Chapter 28 Direct Current Circuits

Chapter 28 Direct Current Circuits
Electromotive Force Resistors in Series and Parallel Kirchhoff’s Rules RC Circuits Norah Ali Al- Moneef

28.1 Electromotive Force Sources of emf
The source that maintains the current in a closed circuit is called a source of emf Any devices that increase the potential energy of charges circulating in circuits are sources of emf Examples include batteries and generators SI units are Volts The emf is the work done per unit charge All sources of emf have what is known as INTERNAL RESISTANCE (r) to the flow of electric current. The internal resistance of a fresh battery is usually small but increases with use. Thus the voltage across the terminals of a battery is less than the emf of the battery Norah Ali Al- Moneef

Direct Current When the current in a circuit has a constant direction, the current is called direct current Most of the circuits analyzed will be assumed to be in steady state, with constant magnitude and direction Because the potential difference between the terminals of a battery is constant, the battery produces direct current The battery is known as a source of emf Norah Ali Al- Moneef

emf and Internal Resistance
A real battery has some internal resistance Therefore, the terminal voltage is not equal to the emf All sources of emf have what is known as INTERNAL RESISTANCE (r) to the flow of electric current. The internal resistance of a fresh battery is usually small but increases with use. Thus the voltage across the terminals of a battery is less than the emf of the battery r  + - Norah Ali Al- Moneef

The schematic shows the internal resistance, r
The internal resistance of the source of emf is always considered to be in a series with the external resistance present in the electric circuit The terminal voltage is ΔV = Vb-Va ΔV = ε – Ir For the entire circuit, ε = IR + Ir The internal resistance of the source of emf is always considered to be in a series with the external resistance present in the electric circuit The terminal voltage is ΔV = Vb-Va ΔV = ε – Ir For the entire circuit, ε = IR + Ir Norah Ali Al- Moneef

 - Ir = V = IR Terminal voltage = V V = IR
the relationship between Emf, resistance, current, and terminal voltage Circuit model looks like this: I Terminal voltage = V V = IR r R  = I r + I R   = I (r + R) I = /(r + R)  - Ir = V = IR The terminal voltage decrease =  - Ir as the internal resistance r increases or when I increases. If r =0 ( no internal resistance ) ( ideal battery )  = V = I R I = V / R I =  / R Norah Ali Al- Moneef

Power P = I Δ V the total power output Iε of the battery
is delivered to the external load resistance in the amount I 2R and to the internal resistance in the amount I 2r. Power dissipated by a resistor, is amount of energy per time that goes into heat, light, etc. Total power of emf Power dissipated as joule heat in: Load resistor Internal resistor Potential gain in the battery Potential drop at all resistances In an old, “used-up” battery emf is nearly the same, but internal resistance increases enormously Norah Ali Al- Moneef

R is called the load resistance
ε = ΔV + Ir = IR + Ir ε is equal to the terminal voltage when the current is zero – open-circuit voltage I = ε / (R + r) R is called the load resistance The current depends on both the resistance external to the battery and the internal resistance When R >> r, r can be ignored Power relationship: I ε = I2 R + I2 r When R >> r, most of the power delivered by the battery is transferred to the load resistor Norah Ali Al- Moneef

At V=10V someone measures a current of 1A through the below circuit
At V=10V someone measures a current of 1A through the below circuit. When she raises the voltage to 25V, the current becomes 2 A. Is the resistor Ohmic? a) YES b) NO Norah Ali Al- Moneef

Example What are voltmeter and ammeter readings? Norah Ali Al- Moneef

example From the figure find the current and the voltage across the internal resistor the voltage across the internal resistor (lost voltage): v = Ir = 1.14 amps × 0.5 ohms = 0.57 volts out the terminal voltage: Terminal voltage = emf - lost voltage = = volts Norah Ali Al- Moneef

