Download Grade 9 Academic Science – Unit 4 Electricity

Grade 9 Academic Science – Unit 4 Electricity
Series and Parallel Circuits
Section 13.10 Pages 571-575
SERIES CIRCUIT
 One path for current to flow
 When two lights are connected in series, the
brightness of each lamp decreases. Why? As more
loads (…in this case, lights…) are added, more
RESISTANCE to the flow of current is built up. With
more resistance, less current flows into each light. Thus,
brightness goes down as more loads are added.
 See Sample Problem #1 on Page 571.
Example
 A circuit diagram has two lights in series. The total
voltage in the circuit is 50 V. The total resistance (R) of the
two lights is 10  (Recall:  = ohms). What is the current (I) in the circuit? HINT: Use
Ohm’s Law formula I = V/R
Solution
I = 50 / 10
I=5A
The current through each light is 5 amps

Another circuit diagram has four lights arranged in a series circuit. The total voltage is
the same as the first circuit (i.e., V = 50 V). Total resistance has increased to 25 .
What is the current on this circuit?
Solution
I = 50 / 25
I=2A
The current through each light is 2 amps.
Thus, the lights in the second circuit would not burn as brightly.
Task
 Draw a circuit diagram with five lights in series. The Potential Difference (i.e., total
voltage) in the circuit is 120 V. The total resistance (R) of the lights is 40  (Recall:  =
ohms). What is the current through each lamp? HINT: Use Ohm’s Law formula I = V/R
Voltage across a Load in Series
 …Not enough? To add more, look at the POTENTIAL ENERGY of the circuit. A battery
converts chemical energy into electric potential energy. As the electron leave the battery,
they establish flow (i.e., they push each other along). As the electrons move, the electric
potential energy becomes kinetic energy.
 When the electron passes through one light, the electric potential energy is converted
into heat and light (…kinetic energy…). When the electron re-enters the battery, it is
recharged (i.e., given more energy) which keeps the movement of electrons going (inand-out). Since voltage is related to potential energy, the voltage drop across one load
(light) equal the voltage drop across the battery.
 If two lights occur on the series circuit, only half the potential electric energy is converted
into heat and light in each light. Why? The battery can only supply so much energy, and
to keep the current flowing, the energy must be distributed across all the loads. The
voltage drop across the light is half the voltage drop across the battery.


If more loads are added, less potential energy is converted to light and heat at each load.
The voltage drop across each load decreases.
The formula is
Vload = Vsource / Number of Loads
Example
 See Sample Problem #2 on Page 572
 A series circuit contains six identical lamps. The potential difference of the battery is
120 V. What is the potential difference across each lamp?
Solution
Vload = Vsource / # of loads
Vload = 120 / 6
Vload = 20 V
The potential difference across each lamp is 20 volts
Practice
 Twenty identical lights are connected in series. A 10 V battery is hooked up as the power
supply. What is the potential difference across each light?

Potential difference at the source is 200 V. Five very large and identical floodlights are
connected in a series circuit to the voltage source. What is the potential difference
across each light?
PARALLEL CIRCUIT
 More than one path for current to flow. The
number of paths is determined by the number of
loads that are connected in parallel.
 When two lights are connected in parallel, the brightness
of each lamp remains the same.
 Why? The electrons leaving the battery are split
between the paths. Thus, the quantity of electrons
leaving the battery is greater than the quantity of the
electrons moving through each load.

The formula is
Iload = Isource / Number of Loads
where Iload is the current (amps) through each identical load, Isource is the current coming from
the energy source, and number of loads is the total number of identical loads in the entire
circuit
Example
 See Sample Problem #3 on Page 573
 There are five loads. Each load has a separate path in a parallel circuit. Total Resistance
is 100 . The potential difference of the energy sources is 20V. What is the current
through each light?
Solution
Isource
=V/R
= 20 / 100
= 0.2 A
Iload
= Isource / # of loads
= 0.2 / 5
= 0.04 A
The current through each lamp is 0.04 amps
Practice

Four identical lights are connected in parallel. Total resistance is 10 . A 10 V battery is
hooked up as the power supply. What is the current across each light?

Ten identical lights are connected in parallel. Total resistance is 100 . A 5000 V
battery is hooked up as the power supply. What is the current across each light?
HOMEWORK
 Page 575 Questions 2-7