28.1

Chapter 28: Fundamentals of Circuits

Source of Electric Potential – the Mechanical Battery (the Van de Graaff ≡ VdG) What determines how much potential difference the motor of the VdG can maintain in order to have a continuous discharge. However, instead of thinking about a sphere, we will consider the situation of two parallel plates. 1. First start with uncharged parallel plates, and transport negative charged from the left

plate to the right plate. The conveyor belt, driven by the motor, exerts a nonelectrical force on the charges (FNE).

2. As the motor transport electrons to the right, the plates exert an electrical force on the electrons (FE) in the opposite direction.

3. Eventually there is enough charge on the plates to make FE = eEplates = FNE. At this point the motor cannot pump out any more charges that it can take in. In other words, the charges leaving the battery is equal to the charges coming into the battery.

Applet PhET Electric Circuit

We see that the function of a battery is to produce and maintain a charge separation (pull electrons out of the positive plate and push them onto the negative plate). The amount of charge separation is limited by and determined by the strength of the motor in a mechanical battery. In a car battery it is the chemical reactions in a chemical battery. The voltage across a battery is given by

NE NEbattery platesV

F d F dE d where emfe e

∆⋅ ⋅

= ⋅ = ≡

The emf (electromotive force, which is bad name) of a battery is a property of the battery. Why is this a terrible name? It is not a force but an energy per unit charge. We will avoid this terminology by just using the abbreviation emf.

emf emf V≡ ≡ ∆ε In a chemical battery the emf is a measure of the chemical energy per unit charge expended by the battery in moving charge through the battery. The emf of a flashlight battery is about 1.5 V, and the associated charge separation is due to chemical reactions in which electrons are reactants or products.

POWER DISSIPATED/DELIVERED One of the most powerful ways to look at “real circuits” is not by measuring current and voltage, but through energy/power. There are two kinds of energy/power: (i) when energy/power is dissipated or absorbed, then this device is called passive; (ii) if a device supplying or delivering energy/power to a circuit, it is then called active.

What determines whether an element is actively adding or absorbing energy into the circuit is the sign of the power function?

1. Passive Power

2. Active Power

28.2

According to conservation of energy, initial final 0i 0fE E E E= → =∑ ∑

However, energy is very difficult to measure in a circuit, so it is customary to measure the power instead. In the language of power, we write the energy going into the circuit must equal the energy going out. That is,

delivered absorbed active passiveP P P P= → =∑ ∑ ∑ ∑

Electrical Power and Energy According to Ohm’s law, the voltage across a resistor is ∆Vab = Ri. The electrical power delivered to the resistor by the circuit is

2for resistors 2 R

loss R R Rusing Ohm's lawP V P V power absorbed by a resistorVi i Ri ( )R

∆= ∆ = ∆→ = =

In this case the potential at a (where the current enters the resistor) is always higher than that at b (where the current exits). When current enters the high-potential terminal of the device, the rate of transfer of electric potential energy into this circuit element.

What becomes of this energy? The moving charges collide with atoms in the resistor and transfer some of their energy to these atoms, increasing the internal energy of the material. Either the temperature of the resistor increases or there is a flow of heat out of it, or both. In any of these cases we say that energy is dissipated in the resistor at a rate of Ri2. Every resistor has a power rating, the maximum power the device can dissipate without becoming overheated and damaged. In practical applications the power rating of a resistor is often just as important a characteristic as its resistance value. Of course, some devices, such as electric heaters, are designed to get hot and transfer heat to their surroundings. But if the power rating is exceeded, even such a device may melt or even explode. Example 28.1 The power rating of a light bulb is the power it dissipates when connected across a 120V potential difference. What is the resistance of (a) a 100W and 60W bulb? (b) How much current does each bulb draw in normal use? (c) Car headlights are either a low or high beam, is there more or less resistance in the high-beam filament?

Solution The bulbs are connected across a 120V potential difference and the power is dissipated according to our power equations.