Example A battery has an emf of 12V and an internal resistance of 0.05Ω. Its terminals are connected to a load resistance of 3Ω. Find: The current in the circuit and the terminal voltage The power dissipated in the load, the internal resistance, and the total power delivered by the battery Norah Ali Al- Moneef

Example A car has a 12-V battery with internal resistance . When the starter motor is cranking, it draws 125 A.What’s the voltage across the battery terminals while starting? Battery terminals Voltage across battery terminals = Norah Ali Al- Moneef

example I = 2 A When current leaves the battery the battery supplies power equal to: ε=12 V R P = Iε = 24W If current were forced to enter the battery, (as in charging it) then it absorbs the same power What does the energy balance look like in this circuit? Resistor: Electrical energy  heat Norah Ali Al- Moneef

28.2 Resistors in Series and Parallel
When two or more resistors are connected end-to-end, they are said to be in series The current is the same in all resistors because any charge that flows through one resistor flows through the other The sum of the potential differences across the resistors is equal to the total potential difference across the combination Norah Ali Al- Moneef

Potentials add ΔV = IR1 + IR2 = I (R1+R2) Consequence of Conservation of Energy The equivalent resistance has the effect on the circuit as the original combination of resistors Norah Ali Al- Moneef

Equivalent Resistance – Series
Req = R1 + R2 + R3 + … The equivalent resistance of a series combination of resistors is the algebraic sum of the individual resistances and is always greater than any of the individual resistors Current has one pathway - same in every part of the circuit Total resistance is sum of individual resistances along path Current in circuit equal to voltage supplied divided by total resistance Sum of voltages across each lamp equal to total voltage One bulb burns out - circuit broken - other lamps will not light Norah Ali Al- Moneef

Resistors in Parallel The potential difference across each resistor is the same because each is connected directly across the battery terminals A junction is a point where the current can split The current, I, that enters a point must be equal to the total current leaving that point I = I 1 + I 2 The currents are generally not the same Consequence of Conservation of Charge Norah Ali Al- Moneef

Equivalent Resistance – Parallel
The inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistance The equivalent is always less than the smallest resistor in the group Norah Ali Al- Moneef

Equivalent Resistance – Parallel, Example
Equivalent resistance replaces the two original resistances Household circuits are wired so the electrical devices are connected in parallel Circuit breakers may be used in series with other circuit elements for safety purposes Norah Ali Al- Moneef

Resistors in Parallel, Final
In parallel, each device operates independently of the others so that if one is switched off, the others remain on In parallel, all of the devices operate on the same voltage The current takes all the paths The lower resistance will have higher currents Even very high resistances will have some currents Household circuits are wired so that electrical devices are connected in parallel Norah Ali Al- Moneef

Equivalent Resistance – Series: An Example
Four resistors are replaced with their equivalent resistance Norah Ali Al- Moneef

Problem-Solving Strategy, 1
Combine all resistors in series They carry the same current The potential differences across them are not the same The resistors add directly to give the equivalent resistance of the series combination: Req = R1 + R2 + … Norah Ali Al- Moneef

Combine all resistors in parallel The potential differences across them are the same The currents through them are not the same The equivalent resistance of a parallel combination is found through reciprocal addition: Norah Ali Al- Moneef

A complicated circuit consisting of several resistors and batteries can often be reduced to a simple circuit with only one resistor Replace any resistors in series or in parallel using steps 1 or 2. Sketch the new circuit after these changes have been made Continue to replace any series or parallel combinations Continue until one equivalent resistance is found Norah Ali Al- Moneef

If the current in or the potential difference across a resistor in the complicated circuit is to be identified, start with the final circuit found in step 3 and gradually work back through the circuits Use ΔV = I R and the procedures in steps 1 and 2 The greater the resistance in a circuit, the lower the current. Norah Ali Al- Moneef

