CIRCUITS - MMSTC PHYSICS · The Current Electric Potential in any circuit can be found by using Ohms Law (Georg Simon Ohm) The amount of Electric Potential (Energy per unit charge)

CIRCUITSCurrent

Resistance

Ohms Law

Power

Series Circuits

Parallel Circuits

Combination Circuits

Circuit Diagram

CIRCUITS

Basic Circuit

High potential

Low potentialSuppliesenergy

uses energy

Electric Current

• Rate of Flow of Electric Charge through a conductor.

• Unit of electric current: the ampere, A. 1 A = 1 C/s.

But, actually it’s the electrons that move through a circuit in the opposite direction of conventional current.

The direction of current in a circuit is described as the direction positive charge move. (+ → −)

Drift Velocity

When Electrons move through a wire, they do not move very fast or very straight.

The bounce off and move around each other.

The average speed of the charges is very slow (measured in centimeters per second)

The average speed of the electron is called drift velocity

Resistance

• Slows the flow of electric charge

• Uses energy in a circiut

• The ratio of voltage to current is called the resistance: 𝑅 =𝑉

𝐼

• Measure in Ohms. (rhymes with “owns”)

• 1𝑉𝑜𝑙𝑡

𝐴𝑚𝑝= 1 Ω (Ohms)

ResistanceThe Resistance of a wire:

• Is Characteristic of the material• (Resistivity)

• Is Directly proportional to Length• L↑, Ω↑

• Is Inversely proportional to Area.• A↑ , Ω↓

• Increases with temperature• T↑, Ω↑

Ohm’s Law

Magic Triangle

𝑉 = 𝐼𝑅

Potential Current Resistance

? 0.5A 12Ω

12V ? 4Ω

6V 1.5A ?

The Current Electric Potential in any circuit can be found by using Ohms Law (Georg Simon Ohm)

The amount of Electric Potential (Energy per unit charge) (V) used by a resistor depends on the Current through the circuit, and the amount of Resistance 𝑉 = 𝐼𝑅

The current depends on the Potential of the Circuit (V (volts)) ,and

the Resistance of the Circuit. 𝐼 =𝑉

𝑅

Solution

𝑉 = 𝐼𝑅 = .5 ∗ 12= 6𝑉

𝐼 =𝑉

𝑅=12

4= 3𝐴

𝑅 =𝑉

𝐼=

6

1.5= 4Ω

Power• Rate at which work is done

• 𝑃 =𝑊𝑜𝑟𝑘

𝑡𝑖𝑚𝑒→ 𝑉 =

𝑊𝑜𝑟𝑘

𝑄→ 𝑊𝑜𝑟𝑘 = 𝑄𝑉 → (𝑠𝑒𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙)

• 𝑃 =𝑊𝑜𝑟𝑘

𝑡=

𝑄𝑉

𝑡=

𝑄

𝑡𝑉 →

𝑄

𝑡= 𝐼 → 𝑃 = 𝐼𝑉

• 𝑃 = 𝐼𝑉 → 𝑉 = 𝐼𝑅 → 𝐼 𝐼𝑅 = 𝑃 = 𝐼2𝑅

• 𝑃 = 𝐼𝑉 → 𝐼 =𝑉

𝑅→ 𝑃 =

𝑉

𝑅𝑉 = 𝑃 =

𝑉2

𝑅

3 ways to calculate power.

𝑃 = 𝐼𝑉 = 𝐼2𝑅 =𝑉2

𝑅Magic Circle

Power: Examples

𝑃 = 𝐼𝑉 = 𝐼2𝑅 =𝑉2

𝑅

Potential Resitance Current Power

12V ? 0.5A ?

? 4Ω 0.75A ?

4V 2Ω ? ?

𝑉 = 𝐼𝑅

Potential Resistance Current Power

𝑅 =𝑉

𝐼= 24Ω

𝑃 = 𝐼𝑉 = 6W

𝑉 = 𝐼𝑅 = 3𝐴 𝑃 = 𝐼2𝑅= 2.25𝑊

𝐼 =𝑉

𝑅= 2𝐴 𝑃 =

𝑉2

𝑅= 8𝑊

Solutions

Use Ohms law, and the Power Equations to Complete the following Table.

