Electricity Unit 1 Physics.

Electricity Unit 1 Physics

INTRODUCTION – ELECTRONS & MATTER
All material is made up of atoms – with Protons, Neutrons & Electrons. The negatively charged electrons revolve around the outside of the atom. When large numbers of electrons move through a conductor, electric current is able to flow.

As different elements have different numbers of electrons in their outer shell, some elements are better conductors of electricity than others. Atoms with an outer shell that are not full are better conductors than atoms with a stable and complete outer shell, they seek electrons/atoms to bond with.

We can categorise elements with these properties as: Conductors Insulators Semiconductors

Some common conductors / semiconductors / insulators

INTRODUCTION – what is electricity?
Video: Electricity (Hila Science Videos)

INTRODUCTION – what is electricity?
Video: Electricity (Bozeman Science Video)

Conventional Current vs electron current
We can represent the current flow in a circuit two ways: Conventional current Electron Current

Electron Current: A flow of current in a wire is a flow of negatively charged electrons. That is, the electrons flow from the negative terminal to the positive in an electric circuit.

However….. Before electrons were discovered, the conventional current theory was used (and is a still widely used and accepted method). Conventional Current, the direction of flow is positive to negative.

direct current vs Alternating current
Direct Current: Refers to circuits where the flow of charge is in one direction only. eg. Power Pack (chargers for electronic goods), Batteries Alternating Current: Refers to circuits where the charge carriers move backward and forward periodically. eg. The electricity that comes out of a power point.

What is current? How is it measured?
Electric Current is the measure of the rate of charge around a circuit. Expressed by: 𝐼= 𝑄 𝑡 where I is the current (Ampere, Amps, A) Q is the amount of charge flowing (Coulombs, C) t is the time (seconds) So, 1 Amp = 1 coulomb passing a point in a second.

Eg1. Find the current flow in a circuit, if 50 coulombs passes a point in 10 seconds? 𝐼= 𝑄 𝑡 = = 𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 = 5 𝐴 Eg2. Find the charge passing through a circuit, if 8 Amps flows through in 2 minutes? 𝐼= 𝑄 𝑡 𝑄=𝐼×𝑡 =8× 2×60 = 8× = 960 𝐶

Eg3. How long must a current of 50mA flow to make 3 Coulombs of charge pass a point in a circuit? 𝐼= 𝑄 𝑡 t= 𝑄 𝐼 = 3 5× 10 −3 =600 𝑠 What is this value in minutes? =10 𝑚𝑖𝑛𝑠

Current, charge & electrons
Recall: Electric Current is the measure of the rate of charge around a circuit. I is the current (Ampere, Amps, A) 𝐼= 𝑄 𝑡 Q is the amount of charge flowing (Coulombs, C) t is the time (seconds) So, 1 Amp = 1 coulomb passing a point in a second. 1 coulomb of charge = the amount of charge carried by 6.24× electrons. Each electron carries a charge of × Coulombs ≅ 1× 10 −19 Coulombs As electrons are negatively charged, their charge is −1× 10 −19 Coulombs

Now do: text book problems questions 1 - 12

electrical Potential energy
So we know that energy is transferred around a circuit by a flow of electrons. Electrical Potential Energy is the potential energy of these charge carriers (electrons) due to their position in a circuit and is measured in Joules. As electrons are forced away from the negative terminal to the positive terminal of an energy source (recall: electron flow notation), potential energy is converted into other forms of energy, so electrical potential energy drops over time.

POTENTIAL DIFFERENCE VOLTAGE
As electrons move away from the negative terminal, their Electrical Potential Energy is converted into other forms (eg. Thermal Energy  Heat). Notice that batteries, mobile phone, computer etc.. get hot as you use them? This can be explained as the circuit losing Electrical Potential Energy as it’s converted to Thermal energy. The amount of electrical potential energy lost by each coulomb of charge in a given part of a circuit is called the Potential Difference or Voltage Drop, measured in Volts.