Example With the switch in this circuit (figure a) open, there is no current in R2. There is current in R1 and this current is measured with the ammeter at the right side of the circuit. If the switch is closed (figure b), there is current in R2. When the switch is closed, the reading on the ammeter (a) increases, (b) decreases, or (c) remains the same. (a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the total current increases. Norah Ali Al- Moneef

Example You have a large supply of lightbulbs and a battery. You start with one light bulb connected to the battery and notice its brightness. You then add one light bulb at a time, each new bulb being added in parallel to the previous bulbs. As the light bulbs are added, what happens (a) to the brightness of the bulbs? (b) to the current in the bulbs? (c) to the power delivered by the battery? (d) to the lifetime of the battery? (e) to the terminal voltage of the battery? Hint: Do not ignore the internal resistance of the battery. (a) The brightness of the bulbs decreases (b) The current in the bulbs decreases (c) The power delivered by the battery increases (d) The lifetime of the battery decreases (e) The terminal voltage of the battery decreases Norah Ali Al- Moneef

Example R1=3 Ohm R2=3 Ohm R3=3 Ohm V=12V R2 & R3 are in parallel
what is the equivalent resistance of all resistors as placed in the below circuit? If V=12V, what is the current I? R3 R1=3 Ohm R2=3 Ohm R3=3 Ohm V=12V R1 R2 I R2 & R3 are in parallel 1/R23=1/R2+1/R3=1/3+1/3=2/3 R23=3/2 Ohm R1 is in series with R23 R123=R1+R23=3+3/2=9/2 Ohm I=V/R=12/(9/2)=24/9=8/3 A V Norah Ali Al- Moneef

Example A tree is decorated with a string of many equal lights placed in parallel. If one burns out (no current flow through it), what happens to the others? a) They all stop shining b) the others get a bit dimmer c) the others get a bit brighter d) the brightness of the others remains the same Before the one light fails: 1/Req=1/R1+1/R2+…+1/Rn if there are 3 lights of 1 Ohm: Req=1/3 I=V/Req Ij=V/Rj (if 3 lights: I=3V Ij=V/1 After one fails: 1/Req=1/R1+1/R2+….+1/Rn-1 if there are 2 lights left: Req=1/2 I=V/Req Ij=V/Rj (if 2 lights: I=2V Ij=V/1) The total resistance increases, so the current drops. The two effects cancel each other R R I V Norah Ali Al- Moneef

Example a person designs a new string of lights which are placed in series. One fails, what happens to the others? a) They all stop shining b) the others get a bit dimmer c) the others get a bit brighter d) the brightness of the others remains the same Assume: If one fails, the wire inside it is broken and Current cannot flow through it any more. Norah Ali Al- Moneef

Example Find the current through the 2- resistor in the circuit.
Equivalent of parallel 2.0- & 4.0- resistors:  Equivalent of series 1.0-,  & 3.0- resistors: Total current is Voltage across of parallel 2.0- & 4.0- resistors: Current through the 2- resistor: Norah Ali Al- Moneef

Equivalent Resistance
Example Three resistors are connected in parallel as shown in the Figure a potential difference of 18.0 V is maintained between points a and b. (A) Find the current in each resistor ,total current ( I ) Equivalent Resistance Norah Ali Al- Moneef

We apply the relationship P = I 2R to each resistor and obtain
(B) Calculate the power delivered to each resistor and the total power delivered to the combination of resistors. We apply the relationship P = I 2R to each resistor and obtain This shows that the smallest resistor receives the most power. Summing the three quantities gives a total power of 198 W. Norah Ali Al- Moneef

Equivalent Resistance – Complex Circuit
Example Equivalent Resistance – Complex Circuit Norah Ali Al- Moneef

Example inthe circuit (a) below to get the equivalent resistance. b ) calculate the power P = I2 R dissipated in each resistor. Norah Ali Al- Moneef

