Download Circuits and Circuit Diagrams

Circuits
&
Circuit
Diagrams
Current- the movement of charge.
Symbol is I, unit is the Ampere (A)
I = q/t
1 A = 1 C/s
Conventional current- from the positive
terminal to the negative terminal.
Drift velocity- the speed in which individual
electrons move within the conductor,
opposite the electric field.
The drift velocity is relatively small.
• Resistance- the opposition to the flow
of charge by a material or device.
• Symbol is R, unit is ohms (W).
• Ohms law R = V/I or V = IR
• Resistance depends on the following
factors:
Length – short wires have less resistance
Area- thick wires have less resistance
Material- different materials have
different conductivity of charge
Temperature- heat slows the flow of
charge.
Circuits and Circuit Diagrams
• Circuit = Complete path along which electrons can
flow
• When drawing circuits…use the following symbols
in your diagrams
• Wire/Conductor
• Resistor
• Battery
• Switch
(light bulbs, fans)
Ohm’s Law
• Resistance = Voltage/ Current
•R=V/I
• Units
– Resistance = ohms (Ω)
– Voltage (Potential Difference) = volt (V)
– Current = ampere (A)
Example #1:
• A 30.0 V battery is connected to
a 10.0 Ω resistor. What is the
current in the circuit?
• I = V/R
• I = 30.0 V / 10.0 Ω
• = 3.00 A
Series Circuits
• Single path for electrons to flow
• If any resistor (bulb) in the circuit is
removed or burnt out, then NO BULBS
will light – the circuit is not complete.
• Electric current is the same through
each device / resistor / bulb. (I = V/Rtotal)
• Total resistance to current in the circuit
is the sum of the individual resistors
along the circuit path (Rtotal = RA + RB)
Series Circuits
• Voltage drop across each device
depends directly on its resistance
(V = I x R)
• Total voltage divides among the
individual electric devices in the
circuit (Rtotal = RA + RB)
Parallel Circuits
• Multiple paths for the current to flow; Branches
• If any resistor (bulb) in the circuit is removed or burnt
out, then the other bulbs will light as long as there is
an unbroken path from the battery through that bulb
and BACK to the battery.
• Total current equals the sum of currents in branches
• As the number of branches is increased, overall resistance of the
circuit is decreased
– think about driving on a 4 lane highway – little resistance to the flow
of traffic
– now consider an accident that blocks three of the lanes…a reduction
to only one lane INCREASED the resistance
– opening all lanes DECREASED the resistance
Combining Resistors
• Total Resistance or Equivalent
resistance in series = sum of
all individual resistances
Rtotal = RA + RB
• Parallel – the inverse of the
total resistance is the sum of
the inverse of the resistors
1/Rtotal = 1/RA + 1/RB
Equations
• Series
I=V/Rtotal
R= R1 + R2 + …..
Parallel
I=V/R
1/Rtotal= 1/R1 + 1/R2 + …
Power
• Measures the rate at which
energy is transferred
• Power = Current x Voltage
• P = IV
• The unit of power is the Watt
Example #2:
• A 6.0 V battery delivers a 0.50 A
current to an electrical motor.
What power is consumed by the
motor?
• P = IV
• P = (0.50 A)(6.0 V)
• P = 3.0 W
Practice…
Req = 2 Ω + 3 Ω = 5 Ω
IA =
IB =
V/Rtotal……9 v / 5 Ω = 1.8 Amps
VA = IA x RA = 1.8 Amps x 2 Ω = 3.6 volts
VB = IB x RB = 1.8 Amps x 3 Ω = 5.4 volts
add these and
you should get
the voltage
supplied by the
battery, 9 volts
Practice…
• Req => 1/ Req = 1/2 + 1/3 = 0.833,
– but remember 1/ Req = 0.833 (rearrange and solve for Req)
– so Req = 1/0.833 = 1.2 Ω
• IA = V/RA = 9.0 V / 2 Ω = 4.5 Amps
• IB = V/RA = 9.0 V / 3 Ω = 3 Amps
• VA = IA X RA = 4.5 Amps x 2 Ω = 9 v
• VB = IB X RB = 3 Amps x 3 Ω = 9 v
In a parallel circuit
these are NOT identical.
these should EACH equal the
voltage being supplied by the
battery, 9.0 volts.