One Volt is defined as the voltage drop that occurs when 1 coulomb of charge transforms one Joule of energy. So we can say that the Voltage drop (V) across a device in a circuit is equal to the amount of energy (E) transformed divided by the amount of charge (Q) that has passed through the device. 𝑉= 𝐸 𝑄 Voltage (V) measured in Volts (V) Energy (E) measured in Joules (J) Charge (Q) measured in Coulombs (C)

Voltage (V) measured in Volts (V) Energy (E) measured in Joules (J) Charge (Q) measured in Coulombs (C) 𝑉= 𝐸 𝑄 Which can be rearranged and represented in the following ways: 𝐸=𝑄𝑉 𝑄= 𝐸 𝑉 𝐸=𝑉𝐼𝑡

Example 1: Use the formulas to solve the following problems: 𝑄= 𝐸 𝑉 𝑉= 𝐸 𝑄 𝐸=𝑄𝑉 𝑉= 𝐸 𝑄 = =1.61 𝑉 𝑉= 𝐸 𝑄 = × 10 −3 = 𝑉 𝐸=𝑄𝑉 = 6× 10 −3 ×500=3 𝐽 𝐸=𝑄𝑉=10×240=2400 𝐽 𝑄= 𝐸 𝑉 = =9 𝐶

Example 2: What is the Potential Difference across a component if 2000 Joules of heat is generated when a current of 10A flows for 15 seconds? 𝐸=𝑉𝐼𝑡 → 𝑉= 𝐸 𝐼𝑡 = 2000𝐽 10𝐴 × 15𝑠 = = 𝑉 So, there has been a drop (or loss) of 13.33V across the component, due to heat.

Measuring VOLTAGE in a circuit
We can measure the voltage in a circuit using a voltmeter. These are connected in parallel with the circuit, as pictured below (we will learn more about series and parallel circuits later)

Measuring CURRENT In a circuit
We can measure the electric current in a circuit using an ammeter. These are connected in series with the circuit, as pictured below.

mA Measuring CURRENT In a circuit For some ammeters we can connect the ammeter to the circuit via 3 different terminals and as such, there are three different scales to display the current reading to cover a range of readings, small or large. The Black terminal is connected toward the negative terminal of the power supply and the red terminal is connected to the positive terminal of the power supply.

Connecting voltmeter and ammeter’s to a circuit
For both voltmeter and ammeters, the black terminal is connected toward the negative terminal of the power supply and the red terminal is connected to the positive terminal of the power supply. Ammeter Setup This is the same as this 

Connecting voltmeter and ammeter’s to a circuit
For both voltmeter and ammeters, the black terminal is connected toward the negative terminal of the power supply and the red terminal is connected to the positive terminal of the power supply. Voltmeter Setup This is the same as this 

Example: If the ammeter is connected to a circuit and gives the following reading (pictured left) – what is the current in the circuit if: The 5 A terminal is connected? The 50 mA terminal is connected? The 500 mA terminal is connected? mA 3.5 A 35 mA 350 mA

Measuring VOLTAGE In a circuit
Example: If the voltmeter is connected to a circuit and gives the following reading (pictured left) – what is the voltage in the circuit if: The 3 V terminal is connected? The 15 V terminal is connected? The 30 V terminal is connected? mA 1.4 V 7 V 14 V

Now do: text book problems questions 14, 15, 16, 17, 18

How do batteries work?

AC vs DC

pOwer delivered by a circuit
Power is the rate of doing work, or the rate at which energy is transformed from one form to another. The rate at which energy is transformed in a circuit, determines its effect on the circuit. eg. The brightness of a light in a circuit is determined by the rate that electrical potential energy is transformed into other forms of energy (heat, light etc.) A circuit with a large amount of power is going light a globe brighter than that with lower power input.

Power is equal to the amount of energy transformed per second. In simpler terms – Power is the rate of energy use. 𝑃= 𝐸 𝑡 This equation tells us that: 1 Watt of Power = 1 Joule of energy transformed per second. Power (P) measured in Watts (W) Energy (E) measured in Joules (J) Time (t) measured in Seconds (s)

𝑃= 𝐸 𝑡 Recall 𝐸=𝑉𝐼𝑡 𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑎𝑦: 𝑃=𝑉𝐼 𝐸=𝑄𝑉 𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑢𝑠𝑒 𝑎𝑛𝑦 𝑜𝑓 𝑡ℎ𝑒𝑠𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑃𝑜𝑤𝑒𝑟: 𝑃= 𝐸 𝑡 = 𝑞𝑉 𝑡 = 𝑞𝐼𝑅 𝑡 = 𝐼 2 𝑅=𝑉𝐼 Power (P) measured in Watts (W) Energy (E) measured in Joules (J) Time (t) measured in Seconds (s) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A)