P1> P2 , P3 , P4 Bulb 1 is the brightest
Example : Four equivalent light bulbs R1 = R2 = R3 = R4 = 4.50  , emf = 9.00 Volts. Find current and power in each light bulb. Which bulb is brightest? Itotal=I1=1.5A,, I2=I3=I4+0.5A., P1=10.1W, P2=P3=P4=1.125W., V2=V3=V4=2.25V P1> P2 , P3 , P Bulb 1 is the brightest Later, if bulb 4 is removed which bulbs get brighter? Dimmer? Still Bulb 1 is the brightest The brightness (intensity) of the bulb is basically the power dissipated in the resistor. For example A 200 W bulb is brighter than a 75 W bulb, all other things equal Norah Ali Al- Moneef

Example What is the current through a?
1 /Req = (1 /5) + (1 /10) (1 /30) Req =30 / (6+3+1) = 3  What is the current through e? What is the current each branch b-d? Same voltage is across each path b: V= IR 30 = Ix I= 6A c: 30= Ix I= 3A d: 30= Ix I= 1A 5 10 30 30v a e Itot = 10A b c d Norah Ali Al- Moneef

Example If all N branches have the same resistance, total resistance is equal to the resistance of one branch divided by the number of branches Total resistance= 1 /Req = (1 /30) + (1 /30) (1 /30) Req = 30 /3 = 10  Total current= V / Req I total = 30 /10 = 3A Current in b= 30 /3 0 = 1 A Norah Ali Al- Moneef

Example If there are only two branches, the total resistance is equal to the product of the resistances divided by the sum of the resistances Total resistance= Req = 12 X 4 /( ) = 3  Norah Ali Al- Moneef

Example What is the: Total resistance? Total current flow?
Current flow through b Current flow through c Current flow through d Voltage between b and d Voltage between c and d Voltage between d and e Norah Ali Al- Moneef

Example Total resistance:
In compound circuits, reduce all parallel parts to a single resistance until you have a simpler series circuit The resistance between a and b is 2  Therefore, total resistance is 4  (2 + 2) Norah Ali Al- Moneef

Example Find the Total current. Total current: V = Itotal Req
the total resistance of the parallel portion of the circuit R1eq = 3 X 6 / ( ) = 2  the total resistance of the series portion of the circuit R eq = = 4  Total current: V = Itotal Req 20v = I X 4  Itot = 5 A Norah Ali Al- Moneef

Example Current flow through b
We need to know the voltage drop across b-d Voltage drop across e-d will be (V= 5A X 2  = 10v) Therefore, voltage drop across each parallel branch (c and b) must be ( Vda =20 V -10V = 10 V) Current flow in b: 10 = I X 3 ; I = 3.33A Current flow in c: 10 = I X 6 ; I= 1.67A Norah Ali Al- Moneef

Example 2 6 12 3. 20 4. Norah Ali Al- Moneef

Example In the following circuit with source voltage V and Total current I, which resistor will have the greatest voltage across it? The resistor with the largest resistance (30 ) Which resistor has the greatest current flow through it? Same for all because series circuit If we re-ordered the resistors, what if any of this would change? Nothing would change Norah Ali Al- Moneef

Example If we added a resistor in series with these, what would happen to the total resistance, total current, voltage across each resistor, and current through each resistor? Total resistance would increase Total current would decrease Voltage across each resistor would decrease (All voltage drops must still sum to total in series circuit; Kirchhoff’s law of voltages) Current through each resistor would be lower (b/c total current decreased, but same through each one) Norah Ali Al- Moneef

Example In the following circuit with source voltage V and Total current I, which resistor will have the greatest voltage across it? All the same in parallel branches Which resistor has the greatest current flow through it? The “path of least resistance” (10) What else can you tell me about the current through each branch They will sum to the total I (currents sum in parallel circuits; Kirchhoff’s law of current) Norah Ali Al- Moneef

Example If we added a resistor in parallel with these, what would happen to the total resistance, total current, voltage across each resistor, and current through each resistor? Total resistance would decrease Total current would increase Voltage across each resistor would still be V Current through each resistor would be higher and would sum to new total I Norah Ali Al- Moneef

