Download Circuits and Circuit Diagrams


Circuit = Complete path where
electrons can flow

Circuit diagram symbols:

Wire/Conductor

Resistor

Battery

Switch
(light bulbs, fans)



Resistance = Voltage/ Current
R=V/I
Units
Resistance = ohms (Ω)
 Voltage = volt (V)
 Current = ampere (A)





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




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)
Current stays the same across all resistors.


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.

Multiple paths for the current to flow; Branches

If any resistor (bulb) in the circuit is removed or burnt

Total current equals the sum of currents in branches

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.
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
The inverse of the total resistance
is the sum of the inverse of the
resistors
1/Rtotal = 1/RA + 1/RB

Series
I=V/Rtotal
R= R1 + R2 + …..
Parallel
I=V/R
1/Rtotal= 1/R1 + 1/R2 + …
Measures the rate at which
energy is transferred
 Power = Current x Voltage
 P = IV
 The unit of power is the Watt





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
Rtotal = 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
Rtotal=
IA =
IB =
VA =
VB =
Rtotal=
IA =
IB =
IC =
VA =
VB =
VC =
Rtotal = 1/ Rtotal = 1/2 + 1/3 = 0.833,
 but remember 1/ Rtotal = 0.833 (rearrange and solve for Rtotal)
 so Rtotal = 1/0.833 = 1.2 Ω
IA
= V/RA = 9.0 V / 2 Ω = 4.5 Amps
In a parallel circuit
these are NOT identical.
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
these should EACH equal the
voltage being supplied by the
battery, 9.0 volts.
Rtotal =
IA =
IB =
VA =
VB =
Rtotal =
IA =
IB =
IC =
VA =
VB =
VC =