Question: which light bulb has the highest resistance?

a. Since there is no current given in the problem, we use the voltage form of the power to determine the resistance of the 100W-bulb:

2 2 2solving for R

100W 100WV V (120 V)P R 144 RR P 100 W

∆ ∆= → = = = Ω =

To determine the resistance of the 60W-bulb, we repeat the process: 2 2 2

solving for R60W 60W

V V (120 V)P R 240 RR P 60 W

∆ ∆= → = = = Ω =

b. The currents through these light bulbs, according to Ohm’s law, are

100W 100W 60W 60WV 120 V 120 Vi 0.830 A i and i 0.500 A i

R 144 240 ∆

= = = = = = =Ω Ω

As expected, the 60W-bulb has more resistance than the 100W-bulb, so it draws less current.

28.3

c. Following this example, the high-beam filament must have a lower resistance to maximize power.

Interest Points 1. Household Wires: A 100m length of 12-gauge copper wire, the size usually used in

household wiring, has a resistance at room temperature of about 0.5Ω. As we just saw, a 100W-bulb has resistance of 144Ω. If the same current flows in both the copper wire and the light, the potential difference is much greater across the light bulb, and much more potential energy is lost in the light bulb. This lost energy is converted by the light bulb filament into light and heat. You don’t want your household wires to glow white-hot so as to not have an electrical fire, so its resistance is kept low by using wire of low resistivity and large cross-sectional area.

2. Hair Driers: a hair drier draws about 10A or P = ∆Vi = 120V∙10A = 1200 W. This is almost the maximum allowed current by a 15A circuit breaker. However, if you turned on your clothes drier simultaneously, it would certainly trip your circuit breaker. This doesn’t happen because very high-power appliances need special outlets and dedicated circuit breakers.

Internal Resistance of a Battery The potential difference of real sources is not equal to the emf (∆Vemf ≠ ∆Vab) because of the internal resistance Rint of the power supply. As the current moves through the voltage source, it will experience a drop in potential equal to Rinti due to the internal resistance. The potential across the whole battery ∆Vab (between the positive and negative posts) when current is flowing in a circuit is

−∆ ≡ ∆ = ∆ − = ∆

∆ ≡ ∆

ab

intab battery emf emf int

EPE lost across thechemical potentialincrease in PE = qVinternal resistance

ab battery

V V R i R i

V V

V V

For a real source of emf, the terminal voltage equals the emf only of no current is flowing through the source.

Details about car batteries

What is the internal resistance of a car battery? The Rint of a commercial 12V lead battery is only a few thousandths of an Ohm: Rint(car battery) = 0.003Ω. What is the proper voltage for a fully charged car battery? It's not 12V but 12.6V because each cell is 2.1V and there are six cells.

Video Hacking Open A Car Battery

If the battery is 12.5V = 85% charged, 12.4V = 65%, 12.3V = 50%, 12.2V = 35%, and 12.1V = drained

DEMO 9V-battery with a coin

The current in the external circuit is still determined by ∆Vab = ∆Vbattery = RiR: where we must include the Rint of the battery

emfab emf int

int

VV Ri V R i i

R R−

∆∆ = = ∆ → =

+

This means that the current equals the emf divided by the total circuit resistance. You might have thought that a battery always produces the same current, no matter what it’s used in. Not true; it depends on external resistance and internal resistance.

28.4

Internal Resistance of a Battery (i) Wires always have R = 0. (ii) An ideal voltmeter has an internal resistance Rint = ∞, so that the current through the voltmeter is zero. Since no current is diverted into the voltmeter, they measure voltages in parallel to an element. (iii) An ideal ammeter has an internal resistance of Rint = 0 and so has no voltage drop across it. When an ammeter is placed in series with an element, although the current goes through the ammeter, there is no voltage drop across it. I want to apply these to open circuits, complete circuits, and Short circuits.