Example R1=100 ohms R2=100 ohms R3=100 ohms Req= ? Ohms
1 / Req= 1 / R1 + 1 / R2 + 1 / R3 1 / Req = 1/ / /100 1 / Req = = .03 Req = 1/.03 = ohms Norah Ali Al- Moneef

Example A parallel circuit is shown in the diagram above. In this case the current supplied by the battery splits up, and the amount going through each resistor depends on the resistance. If the values of the three resistors are: R 1 = 8 Ω R2= 8 Ω R 3 = 4 Ω 1 / R eq = 1 / / / 4 = 4 / 8 R eq = 2 Ω With a 10 V battery, by V = I R the total current in the circuit is: I = V / R = 10 / 2 = 5 A. The individual currents can also be found using I = V / R. The voltage across each resistor is 10 V, so: I1 = 10 / 8 = 1.25 A I2 = 10 / 8 = 1.25 A I3=10 / 4 = 2.5 A Note that the currents add together to 5A, the total current. 52 Norah Ali Al- Moneef

Example What would V1 read in the illustration? Ohm’s Law states:
Therefore: At this point there is insufficient data because I (amp) is unknown. Using Ohm’s Law to solve for the current in the circuit: Knowing the amount of current we can calculate the voltage drop. Norah Ali Al- Moneef

Example Determine the readings for A1 and A2 in the illustration.
In a series circuit the electricity has no alternative paths, therefore the amperage is the same at every point in the circuit. The current in the circuit is determined by dividing the voltage by the circuit resistance. Norah Ali Al- Moneef

Example In a parallel circuit the amperage varies with the resistance.
Determine the readings for amp meters A1 and A2 in the circuit. In a parallel circuit the amperage varies with the resistance. In the illustration, A1 will measure the total circuit amperage, but A2 will only measure the amperage flowing through the 6.3 Ohm resistor. To determine the total resistance of the circuit must be Norah Ali Al- Moneef

Example When the total resistance is known, the circuit current (Amp’s) can be calculated. Total current is: A1= A When the circuit current (Amp’s) is known, the current for each branch circuit can be calculated. Branch current is: A2 = 1.9 A Norah Ali Al- Moneef

Example Determine the readings for the two volt meters in the illustration. Volt meter one (V1) will read source voltage: Volt meter two (V2) will read the voltage in the circuit after the 2.3  resistor. To determine this reading, the voltage drop across the resistor must be calculated. Before the voltage drop across the resistor can be calculated, the circuit current must be determined. 12 V To calculate the circuit current the total circuit resistance must be known. Norah Ali Al- Moneef

Example Volt meter 1 will read source voltage: 12 V
Circuit current is: The voltage drop caused by the 2.3 Ohm resistance is: The voltage remaining in the circuit is: Volt meter 1 will read source voltage: 12 V Volt meter 2 will read 3.6 V. Norah Ali Al- Moneef

Example Determine the readings for the amp meters in the illustration.
The first amp meter will read circuit amps and the second one will measure the current in that branch of the circuit. To determine circuit amperage, the total circuit resistance must be known. This was calculated in the previous slide as 3.29 . Total circuit current was also calculated as 3.65 A. A1 = 3.65 A To determine the reading for A2, the current flowing in that part of the circuit must be calculated. In the previous slide the voltage left after the 2.3  resistor was 3.61 V. A2 = 0.62 A Norah Ali Al- Moneef

Example When a piece of wire is used to connect points b and c in this figure, the brightness of bulb R1 (a) increases, (b) decreases, or (c) stays the same. The brightness of bulb R2 (a) increases, (b) decreases, or (c) stays the same. R1 becomes brighter. Connecting a wire from b to c provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to just R1. Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The current passing through bulb R1 increases, causing this bulb to glow brighter. Bulb R2 goes out because essentially all of the current now passes through the wire connecting b and c and bypasses the filament of Bulb R2. Norah Ali Al- Moneef