Power• The electric company measures kilowatt hours → kWh

• kWh = power * time = Work

• Work required to keep your household running.

• 1 kWh= 1000W * 3600s = 3.6*106 J

Resistors in Series A series connection has a single path from the battery, through each circuit element in turn, then back to the battery.

• Each Resistor in the Circuits uses a certain amount of energy. The greater the resistance the more energy it uses.

• Energy used is referred to as Voltage Drop.

• The sum of the voltage drops across the resistors equals the battery voltage (Total Potential of the Circuit)

• 𝑉𝑏𝑎𝑡𝑡𝑒𝑟𝑦 = 𝑉1+ 𝑉2+ 𝑉3= 𝐼𝑅1+ 𝐼𝑅2+ 𝐼𝑅3

• The current through each resistor is the same. The charge all follow the same path, so they travel at the same Rate

• 𝑉 = 𝐼𝑅1+ 𝐼𝑅2+ 𝐼𝑅3 = 𝐼 𝑅1+ 𝑅2+ 𝑅3

𝑉 = 𝐼 Σ𝑅 = 𝐼𝑅 𝑒𝑞

𝑹𝒆𝒒 = 𝑹𝟏+ 𝑹𝟐+ 𝑹𝟑

The equivalent resistance of a set of resistors in series is the sum of all the resistors in that series.

Resistors in Series A series connection has a single path from the battery, through each circuit element in turn, then back to the battery.

THINGS TO KNOW ABOUT RESISTORS IN SERIES:

The total Resistance is equal to the sum all the resistors.

𝑹𝒆𝒒 = 𝑹𝟏+ 𝑹𝟐+ 𝑹𝟑

Each resistor will have the same current𝑰 = 𝑰 = 𝑰 = 𝑰

The Voltage across each resistor is different𝑽𝑹 = 𝑰𝑹

The Sum of the volt drops is equal to the total voltage of the circuit.

𝑽𝒃𝒂𝒕𝒕𝒆𝒓𝒚 = 𝑽𝟏+ 𝑽𝟐+ 𝑽𝟑

Determine the following:

• Equivalent Resistance of the circuit

• Current through the resistors

• Voltage Drop across each resistor

• Power dissipated by each resistor

• Power dissipated by the entire circuit

Resistors in Series

6V 6Ω

4Ω

2Ω

Equivalent Resistance of the circuit.𝑅𝑒𝑞 = 𝑅1+ 𝑅2+ 𝑅3

𝑹𝒆𝒒 = 𝟒 + 𝟔 + 𝟐 = 𝟏𝟐𝜴

Current through the resistors

𝑰 =𝑽𝒃𝒂𝒕𝒕𝒆𝒓𝒚

𝑹𝒆𝒒

=𝟔

𝟏𝟐= 𝟎. 𝟓𝑨

Voltage Drop across each resistor𝑉𝑅 = 𝐼𝑅

𝑽𝟒 = 𝟎. 𝟓 ∗ 𝟒 = 𝟐𝑽𝑽𝟔 = 𝟎. 𝟓 ∗ 𝟔 = 𝟑𝑽𝑽𝟐 = 𝟎. 𝟓 ∗ 𝟐 = 𝟏𝑽

2+3+1=6=voltage of battery

Power dissipated by each resistor𝑃 = 𝐼𝑉

𝑷𝟒 = 𝟎. 𝟓 ∗ 𝟐 = 𝟏. 𝟎𝑾𝑷𝟔 = 𝟎. 𝟓 ∗ 𝟑 = 𝟏. 𝟓𝑽𝑷𝟐 = 𝟎. 𝟓 ∗ 𝟏 = 𝟎. 𝟓𝑾

Power dissipated by the entire circuit𝑃 = 𝐼𝑉

𝑷 = 𝟎. 𝟓 ∗ 𝟔 = 𝟑𝑾1+1.5+0.5=3W



• Current through the resistors

• Voltage Drop across each resistor

• Power dissipate by each resistor

• Power dissipate by the entire circuit

Resistors in SeriesEquivalent Resistance of the circuit.𝑅𝑒𝑞 = 𝑅1+ 𝑅2+ 𝑅3

𝑅𝑒𝑞 = 3000 + 10000 + 5000 = 18000 𝛺

Current through the resistors

𝐼 =𝑉𝑏𝑎𝑡𝑡𝑒𝑟𝑦𝑅𝑒𝑞

=9

18000= 0.0005𝐴 = 0.5𝑚𝐴

Voltage Drop across each resistor𝑉𝑅 = 𝐼𝑅

𝑉3 = 0.0005 ∗ 3000 = 1.5𝑉𝑉10 = 0.0005 ∗ 10000 = 5𝑉𝑉5 = 0.0005 ∗ 5000 = 2.5𝑉1.5+5+2.5=9=voltage of battery

Power dissipated by each resistor𝑃 = 𝐼𝑉

𝑷𝟑 = 𝟎. 𝟎𝟎𝟎𝟓 ∗ 𝟏. 𝟓 = 𝟕. 𝟓𝑬 − 𝟒𝑾𝑷𝟏𝟎 = 𝟎. 𝟎𝟎𝟎𝟓 ∗ 𝟓 = 𝟐. 𝟓𝑬 − 𝟑𝑾

𝑷𝟓 = 𝟎. 𝟎𝟎𝟎𝟓 ∗ 𝟐. 𝟓 = 𝟏. 𝟐𝟓𝑬 − 𝟑𝑾Σ𝑃 = 4.5𝐸 − 3

Power dissipated by the entire circuit𝑃 = 𝐼𝑉

𝑷 = 𝟎. 𝟎𝟎𝟎𝟓 ∗ 𝟗 = 𝟒. 𝟓𝑬 − 𝟑𝑾 = 𝟒. 𝟓𝒎𝑾

• A parallel connection splits the current. The sum of the current through each resistor is equal to the total current of the circuit.

𝐼 = 𝐼1+ 𝐼2+ 𝐼3

𝐼 =𝑉

𝑅

𝐼 =𝑉

𝑅1

+𝑉

𝑅2

+𝑉

𝑅3

• Each resistor uses the same amount of energy, so the voltage across each resistor is the same.

𝑉𝑏𝑎𝑡𝑡𝑒𝑟𝑦 = 𝑉1 = 𝑉2 = 𝑉3

• The current through each resistor is determined by the voltage of the circuit and the individual resistance.

𝐼 =𝑉

𝑅1

+𝑉

𝑅2

+𝑉

𝑅3

𝑉

𝑅𝑒𝑞

=𝑉

𝑅1

+𝑉

𝑅2

+𝑉

𝑅3

= 𝑉(1

𝑅1

+1

𝑅2

+1

𝑅3

)

1

𝑅𝑒𝑞

= (1

𝑅1

+1

𝑅2

+1

𝑅3

)

•1

𝑅𝑒𝑞

= (1

𝑅1

+1

𝑅2

+1

𝑅3

)

The inverse of the equivalent resistance of a parallel circuit is equal to the sum if inverse of each individual resistor.

Resistors in Parallel: A parallel connection has a multiple paths from the battery, through each circuit element.

The current from the source splits into different path. Each resistor uses the same amount of energy.

Resistors in Parallel: A parallel connection has a multiple paths from the battery, through each circuit element.

The current from the source splits into different path. Each resistor uses the same amount of energy.

THINGS TO KNOW ABOUT RESISTORS IN SERIES:

The inverse of the equivalent resistance is equal to sum of the inverses of each resistor.

𝟏

𝑹𝒆𝒒

= (𝟏

𝑹𝟏

+𝟏

𝑹𝟐

+𝟏

𝑹𝟑

)

Each resistor will have the same voltage𝑽𝒃𝒂𝒕𝒕𝒆𝒓𝒚 = 𝑽𝟏 = 𝑽𝟐 = 𝑽𝟑

Each resistor will have a different current.

𝑰 =𝑽𝒃𝒂𝒕𝒕𝒆𝒓𝒚

𝑹

Current is equal to the sum all the current through each resistors.