Open Circuit with a Voltage Source Consider the open circuit to the right. The wires to the left of a and to the right of the ammeter A are not connected to anything. What does the voltmeter and ammeter read? Since it is not a complete circuit, there is no current through the battery (i = 0), and therefore, no voltage drops across Rint. The voltmeter reads the emf of the battery (Vab = 12V). The only time that you will read the terminal voltage of 12V on a car battery is when there is no current flowing through it. Aside: car battery is a type of rechargeable battery (lead-acid) known as an SLI battery (starting, lighting, ignition), which powers the starter motor, the lights, and the ignition system of a vehicle’s engine.

A Voltage Source in a Complete Circuit Consider the complete circuit on the right. What are the voltmeter and ammeter readings now? The current i through resistor R = 4Ω is

emf

int

V 12Vi 2.0AR R (4 2)

∆= = =

+ + Ω Since wires and ammeters have zero internal resistance, there is no voltage drop across them. The only voltage drop is across the internal resistance and the 4Ω resistor:

4 2 abV V V4 2 8.0V, 2 2 4.0V and 8.0VΩ Ω∆ ∆ ∆= ⋅ = = ⋅ = =

What this means is that the battery is only supplying 8V to the circuit because the internal resistance is “robbing” energy (voltage) from the battery’s emf. However, in circuits speak in the language of power. How much power is delivered by the battery?

=∆= = ⋅ = =

Sbattery emf S batteryi 2AVP i 12V 2A 24 W P

What is the power delivered by the battery? = − = − = − = =2

delivered battery int S int deliveredP P P 24W i R 24W 8W 16 W P That is, the amount of power being delivered from the battery to the 4Ω-resistor is

2 2delivered absorbed dissipatedP P i R (2A) (4 ) 16 W P= = = Ω = =

In summary, the rate at which the battery supplies energy is 24W, of which 8W is dissipated in the battery’s internal resistor and 16W is delivered to the external resistor.

Short Circuiting the Battery If the 4Ω resistor is replaced with a wire (Rwire = 0), what would a DMM read? The current and power out of the battery through the short (or Rint) is

∆∆

∆

= → = = =Ω

= = ⋅ = = =int

emfemf int SC SC

int

battery emf SC battery R

VV

V

12VR i i 6AR 2

P i 12V 6A 72 W P P

Warning: a short circuit can be an extremely dangerous situation. An automobile battery or a house-hold power line has very small internal resistance (much less than in these examples), and the short-circuit current can be great enough to melt a small wire and cause a storage battery to explode. Don’t try it! In real batteries, the internal resistance is

28.5

around 20mΩ where ALL of the converted energy from the source is dissipated within the battery. Car battery that is shorted, will have current and power of

−= = → = ⋅ =× Ω loss312Vi 600 A P 12V 600A 7,200W

30 10

I am not aware of car batteries being able to supply this much current, however, this can put out hundreds of amps in millisecond worth of time. My point is that this much power delivered by short-circuiting battery will either quickly ruined your battery or it may explode.

DEMO 9V-battery with a coin across the terminals (internal resistance is 1.5Ω) Final Remark The principal difference between a fresh flashlight battery and an old one is not in the emf (decreased slightly with use) but in its Rint, which may increase from less than 1Ω to more than 1000 Ω or more. Similarly, a car battery can deliver less current to the starter motor on a cold morning than when the battery is warm, not because the emf is appreciably less but because Rint is temperature dependent, increasing with decreasing temperature. Cold-climate dwellers take a number of measures to avoid this loss and use battery warmers.

Question: We have studied so far that electric field inside a conductor if no charge is placed inside is zero. But we know that every conductor has only a limited number of electrons. What happens when ALL the electrons have aligned to cancel the field inside conductor under application of external field and then we increase the field a little bit more certainly there are no more electrons to cancel the field and this additional field must be present inside the conductor now? I know that practically under electric fields of magnitude that can drag out all electrons from the conductor's body to surface field emission and/or electrical breakdown would be taking place, But is this phenomenon theoretically possible?