Example With the switch in this circuit (figure a) closed, no current exists in R2 because the current has an alternate zero-resistance path through the switch. Current does exist in R1 and this current is measured with the ammeter at the right side of the circuit. If the switch is opened (figure b), current exists in R2. After the switch is opened, the reading on the ammeter (a) increases, (b) decreases, (c) does not change. (b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases. Norah Ali Al- Moneef

Example What is the value of R? IR = V / R =. 8 / 12 = .667A
1 / R eq = 1 / / / R = R / 150 R R eq = R / R R eq = 8 / = 4 Ω = R / R 4 ( R ) = 150 R 100 R = 150R R = 12 Ω IR = V / R =. 8 / 12 = .667A Norah Ali Al- Moneef

Example What is the Value of R? R = Δ V / I
= 5 .0 V / ( 100 X )A R = 25Ω Norah Ali Al- Moneef

Example What is the current through the circuit?
What are the values of potential at points a-f? I: 1A a:0V b:12V c:9V d: 6V e: 0V f: 0V Norah Ali Al- Moneef

28.3 Kirchhoff’s Rules Statement of Kirchhoff’s Rules
There are ways in which resistors can be connected so that the circuits formed cannot be reduced to a single equivalent resistor Two rules, called Kirchhoff’s Rules can be used instead Statement of Kirchhoff’s Rules Junction Rule The sum of the currents entering any junction must equal the sum of the currents leaving that junction A statement of Conservation of Charge Loop Rule The sum of the potential differences across all the elements around any closed circuit loop must be zero A statement of Conservation of Energy Norah Ali Al- Moneef

the Junction Rule I1+I2+I3=I4+I5 I1 = I2 + I3
From Conservation of Charge Diagram b shows a mechanical analog I1 I3 I4 I1+I2+I3=I4+I5 I2 I5 Norah Ali Al- Moneef

Setting Up Kirchhoff’s Rules
Assign symbols and directions to the currents in all branches of the circuit If a direction is chosen incorrectly, the resulting answer will be negative, but the magnitude will be correct When applying the loop rule, choose a direction for transversing the loop Record voltage drops and rises as they occur Norah Ali Al- Moneef

Electrical Work and Power
Resistance R I + - I Higher V1 Lower V2 Current I flows through a potential difference DV Follow a charge Q : at positive end, U1 = QV1 at negative end, U2 = QV2 P.E. Decreases: The speed of the charges is constant in the wires and resistor. What is electrical potential energy converted to? Norah Ali Al- Moneef

This thermal energy means the atoms in the
Electrical resistance converts electrical potential energy to thermal energy (heat), just as friction in mechanical systems converts mechanical energy to heat. This thermal energy means the atoms in the conductor move faster and so the conductor gets hotter. The average kinetic energy of the electrons doesn’t increase once the current reaches a steady state; the electrons lose energy in collisions with the atoms as fast as it is supplied by the field. Norah Ali Al- Moneef

More About the Loop Rule
Traveling around the loop from a to b In a, the resistor is transversed in the direction of the current, the potential across the resistor is –IR In b, the resistor is transversed in the direction opposite of the current, the potential across the resistor is +IR In c, the source of emf is transversed in the direction of the emf (from – to +), the change in the electric potential is +ε In d, the source of emf is transversed in the direction opposite of the emf (from + to -), the change in the electric potential is -ε Norah Ali Al- Moneef

Junction Equations from Kirchhoff’s Rules
Use the junction rule as often as needed, so long as, each time you write an equation, you include in it a current that has not been used in a previous junction rule equation In general, the number of times the junction rule can be used is one fewer than the number of junction points in the circuit Norah Ali Al- Moneef