Simple representation of Some of the FORMULA’s used so far
Power (P) measured in Watts (W) Energy (E) measured in Joules (J) Time (t) measured in Seconds (s) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A) Charge (Q) measured in Coulombs (C)

Power (P) measured in Watts (W) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A) eg1. What is the power rating of a torch if a current of 0.5A flows through it when there is a voltage drop of 9V across the lamp? 𝑃=𝑉𝐼 =9×0.5=4.5𝑊

Power (P) measured in Watts (W) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A) eg2. What is the power rating of a heater if a current of 10A flows through it when there is a voltage drop of 120V across the heating element? 𝑃=𝑉𝐼 =120×10=1200𝑊 𝑜𝑟 1.2 𝑘𝑊

Power (P) measured in Watts (W) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A) eg3. A lamp has a power rating of 80 W. What current flows through the lamp if there is a voltage drop of 120V across the globe? Express your answer in mA. 𝐼= 𝑃 𝑉 = =0.667 𝐴 𝑜𝑟 667 𝑚𝐴

Power (P) measured in Watts (W) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A) eg4. Calculate the voltage of an iPad if it is rated 4 W and draws a current of 500mA. V= 𝑃 𝐼 = 4 500× 10 −3 =8 𝑉

Now do: text book problems questions 19 - 25

Batteries & cells Cells can be represented in a circuit by the symbols: Batteries are one or more electrochemical cells, which are placed together and packed, and such can be represented in a circuit by the symbols: Variable Power Supplies are represented by the symbol:

Batteries & cells Batteries:
How they work/recharge – the car battery:

emf electromotive force
Batteries and Power Packs (like computer chargers, phone chargers etc), provide energy to a circuit. These devices are said to provide an Electromotive Force (EMF). EMF is a measure of energy supplied to the circuit for each coulomb of charge passing through the power supply. EMF is represented by the symbol ε and is measured in Volts.

The amount of energy (E) supplied to the charge passing through a power supply (Battery Cells or Plug-In power packs) is equal to the amount of energy passing given to each coulomb, or EMF (ε), multiplied by the amount of charge passing through the power supply. 𝐸= ε 𝑄 𝑜𝑟 𝐸=εIt 𝑤𝑒 𝑐𝑎𝑛 𝑎𝑙𝑠𝑜 𝑠𝑎𝑦, 𝑃=ε𝐼 This formula is used to determine the rate that the source of the EMF provides energy to the circuit. Energy (E) measured in Joules (J) EMF (ε) measured in Volts (V) Charge (Q) measured in Coulombs (C) Power (P) measured in Watts (W) Current (I) measured in Amps (A)

eg1. A 12V car battery has a current of 3A passing through it. At what rate is it providing energy to the car’s circuit? 𝑃=ε𝐼 =12 𝑉 ×3𝐴 =36 𝑊 The ε in this equation is the same as V. Recall: 𝑃=𝑉𝐼

eg2. What is the EMF of a battery in a circuit that provides 18 J of energy and 2 C of charge? ε= 𝐸 𝑄 = 18 2 =9 𝑉 The ε in this equation is the same as V. Recall: 𝑉= 𝐸 𝑄

Now do: text book problems questions 14 , 26, 27

RESISTANCE The Resistance (R) of a device is a measure of how difficult it is for a current to pass through it. The higher the value of the resistance, the harder it is for the current to pass through the device. Resistance is measured in Ohms, given by the symbol Ω Circuit Representations of resistors:

RESISTANCE The Resistance (R) of a substance is the ratio of the Voltage Drop (V) across it, to the current (I) flowing through it. Given by the equation: 𝑅= 𝑉 𝐼 Resistance (R) measured in Ohms (Ω) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A)

RESISTANCE eg1. Find the resistance of the resistor in the following circuit if a current of 2A flows through it. 𝑅= 𝑉 𝐼 = Ω Resistance (R) measured in Ohms (Ω) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A)

RESISTANCE eg2. Find the resistance of the resistor in the following circuit if a current of 20mA flows through it. 𝑅= 𝑉 𝐼 = × 10 −3 = 600 Ω Resistance (R) measured in Ohms (Ω) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A)

