• Using IB symbols - sketch a series circuit including
•
battery
•
lamp
•
heating element
•
wires with 1 switch
• The current direction real & conventional.
• Show the current measurement with an ammeter
symbol.
• Show the measurement of p.d. around bulb and battery
with the voltmeter.
Series Circuits
Series circuits are a chain of components
connected in a circle providing only 1 path
for current.
By conservation of charge, &
because there is only one route, the
current (I) is the same everywhere.
I1 = I2 = I3 …
1, 2, 3 represent
circuit components.
Resistance on a series circuit.
As more resistors are added Rtot increases.
The total/ equilvalent resistance is:
R1 + R2+ R3 = Req.
Voltage around resistors.
The eq V is the sum of the p.d.’s across
each resistor on the circuit.
V1 + V2+ V3 ~ VT.
Apply Ohm’s Law to individual
components on circuit
The p.d. or V across each resistor is always:
V1 = IR1.
V2 = IR2.
V3 = IR3.
Add each device voltage to find the VT.
The eq (total) resistance is the sum.
R1 + R2 + R3 = Req.
I
R1
R2
R3
Since there is only one path for the
charges to follow if one conductor
(resistor) is disconnected, the circuit is
broken.
The current flow stops.
Since the devices must share the
voltage, as more are added the
energy of the charges decreases.
The voltage “divides” between
resistors.
Energy Supply in Series
As you add more cells, the total
voltage adds.
Parallel Circuits
Parallel Circuits are connected so
that the current reaches a fork or
junction & divides.
Since each resistor is connected
across the voltage, the voltage is
equal everywhere, and equal to
the battery voltage.
V1 = V2 =V3 = Vtot
Current (I), when the charges reach a junction,
they divide. The total current in the circuit
= S individual currents in each branch.
Itot = I1 + I2 + I3 …
Since I = V, & all components have same V, then
R
Itot =
V +
R1
V +
R2
V
R3.
Resistance is a reciprocal
relationship.
1/Req = 1/R1 + 1/R2 + 1/R3 …
Where Req is the equivalent or
total resistance.
As you add resistors, the total
equivalent resistance goes
down, the total current goes
up.
The battery or source provides
unlimited current!!
Since parallel circuits offer
more than one path for the
charge to flow, individual parts
can be disconnected and
charge will still flow through
other branches.
Since the voltage is equal on
each branch, adding more
branches does not reduce the
energy each branch receives.
Add more bulbs, the others
stay bright!
Ammeters measure current & are connected
in series on the circuit.
Current must flow through them.
Voltmeters measure potential difference
& are connected in parallel around the
component to be measured.
Ex 1: A 9V battery is connected in series to
2 bulbs: 4W, & 2W.
• A) Sketch the diagram with the proper
symbols. Show real and conventional current
flow direction.
B) Sketch a voltmeter reading the voltage
at the 2 W resistor and 2 ammeters: one
reading the total current and one reading
the current at the 4 W resistor.
• C) Find the equivalent or total resistance
on the circuit.
D) Find the total current in the circuit.
E) What is the voltage in each resistor?
F) What is the current flowing through
the 4 W resistor?
Ex 2: A 9V battery is connected in parallel to 2
bulbs: 4W, & 2W.
A) Sketch the diagram with the proper symbols.
Show real and conventional current flow
direction.
B) Sketch a voltmeter reading the voltage at the 2
W resistor and 2 ammeters: one reading the total
current and one reading the current at the 4 W
resistor.
B) Find the equivalent or total
resistance on the circuit.
C) Find the total current in the
circuit.
D) What is the voltage in each
branch?
E) Find the current in each branch.
F) Add the currents from each branch
together. How do they relate to the total
current?
G) Now add a 3 W bulb to the circuit.
Recalculate the equivalent resistance.
H) Recalculate the current.
Ex: 3 bulbs are in parallel connected to 6V. R1 =
2W, R2 = 3 W, R3 = 4W.
Which equation is best to calculate power?
Find the power in each bulb.
What is relative brightness?
What if the 2W is removed? What happens to
the brightness of the remaining bulbs?
Hwk Kerr
read 137 – 141 do pg
147 #1, 4, 6, 8, 9, 12, 13, 15, 22.
Power
• The bulb brightness in each resistor
is dependant on the power in each
resistor.
• Since P = VI or P = I2R or P = V2/R,
we can calculate the power and
deduce the brightness.
Ex: 2 bulbs in series connected to 6V
source. R1 = 2W, R2 = 4 W.
Find the power in each.
What is relative brightness?
What if the 2W is removed? What is
the power in the remaining bulb?
For series use P =
2
I R.
Why?
For parallel I would use P = V2/R
Combo circuits
Use of Meters
Ammeters measure current so circuit
current must flow through meter.
Connect meter in series to measure
current.
Ideal ammeter has zero resistance.
Voltmeters measure p.d. across
resistors so must be connected in
parallel.
Ideal voltmeter has infinite
resistance.
Kirchoff’s Laws
The current entering junction = current
exiting. Application of “conservation of
charge”.
Total Voltage = sum of all partial voltages
on circuit. Application of conservation of
energy.
6V
2V
4V
Fuses – appliances are rated for the
power they can safely dissipate.
That implies a certain current & voltage
(power).
Fuses should be chosen to have a current
rating a bit higher than the one for
which the resistor is designed.
Should fuses and circuit breakers be
connected in parallel or in series?
Why?
If a 60 W bulb is connected to a 120 V
source, the current is:
P = VI
I=
I = P/V.
60 J/s = 0.5 C/s .5A.
120 J/C
Fuse should be ~ .6-1 A.