Loop Equations from Kirchhoff’s Rules
The loop rule can be used as often as needed so long as a new circuit element (resistor or battery) or a new current appears in each new equation You need as many independent equations as you have unknowns Norah Ali Al- Moneef

Problem-Solving Strategy – Kirchhoff’s Rules
Draw the circuit diagram and assign labels and symbols to all known and unknown quantities Assign directions to the currents. Apply the junction rule to any junction in the circuit Apply the loop rule to as many loops as are needed to solve for the unknowns Solve the equations simultaneously for the unknown quantities Check your answers Norah Ali Al- Moneef

Example The emfs and resistances in the circuit have the following values Starting at point a lets apply the loop rule counter-clockwise and tally the voltages along the way Norah Ali Al- Moneef

What is the potential difference between the terminals of battery 1 ??
This is the equivalent of asking what the potential difference is between points b and a. So lets travel from point b to point a clockwise Vb – i r1 + ε1 = Va Va - Vb = - i r1 + ε 1 = -(0.24A) (2.3 ) + 4.4V = 3.8V Norah Ali Al- Moneef

Example. Multiloop Circuit
Find the current in R3 in the figure below. Loop 1: Loop 2: Norah Ali Al- Moneef

Example Label the 3 branch currents I2, I3, and I4.
Since VAB across all 3 branches is the same and is known: V4 = I4R4 = 5A X (4Ω) = 20 Volts, the currents and ε can be readily solved. V4 = I4R4 = 5A (4W) = 20 Volts I3 = V3 / R3 = 4V / 3W = 1.33 A At junction B, S I in = S I out I4 = I2 + I3 ; I2 = I4 – I3 = 5A - 1.3A I2 = 3.7A Loop #1: S Vi =0 - e + I2R2 + I4R e = 3.7A (2W) + 5A(4W) = 27.4 V Norah Ali Al- Moneef

Example ε = -5 V Find r and ε . I in = I out
I = 1 A + 2A = 3 A Loop 1: - ε +1 A X 1Ω -2A X 3 Ω = 0 ε = -5 V Loop 3 : 12 – 3 A Xr – 2 A X 3 Ω = 0 r = 6 V / 3 A = 2 Ω Norah Ali Al- Moneef

Example Given V=12V, what is the current through and voltage over each resistor R eq = 3 +( 3X 3 / ) = 3 + ( 9 / 6 ) = 4.5 Ω 1) I1= 12 / ( 9 /2 ) = 8/3 A Kirchhof 2 2) V-I1R1-I2R2= I1-3 I2=0 3) V-I1R1-I3R3= I1-3I3=0 4) 0-I3R3+I2R2= I3+3I2=0 Kirchhoff I 5) I1-I2-I3= I1=I2+I3 & 2) I2=0 so =3I2 and I2=4/3 A 1) & 3) I3=0 so =3I3 and I3=4/3 A Use V=IR for R V1=8/3X3= 8 V for R V2=4/3X3= 4 V for R V3=4/3X3= 4 V I3 R3 I1 R1 I2 R2 I=I1 I=I1 V R1=3 Ω R2=3 Ω R3=3 Ω V=12V Norah Ali Al- Moneef

Example which of the following cannot be correct?
a) V-I1R1-I3R3-I2R2=0 b) I1-I3-I4=0 c) I3R3-I6R6-I4R4=0 d) I1R1-I3R3-I6R6-I4R4=0 e) I3+I6+I2=0 I1,R1 I4 R4 V I3 R3 I6 R6 I2 R2 I5 R5 NOT a LOOP Norah Ali Al- Moneef

Example no, this is only TRUE in the case of 1)
consider the circuit. Which of the following is/are not true? If R2=R3=2R1 the potential drops over R1 and R2 are the same for any value of R1,R2 and R3 the potential drop over R1 must be equal to the potential drop over R2 The current through R1 is equal to the current through R2 plus the current through R3 (I1=I2+I3) R3 R1 R2 I V if R2=R3=2R1 then 1/R23=1/R2+1/R3=1/R1 so R23=R1 and I1=I23 and potential of R1 equals the potential over R23 and thus R2 and R3. THIS IS TRUE no, this is only TRUE in the case of 1) true: conservation of current. Norah Ali Al- Moneef