RESISTANCE eg3. Find the Voltage across the resistor in the following circuit if a current of 50mA flows through it. 𝑉=𝐼𝑅 = 50× 10 −3 × 2× 10 3 = 100 𝑉 Resistance (R) measured in Ohms (Ω) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A)

RESISTANCE eg4. Find the Current flow in the following circuit. Express your answer in mA. 𝐼= 𝑉 𝑅 = 30 10× 10 3 = 3 𝑚𝐴 Resistance (R) measured in Ohms (Ω) Voltage (V) measured in Volts (V) Current (I) measured in Amps (A)

Now do: text book problems question 28

OHM’s Law Recall Ohm’s Law:
When we plot the graph of Voltage (V) vs Current (I), obeying Ohm’s Law, this gives us a linear graph. The resistance is given by the gradient of the line. This tells us that under constant physical conditions (ie. Constant temperature), the resistance is constant for all currents that pass through it. We say that a component/device that produces this graph is called Ohmic device – if not, we say it’s non-Ohmic. OHM’s Law

RESISTANCE The Ohm is named in honour of Georg Simon Ohm (1787-1854).
Ohm was a German Physicist who investigated the effects of different materials in electric circuits. His studies found: The resistance of a conductor relies on 4 factors: The length (l) of the conductor – The longer the length, the greater the resistance The cross sectional area (A) – The greater the Area, the easier current flows through – so the smaller the resistance will be. The type of material that its made from – some materials are better conductors of electricity than others – the property that contributes to its resistance is its resistivity (𝜌) 4) Temperature of the wire

RESISTANCE The resistance of a conductor relies on factors:
Length (l), measured in meters (m) Cross sectional area (A), measured in metres squared (𝑚 2 ) Resistivity (𝜌), measured in ohm metres (Ωm) We can use these to find the resistance that a material provides: 𝑅= 𝜌𝑙 𝐴 Temperature effects the resistance, however we are ignoring it in our calculations in the text. See page 39 table 2.1 for further.

RESISTANCE eg1. Use the table below to find the resistance of a copper wire at temperature 18˚, with length 5m, cross-sectional Area of 1 × 10 −8 𝑚 2 . 𝑅= 𝜌𝑙 𝐴 = (1.78× 10 −8 Ω𝑚)×(5𝑚) 1× 10 −8 𝑚 2 =8.9 Ω Length (l) measured in metres (m) Cross-Sectional Area (A) measured in 𝑚 2 Resistivity (𝜌) measured in Ohm metres (Ωm)

RESISTANCE eg2. Use the table below to find the resistance of a silver wire at temperature 100˚, with length 50m, cross-sectional Area of 1 × 10 −8 𝑚 2 . 𝑅= 𝜌𝑙 𝐴 = (2.13× 10 −8 Ω𝑚)×(50𝑚) 1× 10 −8 𝑚 2 =106.5 Ω Length (l) measured in metres (m) Cross-Sectional Area (A) measured in 𝑚 2 Resistivity (𝜌) measured in Ohm metres (Ωm)

RESISTANCE Note: Cross-Sectional Area of a wire = 𝑟 2 𝜋 = 𝑑 2 𝜋 4 eg3. Use the table below to find the resistance of a carbon wire at temp 18˚, with length 25m, diameter of 2 × 10 −4 m. Give your answer in 𝑘Ω to 1 decimal place. 𝑅= 𝜌𝑙 𝐴 = (5× 10 −5 Ω𝑚)×(25𝑚) × 10 −8 𝑚 2 = Ω =39.8𝑘Ω A= 𝑑 2 𝜋 4 = 𝜋 (2× 10 −4 ) 2 4 =3.1416× 10 −8 𝑚 2 Length (l) measured in metres (m) Cross-Sectional Area (A) measured in 𝑚 2 Resistivity (𝜌) measured in Ohm metres (Ωm)

Note: Cross-Sectional Area of a wire = 𝑟 2 𝜋 = 𝑑 2 𝜋 4

RESISTors Resistors are generally made from the semiconductor Carbon.
Variable Resistors Potentiometer or “Pot” Trim Pot

RESISTors Resistors have 4 - 6 bands of colour on them.
We can read the value of the resistance on each resistor by reading and interpreting their colour bands. We will learn how to read 4-colour band resistors.