𝑰 = 𝑰𝟏+ 𝑰𝟐+ 𝑰𝟑



• Current through the circuit

• Current across each resistor


Resistors in ParallelEquivalent Resistance of the circuit:

1

𝑅𝑒𝑞

= (1

𝑅1

+1

𝑅2

+1

𝑅3

)

1

𝑅𝑒𝑞

=1

4+

1

6+

1

8=

13

24→

1

𝑅𝑒𝑞

=13

24→

𝑅𝑒𝑞

1=

24

13= 1.85 Ω

Or 𝑅𝑒𝑞 =1

4+

1

6+

1

8

− 1 = 1.85Ω

Current through the circuit: 𝐼 =𝑽𝒃𝒂𝒕𝒕𝒆𝒓𝒚

𝑅𝑒𝑞

𝐼 =12

1.85= 6.5𝐴

Current across each resistor: 𝐼 =𝑉

𝑅1

𝐼4 =12

4= 3𝐴

𝐼6 =12

6= 2𝐴

𝐼8 =12

8= 1.5𝐴

Σ𝐼 = 6.5𝐴

Power dissipated by each resistor: 𝑃 = 𝐼𝑉𝑃4 = 3 ∗ 12 = 36𝑊𝑃6 = 2 ∗ 12 = 24𝑊𝑃8 = 1.5 ∗ 12 = 18𝑊

Σ𝑃 = 78𝑊

Power dissipated by the entire circuit: 𝑃 = 𝐼𝑉𝑃 = 6.5 ∗ 12 = 78𝑊



• Current through the circuit

• Current across each resistor

• Power dissipated by each resistor


Resistors in ParallelEquivalent Resistance of the circuit:

1

𝑅𝑒𝑞

= (1

𝑅1

+1

𝑅2

+1

𝑅3

)=16

10000→ 𝑅𝑒𝑞 =

10000

16= 625Ω

𝑅𝑒𝑞 =1

10000+

1

2000+

1

1000

− 1 = 625Ω

Current through the circuit: 𝐼 =𝑽𝒃𝒂𝒕𝒕𝒆𝒓𝒚

𝑅𝑒𝑞

𝐼 =9

625= 0.0144 𝐴

Current across each resistor: 𝐼 =𝑉

𝑅1

𝐼10𝑘 =9

10000= 0.0009𝐴

𝐼2𝑘 =9

2000= 0.0045𝐴

𝐼1𝑘 =9

1000= .009𝐴

Σ𝐼 = 0.0144𝐴

Power dissipated by each resistor: 𝑃 = 𝐼𝑉

𝑃10𝑘 = .0009 ∗ 9 = 0.0081𝑊𝑃2𝑘 = 0.0045 ∗ 9 = 0.0405𝑊𝑃1𝑘 = 0.009 ∗ 9 = 0.081𝑊

Σ𝑃 = 0.1296𝑊

Power dissipated by the entire circuit:𝑃 = 𝐼𝑉

𝑃 = 0.0144 ∗ 9 = 0.1296𝑊

Complex Circuits Example #1

Determine the following:• Equivalent Resistance of the circuit• Total Current through the circuit• Voltage Drop across each resistor• Current through each resistors

• Equivalent Resistance of the circuit• Follow the current around the circuit. Out of the

positive end of the battery , thru R1, then the current splits as it goes through R2 & R3 which are connected in parallel. The current come back together and goes through R4.

• The two parallel resistors can be considered one smaller resistor in series with the other.

R4

R1

R2 & R3

Determine the Req of the parallel resistors:

1

𝑅𝑒𝑞

= (1

𝑅1

+1

𝑅2

)→ 𝑅𝑒𝑞 =1

30+ 1

50

− 1 = 18.75Ω



20Ω

20Ω

18.75Ω

Now treat it as a simple series circuit to find the total resistance of the circuit.

𝑅𝑒𝑞 = 𝑅1+ 𝑅2+ 𝑅3

𝑅𝑒𝑞 = 20 + 18.75 + 20 = 58.75 𝛺

𝑹𝒆𝒒 = 𝟓𝟖. 𝟕𝟓 𝜴

Total Current through the circuit• Find total current using Ohms Law. Apply the

total Voltage and the total resistance:

𝑉 = 𝐼𝑅 → 𝐼 =𝑉

𝑅=

10

58.75= 0.17𝐴



Still treating the circuit as a simple series circuit.We know, current remains the same in each series resistor.

