + i3 R3  0
- i3 R3 - i2 R2 -  2  0

Solution:
i1 
i2 
 1 R2 +  1 R3 -  2 R3
R1 R2 + R2 R3 + R1 R3
 1 R2 -  2 R3 -  2 R1
i3 
-  2 R1 -  1 R2
R1 R2 + R2 R3 + R1 R3
R1 R2 + R2 R3 + R1 R3
October 24, 2007
RC Circuits – Charging a Capacitor
RC circuits: time-varying currents, switch to a
 Start with,
q
 - iR -  0
Loop rule

C

Then,
i
 - iR -
dq
dt
Substituting and rearranging,
 Boundary condition,

q (t  0)  0;


Therefore,
Integrating,
i (t  0) 

R
R
;
dq q
+ 
dt C
q(max)  C ;
i  0;
dq 
q
 dt R RC
dq C
q
q - C

dt RC RC
RC
q
t
dq
1

0 q - C RC 0 dt
q
0
C
dq
1
dt
q - C
RC
t
 q - C 
ln 
RC
 - C 
October 24, 2007
RC Circuits – Charging a Capacitor

Charge,

For charging current,
q(t )  C (1 - e -t / RC )
i (t ) 
dq (t )  -t / RC
 e
dt
R

A capacitor that is being charged initially acts like
ordinary connecting wire relative to the charging current.
A long time later, it acts like a broken wire.

Potential difference,
VC (t ) 
 - iR -
q
0
C
q (t )
  (1 - e -t / RC )
C
t = 0: q = 0, Vc = 0, i = /R;
 t => : q = c, Vc = , i = 0;
 t = RC: q = c(1-e-1) = 0.632c; i = /Re-1 = 0.368 /R

October 24, 2007
Unit of RC
5. If  = RC, what it the unit of  ?
A.
B.
C.
D.
E.
F
C/A
C/V
VF/A
s
(ohmfarad)
(coulomb per ampere)
(ohmcoulomb per volt)
(voltfarad per ampere)
(second)
[] = [RC] =[(V/i)(Q/V)]=[Q/Q/t]=[t]
October 24, 2007
RC Circuits – Discharging a Capacitor
RC circuits: time-varying currents, switch to b
 Start with,
q
- - iR  0
Loop rule
C

dq
1
dt
q
RC
dq q
-R

dt C

Then,

Boundary condition,

Therefore,

Hence,
q
 - iR -
q
0
C
q(t  0)  q0
t
dq
1

q q RC 0 dt
0
q(t )  q0 e -t / RC
q
t
ln    RC
 q0 
i (t ) 
q
dq(t )
 - 0 e -t / RC
dt
RC
t = 0: q = q0 = CV0, i = q0/RC;
 t => : q = 0, i = 0;

October 24, 2007
Resistors in series and parallel
6.
A.
Consider the circuit shown and assume the battery has
no internal resistance. Just after the switch is closed,
what is the current in the battery?
0.
D.
/2R.
2/R.
/R.
E.
impossible to determine
B.
C.
C

R
October 24, 2007
R
Resistors in series and parallel
7.
A.
Consider the circuit shown and assume the battery has
no internal resistance. After a very long time, what is
the current in the battery?
0.
D.
/2R.
2/R.
/R.
E.
impossible to determine
B.
C.
C

R
October 24, 2007
R
Summary









An emf device does work on charges to maintain a potential difference between
its output terminals.
dW

Kirchhoff’s rules:
dq
Loop rule. The algebraic sum of the changes in potential encountered in a
complete traversal of any loop of a circuit must be zero.
Junction rule. The sum of the current entering any junction must be equal to
the sum of the currents leaving that junction.
n
Series resistances: when resistances are in series, they have the same current. Req   Ri
i 1
The equivalent resistance that can replace a series combination of resistance is
n
Parallel resistance: when resistances are in parallel, they have the same
1
1


potential difference. The equivalent resistance that can replace a parallel
Req
i 1 Ri
combination of resistance is,

Single loop circuits: the current in a single loop circuit is given by i  R + r
Power: when a real battery of emf and internal resistance r does work on the
charges in a current I through the battery,
P  iV  i( - ir )  i - i 2 r
RC Circuits: when an emf is applied to a resistance R and capacitor C in series,
dq (t )  -t / RC
q(t )  C (1 - e -t / RC )
i (t ) 
 e
dt
R
RC Circuits: when a capacitor discharges through a resistance R, the charge
decays according to
q(t )  q0 e -t / RC
dq (t ) q0 -t / RC
And the current is
i (t ) 

e
dt
RC
October 24, 2007