RESISTors eg1. Find the resistance of a resistor with the colours (in order): Blue, Green, Red, Silver. 6 5 × Ω ± 𝟏𝟎% =6500 Ω ±10% So measuring the resistor would show a resistance between to , so between 5850 – 7150 Ω

RESISTors 1 × 10 3 Ω =10,000 Ω ±5% 𝑜𝑟 10𝑘Ω ± 5% ± 𝟓%
eg2. Find the resistance of a resistor with the colours (in order): Brown, Black, Orange, Gold. 1 × Ω ± 𝟓% =10,000 Ω ±5% 𝑜𝑟 𝑘Ω ± 5% So measuring the resistor would show a resistance between to , so between Ω

RESISTors eg3. Find the resistance of a resistor with the colours (in order): Red, Black, Brown, Silver. 2 × Ω ± 10% =200 Ω ±10% So measuring the resistor would show a resistance between to , so between Ω

Now do: text book problem question 29

Now do: text book problems question 29 Now Do: Q (and any other questions still remaining from chapter 2 on your work record)

sErIes vs parallel circuits
Series Parallel

sErIes circuits Resistance
The total resistance is equal to the sum of all individual resistors. 𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑅 1 + 𝑅 2 + 𝑅 3 … Current The current flowing in all parts of the circuit is the same, found using Ohm’s Law: 𝐼= 𝑉 𝑅 𝑇

sErIes circuits eg1. What is the total resistance of the circuit?
𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑅 1 + 𝑅 2 + 𝑅 3 = =30Ω What is the Current in the circuit? 10Ω 10Ω 9V 𝐼= 𝑉 𝑅 𝑇 = =0.3𝐴 10Ω

sErIes circuits eg2. What is the total resistance of the circuit?
𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑅 1 + 𝑅 2 + 𝑅 3 = =360 Ω What is the Current in the circuit? 40Ω 100Ω 12V 𝐼= 𝑉 𝑅 𝑇 = 12 𝑉 360 Ω = 𝐴 =33 𝑚𝐴 220Ω

KIrCHOFF’s second law 15V
This Law is an application of the conservation of electrical energy around a circuit. It can be stated as: In any closed loop of a circuit, the sum of the voltage drops must equal the sum of the EMF’s in that loop. The following slides show how this impacts the voltage throughout a series circuit…

The voltage drops across each resistor.
The sum of the voltage drops is equal to the applied voltage. I = 0.5A 15V 0V 5V Use this to calculate the current: I= 𝑉 𝑅 = =0.5𝐴 Then use Ohm’s law to find the voltage drop across each resistor 𝑉=𝐼𝑅=0.5×10=5 𝑉 eg1. How can we find the voltage drop across each resistor? Find the value of 𝑅 𝑇𝑜𝑡𝑎𝑙 in the circuit. 𝑅 𝑇 = 𝑅 𝑇 =30Ω

7.5V Voltage Drop I = 0.05A eg2. Find the voltage drop across each resistor. Find the value of 𝑅 𝑇𝑜𝑡𝑎𝑙 in the circuit. 𝑅 𝑇 = 𝑅 𝑇 =180Ω 9V 0V 5V Then use Ohm’s law to find the voltage drop across each resistor 𝑉 30Ω =𝐼𝑅=0.05×30=1.5 𝑉 𝑉 50Ω =𝐼𝑅=0.05×50=2.5 𝑉 𝑉 100Ω =𝐼𝑅=0.05×100=5 𝑉 Use this to calculate the current: I= 𝑉 𝑅 = =0.05𝐴

9V Current is the same throughout the circuit, I b =1 A
Voltage drop across 𝑅 2 : V=IR=1×60=60V Voltage drop across 𝑅 1 : 100V-60V = 40V 𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑉 𝐼 = =100Ω 𝑅 1 =100Ω−60Ω=40Ω

Now do: text book problems Chapter three question 3-9

KIrCHOFF’s current law
Kirchoff’s Current Law (Kirchoff’s First Law) tells us that electric charge in a circuit is conserved. This means that current flow in and out of junctions in the circuit is equal. Total Current In = Total Current Out ie. For the junction of wires pictured, Kirchoff’s Law tells us that: 𝐼 𝑎 + 𝐼 𝑐 = 𝐼 𝑏 + 𝐼 𝑒 + 𝐼 𝑑 eg1. If 𝐼 𝑎 =1𝐴 𝐼 𝑐 =2.5 𝐴 𝐼 𝑏 =0.5𝐴 𝐼 𝑒 =1.5𝐴 𝐼 𝑑 = ? 1+2.5= 𝐼 𝑑 3.5=2.0+ 𝐼 𝑑 ∴ 𝐼 𝑑 =1.5 𝐴