R1=20ΩI=0.17A

R2,3=18.75ΩI=0.17A

R4=20ΩI=0.17A

Use Ohms Law to calculate voltage across each resistor.

𝑉 = 𝐼𝑅𝑉1 = 𝐼𝑅 = .17 ∗ 20 = 3.4𝑉𝑉2,3 = .17 ∗ 18.75 = 3.2𝑉𝑉4 = 𝐼𝑅 = .17 ∗ 20 = 3.4𝑉

3.4+3.2+3.4=10V



We already calculated the current through resistors 1 &4

We now have to treat resistors 2 & 3 as two separate resistors parallel to each other that divide the current

.

Use Ohms Law to calculate current across each resistor.

𝐼 =𝑉

𝑅, 𝐼2 =

3.2

30= 0.11𝐴, 𝐼3 =

3.2

50= 0.06𝐴

0.11 + 0.06=0.17A (checks out)

V1=3.4I=0.17A

V1=3.4I=0.17A

V2=3.2 V3=3.2

V2,3=3.2V V23=3.2V

Important to recall: Each will have the same voltage.The current will depend on the resistance

Complex Circuits Example #2Determine the following:• Equivalent Resistance of the circuit• Total Current through the circuit• Voltage Drop across each resistor• Current through each resistors

• Follow the current around the circuit. Out of the positive end of the battery , It splits before going through parallel resistors R1 and R2, comes back together before splitting again as it goes through R3 & R4 which are also connected in parallel.

• The two sets of parallel resistors can each be considered one smaller resistor in series with the other.

• Determine the equivalent resistance of each parallel pair.

1

𝑅𝑒𝑞

= (1

𝑅1

+1

𝑅2

)→ 𝑅𝑒𝑞 =1

100+ 1

250

− 1 = 71.4Ω

1

𝑅𝑒𝑞

= (1

𝑅3

+1

𝑅4

)→ 𝑅𝑒𝑞 =1

350+ 1

200

− 1 = 127.3Ω

• The two sets of resistors can each be considered one smaller resistor in series with the other.

71.4Ω

127.3Ω

Now Find the sum of the series portions𝑅𝑒𝑞 = 𝑅1,2

+ 𝑅3,4

𝑅𝑒𝑞 = 71.4 + 127.3 = 199 𝛺


𝑹𝒆𝒒 = 𝟕𝟏. 𝟒𝜴

𝑹𝒆𝒒 = 𝟏𝟐𝟕. 𝟑𝜴

Total Current through the circuit• Find total current using Ohms Law.

• Apply the total Voltage: 𝑉 = 24𝑉

• and the total resistance: 𝑅𝑒𝑞 = 199Ω

𝑉 = 𝐼𝑅 → 𝐼 =𝑉

𝑅=

24

199= 𝟎. 𝟏𝟐𝑨

Voltage Drop across each resistor• Use the total current and equivalent

resistance of each parallel pair to find the voltage drop across each pair

𝟕𝟏. 𝟒𝜴

𝟏𝟐𝟕. 𝟑𝜴

𝐼 = 0.12𝐴𝑉 = 𝐼𝑅

𝑉1,2 = 0.12 ∗ 71.4 = 𝟖. 𝟔𝑽

𝑉3,4 = 0.12 ∗ 127.3 = 𝟏𝟓. 𝟑𝑽

8.6+15.3=23.9=24V


We now have to treat each pair resistors as two separate parallel circuits with different voltages.