Example A B C D circuit 1: I=12/(2R)=6/R VA=12-6/R*R=6V circuit 2: VB=12V circuit 3: I=12/(3R)=4/R VC=12-4/R*R=8V VD=12-4/R*R-4/R*R=4V 1 2 3 12V 12V 12V At which point (A,B,C,D) is the potential highest and at which point lowest? All resistors are equal. highest B, lowest A highest C, lowest D highest B, lowest D highest C, lowest A highest A, lowest B Norah Ali Al- Moneef

Example what is the current through and voltage over each R?
R1=R2=R3= 3 Ohm V1=V2= 12 V R1 R3 R2 I2 apply kirchhoff’s rules 1) I1+I2-I3= (kirchhoff I) 2) left loop: V1-I1R1+I2R2= so I1+3I2=0 3) right loop: V2+I3R3+I2R2=0 so I3+3I2=0 4) outside loop: V1-I1R1-I3R3-V2=0 so –3I1-3I3=0 so I1=-I3 combine 1) and 4) I2=2I3 and put into 3) 12+9I3=0 so I3= 4/3 A and I1= - 4/3 A and I2= 8/3 A V1 V2 Norah Ali Al- Moneef

Example what is the power dissipated by R3? P=VI=V2/R=I2R
R1= 1 Ω , R2= 2 Ω , R3= 3 Ω , R4= 4 Ω V= 5V R1 I2 R2 I=I1 I=I1 We need to know V3 and/or I3. Find equivalent R of whole circuit. 1/R23=1/R2+1/R3= 1/2+1/3= 5/ R23= 6/5 Ω R1234= R1+R23+R4= 1+6/5+4=31/5 Ω I=I1=I4= V / R1234= 5 / (31/5) I = 25 / 31 A Kirchhoff 1: I1=I2+I3= 25 / Kirchhoff 2: I3R3-I2R2=0 so I3-2I2= I2= 3 / 2 I3 Combine: 3/2 I3+I3= 25 / 31 so 5 / 2 I3=25 / I3=10/31 A P=I2R so P=(10 / 31)2 X 3=(100/961)X3= W V Norah Ali Al- Moneef

Example R4 I4 What is Kirchhoff I for ? I1+I2-I3-I4= b) I1+I2+I3+I4= c) I1-I2-I3-I4=0 What is Kirchhoff II for the left small loop ( with R4 and R1?) a) I4R4+I1R1= b) I4R4- I1R1= c) I4R4+I1R1-V=0 I3 R3 I1 R1 I2 R2 V What is Kirchhoff II for the right small loop (with R2 and R3)? I3R3+I2R2= b) I3R3-I2R2= c) I3R3-I2R2+V=0 What is Kirchhoff II for the loop (with V,R4 and R3)? a) V-I4R4+I3R3= b) V+I4R4-I3R3= c) V-I4R4-I3R3=0 Norah Ali Al- Moneef

 - Ir = V = IR Summary: Electromotive Force (  )
= I r + I R   = I (r + R) I = /(r + R)  - Ir = V = IR Terminal voltage = V = IR The terminal voltage decrease =  - Ir as the internal resistance r increases or when I increases .If r =0 ( no internal resistance ) ( ideal battery )  = V = I R I = V / R I =  / R Norah Ali Al- Moneef

Norah Ali Al- Moneef

Resistors connected in a circuit in series or parallel can be simplified using the following:
Series connection Parallel connection 27-1 Resistors in a circuit Norah Ali Al- Moneef

Chapter 28 Direct Current Circuits

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Chapter 28 Direct Current Circuits

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