KIrCHOFF’s current law
eg2. Using this law analyse the junction of wires pictured. For the wire marked 𝐼 ; What amount of current runs through this branch? Is the current going in or out of the junction? Total Current In = Total Current Out Current IN = 1 A + 4 A = 5 A Current OUT = 1.3 A A = 3.8 A Unknown 𝐼 = 5 A – 3.8 A = 1.2 A OUT

eg3. In the pictured circuit, find the current at point a, b, c, d, e
Total Current In = Total Current Out a Junction A: 15.3 = a a = = 7.4 mA b Junction B: = b b = 15.3 mA c Junction C: 15.3 = c c = 15.3 – 2.1 = 13.2 mA d Junction D: 13.2 = d d = 13.2 – 6.5 = 6.7 mA e = 2.1 mA f Junction E: ( ) = f f = 15.3 mA

Resistance The total resistance found by the formulas -
parallel circuits Resistance The total resistance found by the formulas - For 2 resistors: For any number of resistors: 𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑅 1 × 𝑅 2 𝑅 1 + 𝑅 𝑅 𝑇𝑂𝑇𝐴𝐿 = 1 𝑅 𝑅 𝑅 3 𝑒𝑡𝑐… Voltage The voltage flowing in each branch of the circuit is equal to the EMF of the battery – the voltage drop across each resistor is the same. Current Total current is equal to the sum of all of the currents in each branch. Ohm’s Law can be applied to resistors in each branch to find the current in each branch.

parallel circuits eg1. What is the total resistance of the circuit?
10Ω eg1. What is the total resistance of the circuit? 𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑅 1 × 𝑅 2 𝑅 1 + 𝑅 2 = 10× = =5Ω What is the Current in the circuit? 10Ω 𝐼 𝑇𝑂𝑇𝐴𝐿 = 𝑉 𝑅 𝑇 = 10 𝑉 5 Ω =2𝐴 10V In each branch: 𝐼= 𝑉 𝑅 = 10 𝑉 10 Ω =1𝐴

50Ω eg2. What is the total resistance of the circuit? 𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑅 1 × 𝑅 2 𝑅 1 + 𝑅 2 = 50× = =40Ω What is the Current in the circuit? 200Ω 𝐼 𝑇𝑂𝑇𝐴𝐿 = 𝑉 𝑅 𝑇 = 10 𝑉 40 Ω =0.25 𝐴 =250 𝑚𝐴 In each branch: 𝐼 𝑅1 = 𝑉 𝑅 = 10 𝑉 200 Ω =0.05𝐴=50 𝑚𝐴 𝐼 𝑅2 = 𝑉 𝑅 = 10 𝑉 50 Ω =0.2𝐴=200 𝑚𝐴 10V

3kΩ eg3. What is the total resistance of the circuit? 𝑅 𝑇𝑂𝑇𝐴𝐿 = 𝑅 1 × 𝑅 2 𝑅 1 + 𝑅 2 = 1000× = =750 Ω What is the Current in the circuit? 1kΩ 𝐼 𝑇𝑂𝑇𝐴𝐿 = 𝑉 𝑅 𝑇 = 24 𝑉 750 Ω =0.032 𝐴 =32 𝑚𝐴 In each branch: 𝐼 𝑅1 = 𝑉 𝑅 = 24 𝑉 1000 Ω =0.024𝐴=24 𝑚𝐴 𝐼 𝑅2 = 𝑉 𝑅 = 24 𝑉 3000 Ω =0.008 𝐴=8 𝑚𝐴 24V

Now do: text book problems Chapter three question 2, 10-16

Electricity Unit 1 Physics.

Similar presentations

Presentation on theme: "Electricity Unit 1 Physics."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Electricity Unit 1 Physics.

Similar presentations

Presentation on theme: "Electricity Unit 1 Physics."— Presentation transcript:

Similar presentations

About project

Feedback