Electric Currents
Physics Department, New York
City College of Technology
Key words








Electric battery
Electric current
Resistance
Ohm’s law
Resistivity
Electric energy
Electric power
Direct current



Alternating current
Average power
rms values
Electric battery

A battery produces electricity by
transforming chemical energy into
electrical energy.
Electric battery

In a diagram of a circuit, it is denoted
by the symbol
+

Electric current

Electric current is a flow of charge
Electric current

Current is defined as




Q
I
t
I is current
∆Q is the amount of charge that passes
through the conductor at any location
Δt is the time interval
The unit of I is ampere (A)
Direction of electric current
Electrical resistance

Defined as




V
R
I
R is the resistance of a wire
V is the potential difference applied across
the wire
I is the current
The symbol of a resistor is
Electron motion model

Disc 17, #22
Ohm’s law



Resistance R is a
constant
independent of V,
i.e., V  IR
Experimentally
found by Ohm
Generally holds in
metal conductors in
room temperature
Ohm’s law



Disc 17, #19
Disc 17, #20
Disc 17, #21
Resistivity

It is experimentally found that



L
R
A
ρ is resistivity and depends on the material
L is the length of a wire
A is the cross-sectional area
Resistivity
Example #1

A flashlight bulb draws 300mA
from its 1.5-V battery.
(a) What is the resistance of
the bulb?
(b) If the battery becomes
weak and voltage drops to
1.2V, how would the current
change?
Example #1—continued
(a)
(b)
300 mA  0.3 A
V 1.5V
R 
 5.0
I 0.3 A
V 1.2V
I 
 0.24 A  240 mA
R 5.0
Example #2

Connect stereo to speakers.
(a) If each wire must be
20m long, what diameter
copper wire should be used
to keep the resistance less
than 0.10 Ω? (b) If the
current is 4.0A, what is the
potential difference across
each wire?
Example #2—continued
(a)
 for copperis 1.68108   m
L (1.68108   m)(20m)
A  
 3.4 106 m 2
R
0.10
A  r , r 
2
A

 1.04103 m  1.04m m,
d  2r  2.08m m.
(b)
V  IR  (4.0 A)(0.1)  0.4V
Electric energy
Energy transformation between electric and
other forms are very common




Motors transform electric energy into mechanical
energy
In electric heaters and hair dryers, electric
energy is transformed into thermal energy
In a lightbulb, electric energy is transformed into
light and thermal energy
Electric power

The electric power transformed by any device
is P  IV





P is electric power
I is current
V is potential difference
The unit of electric power is Watt (W)
For a resistance R, we have
P  IV  I ( IR)  I 2 R
V
V2
P  IV  ( )V 
R
R
Example #3

Calculate the
resistance of a 40-W
automobile headlight
designed for 12V.
Example #3—continued
P  40W ,V  12V
V 2 (12V ) 2
R

 3.6
P (40W )
Alternating current

The voltage and current produced by an
electric generator are sinusoidal
V  V0 sin 2ft  V0 sin t
V V0
I   sin t
R R



V0 is the peak voltage
I0 is the peak current
The frequency f is the number of complete
oscillations per second, and   2f
DC and AC
AC power


The power transformed in a resistance
R at any instant is P  I 2 R  I 02 R sin 2 t
The average power is calculated as
1 2
P  I0 R
2
or
1 V02
P
2 R
AC power—continued
rms (root-mean-square)
values

rms values and peak
values
I rms  I 
2
Vrms  V 2 
I0
2
V0
2
 0.707I 0
 0.707V

The average power
in rms values:
P  I rms Vrms
1 2
2
P  I 0 R  I rms
R
2
2
2
V
V
1 0
P
 rms
2 R
R
Example #4

(a) Calculate the
resistance and the
peak current in a
1000-W hair dryer
connected to a 120-V
line. (b) What if it is
connected to a 240-V
line in Britain?
Example #4
(a)
I rms
P
1000W


 8.33A
Vrms
120V
I 0  2I rms  11.8 A
Vrms 120V
R

 14.4
I rms 8.33A
(b)
2
Vrms
(240V ) 2
P

 4000W
R
14.4