Answer: We would require an extremely high intensity electric field to make the conductor “run out” of electrons. As we know, the external electric field is countered by electrons arranging themselves, thereby creating another electric field (keep this in mind, since it means that at the very moment the field is applied, it does in fact penetrate the conductor, but this changes very fast). When the field magnitude is very high, the affects you spoke of can theoretically happen. At the very least though, the intrinsic electric field established by the movement of electrons will stay constant Resistors Series and Parallel Combinations The fundamental nature of how electrical components (C, R, L) combine will either increase/decrease length or increase/decrease area, depends on either adding them in series or parallel. If components are added in series, this combination effectively increases length, whereas components added in parallel increase the effective area. By looking at the resistance equation, the geometry portion has resistance directly proportional to length and inversely proportional to area:

L LRA A

= ρ ∝

Adding resistors in series: since the resistance if directly proportional to the length, adding resistors in series effectively increases the length. That is, the length of one resistor R1 compared to the length of several resistors (R1, R2, R3) is shorter. Adding several resistors in series then increases the length and the overall value of the resistance increases linearly:

series 1 2 3 series 1 2 3R L R R R R where R R ,R ,R ,∝ → = + + >

28.6

Adding resistors in parallel: resistance is inversely proportional to area, so adding resistors in parallel effectively increases the area. That is, the area of one resistor R1 compared to the area of several resistors (R1, R2, R3) is smaller. Adding several resistors in parallel then increases the area and the overall value of the resistance decreases inversely proportional to the area:

parallel 1 2 3parallel 1 2 3

1 1 1 1 1R where R R ,R ,R ,A R R R R

∝ → = + + <

When two or more resistors are combined in series or parallel in a circuit, the combination of the (i) resistance equation and the (ii) geometrical nature of series or parallel setups, automatically defines how the resistance, voltage, current and power behave. Below is a summary of series and parallel resistors.

Conceptual Understanding of Series and Parallel Lightbulbs To observe the behavior of series and parallel resistors, several demonstrations will be shown of combination of lightbulbs in (i) series, (ii) parallel, (iii) one in series with two in parallel and (iv) one in parallel with two in series.

For lightbulbs, a measure of brightness is the power. Let’s analyze these four circuits with no calculations but derive the brightness of each bulb via the above table for resistance, voltage, current, and power.

Series Parallel

Resistance ≡ R

Adding in series INCREASES the resistance increases

series 1 2R R R= + + where Rseries > R1, R2, …

Adding in parallel DECREASES the resistance

parallel 1 2 3

1 1 1 1

R R R R= + +

where Rparallel < R1, R2, …

Voltage ≡ ∆V

Voltages are DIVIDED according to

nn branch

Series

RVDR : V V

R= ∆

branch 1 2KVL : V V V∆ = ∆ + ∆ +

Voltages are NOT divided and are the same across parallel resistors:

Branch 1 2V V V∆ = ∆ = ∆ =

Current ≡ i

Currents are NOT divided and are the same branch 1 2i i i= = =

Series resistors have SMALLER branch currents.

Currents are DIVIDED according to parallel

n branchn

RCDR : i i

R=

branch 1 2KCL : i i i= + + Parallel resistors have a LARGER branch current.

Power ≡ ∆V∙ i

Smaller voltage, higher resistance implies smaller current and power. The battery does not have to work as hard to maintain the lower current output.

Larger voltage, lower resistance implies larger current and power. The battery has to work harder to maintain the higher current output.

28.7

DEMO Series versus Parallel Lightbulbs

Consider the two circuits:

Find the current, voltage drop, power delivered to each bulb and the power delivered to the entire circuit.

