Syllabus pointsCircuit analysis and design involve calculation of the
potential difference across, the current in, and the power supplied to, components in series, parallel, and series/parallel circuits
This includes applying the relationships for series components:
πΌ = ππππ π‘πππ‘ππ‘ = π1 + π2 +β―πππ π‘ = π 1 + π 2+. . . π π
Learning goals
Be able to use the equations below to determine potential difference, current, power and resistance in series circuits
πΌ = ππππ π‘πππ‘ππ‘ = π1 + π2 +β―πππ π‘ = π 1 + π 2+. . . π π
Apply your knowledge of electrical circuits to everyday situations to explain how they work
Series circuit The flow of charge (current) must flow through
devices sequentially
In a series circuit the current only has one path to follow
Current in a series circuit Current in a series circuit is the same everywhere
πΌπ = πΌ1 = πΌ2 = πΌ3β¦
Same number of charges passing each point per second
Charge does not pile up, or get βused upβ by resistors
The actual amount of current varies inversely with the amount of overall resistance (higher the total resistance, lower the current)
Resistance in series circuit Total resistance of a series circuit is the sum of the
individual resistancesπ π = π 1 + π 2 + π 3β¦
More resistors = more overall resistance
Potential Difference in a series circuit Total potential difference of a series circuit is the sum of the
individual potential differencesππ = π1 + π2 + π3β¦
The total potential difference is shared between the components in series (but not shared equally, depending on resistance of components)
Example 1
A series circuit contains a battery with a voltage of 12 V, and three resistors with resistances of π 1 = 1.00 Ξ© , π 2 = 5.00 Ξ© and π 3 = 15.0 Ξ© .
a) What is the total resistance of the circuit?
b) Calculate the current of the circuit
c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source
Example 1
A series circuit contains a battery with a voltage of 12 V, and three resistors with resistances of π 1 =1.00 Ξ© , π 2 = 5.00 Ξ© and π 3 = 15.0 Ξ© .
a) What is the total resistance of the circuit?
Example 1
A series circuit contains a battery with a voltage of 12 V, and three resistors with resistances of π 1 =1.00 Ξ© , π 2 = 5.00 Ξ© and π 3 = 15.0 Ξ© .
b) Calculate the current of the circuit
Example 1A series circuit contains a battery with a voltage of 12 V, and three resistors with resistances of π 1 =1.00 Ξ© , π 2 = 5.00 Ξ© and π 3 = 15.0 Ξ© .
c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source
Example 2For the circuit pictured below calculate the:
a) total resistance
b) current
c) current through each resistor
d) voltage drop across each resistor
e) power dissipated in each resistor.
Example 2For the circuit pictured below calculate the:
a) total resistance
Example 2For the circuit pictured below calculate the:
b) current
c) current through each resistor
Example 2For the circuit pictured below calculate the:
d) voltage drop across each resistor
Example 2For the circuit pictured below calculate the:
e) power dissipated in each resistor.
Context β Christmas lights Christmas tree lights used to be connected in a series
circuit. Why have they changed them to parallel circuits?
Series Circuit β sum up Current = constant, same at all points of circuit
Total resistance = sum of individual resistancesπ π = π 1 + π 2+. . .
Potential difference = shared among the components (but not shared equally)
ππ = π1 + π2 + π3β¦
ResourcesAV
Khan Academy - Circuits (part 1) (11:40)
Khan Academy - Circuits (part 2) (11:08)
Khan Academy - Circuits (part 3) (12:21)
Khan Academy - Circuits (part 4) (7:07)
Simulation
PhET β DC circuit construction kit
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