Current & Circuits
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Quiz on Friday (This weeks
Examination #2 one week from
Friday: March 11th.
We discussed current, resistance,
power and the elements of the
microscopic theory of resistivity.
Today we start electric circuits after
a short diversion.
The Spectrum of Conductors
VERY IMPORTANT MATERIAL
Silicon Crystal Lattice
Add an impurity
Effect of Impurities on r
P N P
Silicon Transistor CHIPS
Back to the Mundane
Fun and Frolic With Electric Circuits
Power Source in a Circuit
The ideal battery does work on charges moving
them (inside) from a lower potential to one that is
A REAL Power Source
is NOT an ideal battery
Internal Resistance V
E or Emf is an idealized device that does an amount
of work E to move a unit charge from one side to
By the way …. this is called a circuit!
A Physical Battery
Back to Potential
Change in potential as one circuits
this complete circuit is ZERO!
Represents a charge in space
Consider a “circuit”.
This trip around the circuit is the same as a path
THE CHANGE IN POTENTIAL FROM “a” AROUND
THE CIRCUIT AND BACK TO “a” is ZERO!!
In a real circuit, we can neglect the
resistance of the wires compared to the
We can therefore consider a wire in a circuit to
be an equipotential.
A resistor allows current to flow from a
high potential to a lower potential.
The energy needed to do this is supplied
by the battery.
NEW LAWS PASSED BY THIS SESSION
OF THE FLORIDUH LEGISLATURE.
The sum of the voltage drops (or rises)
as one completely travels through a
circuit loop is zero.
Sometimes known as Kirchoff’s loop
The sum of the currents entering (or
leaving) a node in a circuit is ZERO
TWO resistors again
V iR iR1 iR2
R R1 R2
General for SERIES Resistors
A single resistor can be modeled
ADD ENOUGH RESISTORS, MAKING THEM SMALLER
AND YOU MODEL A CONTINUOUS VOLTAGE DROP.
Take a trip around this circuit.
Consider voltage DROPS:
-E +ir +iR = 0
E=ir + iR
Might as well repeat the parallel thing!
i i1 i2
V V V
V R1 i1 R2 i2 R R1 R2
1 1 1
R R1 R2
R j Rj
1 1 1 50 1
R 20 30 600 12
NOTICE ASSUMED DIRECTION OF
-E1 –i1R1 + i2R2 + E2 +i1R1 = 0
-i3R1 + E2 – E2 –i2R2 =0 NODE
I3 +i2 = i1
In the figure, all the resistors have a resistance of 4.0 and all the (ideal)
batteries have an emf of 4.0 V. What is the current through resistor R?
The Unthinkable ….
Initially, no current
through the circuit
Close switch at (a)
and current begins to
flow until the
capacitor is fully
If capacitor is charged
and switch is switched
to (b) discharge will
Close the Switch
E iR 0
I need to use E for E dq q
Note RC = (Volts/Amp)(Coul/Volt) dt C
= Coul/(Coul/sec) = (1/sec) or
dq q E
dt RC R
This is a differential equation.
To solve we need what is called a
particular solution as well as a
We often do this by creative
“guessing” and thenb matching the
guess to reality.
You may oir may not have studied
this topic … but you WILL!
q q p Ke at
Look at particular solution :
dq q E
dt RC R
When the device is fully charged,dq/dt 0 and
q p CE
When t 0, q 0 and from solution
0 CE K
dq q E
and q CE(1 - e -at )
dt RC R
CE (ae at ) CE(1 - e -at ) E / R
for t 0
CEa 0 E/R
q=CE(1-e-t/RC) and i=(CE/RC) e-t/RC
E t / RC
Discharging a Capacitor
qinitial=CE BIG SURPRISE! (Q=CV)
q q0 e t / RC
dq q0 t / RC
In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase
of thermal energy Eth in the resistor as a function of time t.
(a)What is the electric potential across the battery? (60)
(b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39)
(c) Did you put your name on your paper? (1)
Looking at the graph, we see that the
resistor dissipates 0.5 mJ in one second.
Therefore, the POWER =i2R=0.5 mW
P 0.5 mW
i2 2.38 10 5 amp 2
R 21 Ω
i 4.88 10 3 amp 4.88 ma Voltage drop across the reisitor iR or
V iR 4.8810-3 amp 21 102mV
If the resistance is doubled what is the
power dissipated by the circuit?
R 42 V 102 mV
V 102 10
P i R 0.248mJ