Important to recall: The resistors in each pair will have the same voltage.Their current will depend on the resistance.The total current should be the same in each pair

𝑉1,2 = 𝟖. 𝟔𝑽

𝑉3,4 = 𝟏𝟓. 𝟑𝑽

𝑉1,2 = 𝟖. 𝟔𝑽

𝑉3,4 = 𝟏𝟓. 𝟑𝑽

𝐼1 =𝑉

𝑅1

=8.6

100= 0.0856𝐴

𝐼2 =𝑉

𝑅2

=8.6

250= 0.0344𝐴

𝐼3 =𝑉

𝑅1

=15.3

350= 0.0437𝐴

𝐼4 =𝑉

𝑅4

=15.3

200= 0.0765

Complex CircuitsDetermine the following:• Equivalent Resistance of the circuit• Total Current of the circuit• Voltage Drop across each resistor• Total Current through the circuit

Look for smaller circuits within the big circuit and determine the Req of each smaller circuit

1) the 6 & 12 resistor are

parallel to each other𝑅𝑒𝑞 =

16+ 1

12

− 1 = 4Ω

4Ω2) The parallel circuit is in series with a 4Ω resistor 𝑅𝑒𝑞 = 4 + 4 = 8Ω

4) That circuit is parallel to two series resistors. Determine the Req of the series

𝑅𝑒𝑞 = 5 + 3 = 8Ω

𝑅𝑒𝑞 =18+

18

− 1 = 4Ω

5) Determine the Req of the parallel sets of resistors

The Total Resistance of the circuit is 4Ω

8Ω

8Ω

Complex CircuitsDetermine the following:• Equivalent Resistance of the circuit• Total Current of the circuit• Voltage Drop across each resistor• Total Current through the circuit

The Total Resistance of the circuit is 4Ω

The Total Voltage is the circuit is 12𝑉

Use Ohms Law to calculate current/

𝑉 = 𝐼𝑅 → 𝐼 =𝑉

𝑅

𝐼 =12

4= 3Ω

Complex CircuitsDetermine the following:• Equivalent Resistance of the circuit• Total Current of the circuit• Total Current through the circuit• Voltage Drop across each resistor

Follow the current out of the battery and keep in mind the rules about current a voltage.

1) The current splits at the first junction. Each path in the junction will have the same voltage because the are parallel. (12V)

12V

12V

2) Use the total voltage (12V) across the series circuit and the equivalent resistance (8Ω) to calculate current.

𝐼 =𝑉

𝑅𝑒𝑞

=12

8= 1.5𝐴

Req=8Ω

V=12V

3) Use the current (1.5A) and each resistance to determine the voltage drop across each resistor.

𝑉3 = 1.5 ∗ 3 = 4.5𝑉𝑉5 = 5 ∗ 1.5 = 7.5𝑉

𝐼 = 1.5𝐴

𝑉 = 𝐼𝑅 𝑉 = 𝐼𝑅

𝐼 = 1.5𝐴𝑉 = 4.5𝑉

𝐼 = 1.5𝐴𝑉 = 7.5𝑉

Complex CircuitsDetermine the following:• Equivalent Resistance of the circuit• Total Current of the circuit• Total Current through the circuit• Voltage Drop across each resistor

Follow the current out of the battery and keep in mind the rules about current a voltage.

1) The current splits at the first junction. Each path in the junction will have the same voltage because the are parallel. (12V)

12V

12V

4Ω

2) Use the total voltage (12V) across the series circuit and the equivalent resistance (8Ω) to calculate current. Each series resistor will have the same current.

𝐼 =𝑉

𝑅𝑒𝑞

=12

8= 1.5𝐴12V

𝑉4 = 1.5 ∗ 4 = 6𝑉3) Use the current (1.5A) and each resistance to determine the voltage of the the lone resistor. 𝑉 = 𝐼𝑅

4) Use the Total current (1.5A) and the equivalent resistance (4Ω) of the parallel to determine the voltage of the parallel circuit.

4Ω 𝑉 = 𝐼𝑅𝑒𝑞 𝑉6,12 = 1.5 ∗ 4 = 6V

5) The current splits as it goes through the parallel circuit. The Voltage across each resistor is the same (6V). The voltage and each resistance to find each current.

𝐼 =𝑉

𝑅6V

𝐼12 =6

12= 0.5𝐴

𝐼6 =6

6= 1𝐴

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CIRCUITS - MMSTC PHYSICS · The Current Electric Potential in any circuit can be found by using Ohms Law (Georg Simon Ohm) The amount of Electric Potential (Energy per unit charge)