Since this is a simple circuit, I will provide the results only:

221 2 S battery S S R R S 2

2

2SS battery S S R R S 2

P 32W

V

VV

V 4VSeries i i i 2A P ( i ) P P 2 i R 16WR 2

8Parallel i 8A P ( i ) P P 2 i R 64WR 1

ΩΩ

Ω

Ω

∆

∆∆

→ = ≡ = = = → = ⋅ = + = ⋅ =Ω

→ = = = → = ⋅ = + = ⋅ =

Remarks 1. Compare the currents and voltages for the series versus the parallel circuits:

parallel series

parallel series

parallel series

Voltages : V V 8V 4V (twice as large)

Currents : i i 8A 4A (twice as large)

Power : P P 32W 8W (four times as large)

> → > > → >

> → >

Bulbs connected in parallel will glow more brightly than in the series circuit

2. Note that this extra power dissipated by a parallel circuit comes at a price. The increased power isn’t obtained “for free.” Energy is extracted from the source 4 times more rapidly in the parallel case than in the series case. If the source is a battery, it will be used up 4 times as fast.

3. It turns out that our calculations are not completely correct – light bulbs are not ohmic; they are dependent on temperature. However, the same results hold true.

DEMO lightbulb circuit Example 28.2 What is the voltage v5Ω and the current i6Ω using series and parallel reduction techniques? Solution When dealing with any circuit with numerous resistors, always identify nodes using different colors and label one of the nodes as the ground. Using this technique helps one learn how to “see” series and parallel resistors.

28.8

There are two nodes connecting resistors 6Ω and 12Ω (nodes A and B), so they are in parallel: 6||12 = 6∙12/(6 + 12) = 4Ω. If the circuit is redrawn, and the lines of potential (nodes) are “straighten out”, the circuit looks like the one on the right. Note that there are two main nodes, which are between node A and the ground and as a consequence, there are two sub-circuits that have the same voltage drop: nodes A-B-ground and A-C-ground. Because of this, the branches act independently and can be treated so.

Nodes A-B-ground: the current i6Ω is encased in the first 4Ω resistor and determining the voltage v4Ω leads directly to the current. Using VDR, we get

Ohm' s law 44 6 6

6

v4v 12V 6V i 1 A i4 4 R

ΩΩ Ω Ω

Ω

= = → = = =+

Nodes A-C-ground: the voltage v5Ω is determined via VDR:

4 420 203 3

5v 12V V V 6.67 V v3 5Ω Ω= = = = =

+

Example 28.3 What is the source current iS and the current i6Ω using series and parallel reduction techniques? Solution Using the color scheme for nodes as we did in the previous example problem, the circuit looks as

Note that because of the location of the 30V-source, the voltage for the right side of the 30V-source is the same as the voltage for the left side. These sub-circuits can be analyzed independently (How cool is this!). That being said, the current of the 30V-source must be i30V = iright + ileft. To solve for the indicated current i6Ω of the left side, the sub-circuit Circuit A is remove from the original circuit.

Because there are two nodes (A-B) are connecting two 14Ω resistors, they are in parallel. The same can be said about the two 14Ω resistors because they also have two nodes between them (B-ground). Mathematically, we get

28.9

14 14 6 614 14 7 and 6 6 314 14 6 6

⋅ ⋅= = Ω = = Ω

+ +

and the sub-circuit Circuit B is shown on the right. Use VDR to determine the voltage across the 3Ω resistor in Circuit B,

3 3v 30V 9V

7 3Ω = =+

Returning back to Circuit A, the 3Ω resistor is two parallel 6Ω resistors, so the current through the indicated 6Ω resistor is

33 6 6 6 62

9Vv R i i A i6Ω Ω Ω Ω Ω= → = = =

Next, ileft is determined from Circuit B using Ohm’s law:

left30Vi 3A

(7 3)= =

+ Ω

Focusing on the right side of the circuit, and repeating the same process as with the left, resistor reduction leads to

Using Ohm’s law to get iright,

right30Vi 10A

(1.5 1.5)= =

+ Ω

Applying KCL at the bottom node connecting to the power supply in the original circuit, the source current i30V is

30V 1 2 30Vi i i 3A 10A 13A i= + = + = =

KIRCHHOFF’S LAWS In order to completely understand a circuit, one needs to know the current and voltage of the all of the elements in the circuit. Kirchhoff’s laws are new techniques that allow one to solve more complicated circuits, however, Kirchhoff’s laws do not introduce any new physics.

Definitions: Node (or Junction): a point in a circuit where 2 or more conductors meet Loop: any closed path

How many nodes and loops are in each of the circuits?

n in out

The amount of current entering a node is equal to the amount of current leaving a node i 0 i i= ←→ =∑ ∑ ∑

Kirchhoff's Current Law (KCL) or Conservation of Charge

28.10

Sign Convention: Current INTO a node (+)-signCurrent OUT of a node ( )-sign

→ → −

Use this sign convention to find the current i1:

nn

The algebraic sum of voltages in any loop is ZERO: V 0=∑Kirchhoff's Voltage Law (KVL) or Conservation of Energy

This statement of Kirchhoff’s law is about the nature of conservative forces: the work done about a closed path that ends up at its starting point is always zero: Wclosed = 0.

Sign Conventions:

Problem Solving Strategies Step 1: Pick the current directions. Pick your loop direction to be in the same direction

as the current direction. Define the resistor polarities.

Step 2: Apply Kirchhoff’s laws • Apply KCL at all nodes • Apply KVL to each loop

Important notes: 1. In multi-loop circuits, the number of unknown currents must equal the number of

equations you have to set up. That is, ( ) ( )# of unknown currents in the circuit # of KCL + KVL equations=

2. When solving for currents in a circuit, when using KVL replace all resistor voltages with V = Ri

Step 3: Solve the simultaneous equations for the current.

Hints: Draw a large circuit diagram and label all quantities. The loop method is mathematical – try not to think about it too much. Just do it!

Example The circuit shown contains two batteries, each with an emf and an interval resistance, and two resistors. Find (a) the current in the circuit, (b) the potential difference Vab, and (c) the power output of the emf of each battery.

Solution a. Find the current in the circuit

How many currents are in this circuit? One – that means that there will only be one current to solve for and therefore, one loop according to KVL.

Pick the current and loop directions, and define the resistor polarities

n 1 1n

in out 1 1

KCL i 0 1 3 7 i 0 i 3A

i i i 1 3 7 i 3A

⇒ = ⇒ + − − = ⇒ =

⇒ = ⇒ + + = ⇒ =

∑

∑ ∑

28.11

Apply KVL around the loop and rewrite the resistor voltages using Ohm’s law into currents. Starting at node a:

4 2 7 1 2 3V Vv v v v 0 4i 6V 7i 12V 2i 3i 0

Ω Ω Ω Ω∆ ∆+ + − + + =

+ + − + + =

Solving for i, we get solving for i4i 4 7i 12 2i 3i 0 8 16i i 0.5A+ + − + + = → = → =

b. To find the voltage Vab, note that Vab = Va – Vb so the polarity or signs at points a and b are (+) and (−) respectively. The loop direction will be in the CW direction. Applying KVL, we get

absolving for Vab ab abV 7i 4 4i 0 V 11i 4 V 9.5V− − − = → = + ⇒ =

c. Remember the definitions of passive and active elements. If the current and voltage are in opposite directions, the element is active; whereas if they are in the same direction, the element is passive. The power output of each battery or emf is

11 2

12 2

1

2

V

V

P i 12V A 6 Watts delivering power to the circuit

P i 4V A 2 Watts absorbing power from the circuit

∆

∆

= ⋅ = ⋅ = ⇒

= ⋅ = − ⋅ = − ⇒

Remarks 1. The 12V-emf delivers 6 W of power – so where does it all go?

2

Resistorsgoes into charging V dissipated by the resistances

1 2P P P 6W 2W 4W ∆

= + → = +

2. What happens if the positive terminal of the ∆V1 battery is connected directly to the negative terminal of the ∆V 2 battery? Short out the battery and the sulfuric acid in the battery boils and creates an enormous pressure inside the battery that it explodes.

Example 28.4 (a) What is the current of the indicated 8Ω-resistor using series and parallel reduction techniques? (b) What is the energy transfer (power) of the 12V and 8V-sources (Hint: determine i1 and i2)? Which source(s) is active or passive? Solution a. To determine the current of the indicated 8Ω-resistor, focus on the right side

and apply some resistor reduction:

28.12

Note that the indicated 8Ω resistor is in the 4Ω resistor in Circuit C. Apply VDR to get v4Ω:

4 4v 12V 6.86V

4 3Ω = =+

Moving from Circuit C to Circuit B, this voltage is across the parallel elements in Circuit B. The important point though is that this voltage is across the series two 4Ω resistors. Apply VDR again, the voltage is

4 4v 6.86V 3.43V

4 4Ω = =+

Moving from Circuit B to Circuit A, this voltage is across the two parallel 8Ω resistors in Circuit A. Since parallel resistors have the same voltage, the indicated 8Ω resistor has a voltage of 3.43V and a current of

88 8 8 8 8

8

v 3.43Vv R i 3.43V i 0.429 A 429 mA iR 8

ΩΩ Ω Ω Ω Ω

Ω

= = → = = = = =Ω

b. To determine whether a source is active or passive, the currents i12V and i8V must be determined. That is, the powers are P12V = V12Vi12V and P8V = V8Vi8V. The “easiest” current is i2 = i8V, since we can isolate the current i2 by focusing on the right loop of the original circuit, which I labeled as Circuit D. Assuming that the current i2 goes down, the polarities of 4Ω and 1Ω resistors are then defined. Applying KVL in the clockwise direction,

4 1 2 2 2 24A

12 v v 8 0 12 4i 1i 8 0 i 4A or iupΩ Ω

− + + + = → − + + + = → = − =

This means that current i2 enters the negative node of the 8V-source and is therefore, active:

8V 8V 2 8VP V i 8V ( 5A) 20W P Active= = ⋅ − = − = To determine the power of the 12V-source, the current i12V must be determined, which forces us to focus on Circuit D. The direction of current i1 must be downwards because the voltage drops across the 3Ω and 8Ω resistors is parallel to the 12V-source. Applying Ohm’s law to get the current i1, we get

1 11.09 A12Vi 1.09A idown(3 8)

= = + → =

+ Ω

At node X, applying KCL to get i12V, 12V 1 2i i i 1.09A 1.6A 2.69A= + = + =

The current i12V enters the negative node of the 12V-source and is active:

12V 12V 12V 12WP V i 12V ( 2.69A) 32.3W P Active= = ⋅ − = − =

28.13

Example 28.5 (a) Determine the magnitude and direction of currents i1 and i2. (b) What is the energy transfer (power) of the 12V and 4V-sources? Which source(s) is active or passive?

Solution a. The resistor for the current i1 is in parallel with the 12V-source

and therefore, must have the opposite direction for the current given by

1 13A12Vi 3A idown4

= − = − → =

Ω

To determine the current i2, use series and parallel resistor reduction to simplify the right side of the circuit:

right sideR 4 4 4 4 10= + + = Ω

The circuit with this resistor reduction looks like the one on the right. Applying KVL clockwise around the right loop gives

4 10 2 21.6A

v v 4 0 12 10i 4 0 or idownΩ Ω

− + − = → − + − = =

b. Since the current i2 enters the negative node of the 4V-source, the source is active (supplying energy to the circuit) where its power is

4V 4V 4V 4V 2 4VP V i V i 4V ( 1.6A) 6.4W P Active= = = ⋅ − = − = To determine the power of the 12V-source, the current through the source must first be determined. Apply KCL at node X, both current flow into the node, so

12V 1 2i i i 3A 1.6A 4.6A= + = + = Therefore, the current i12V enters the negative node of the 12V-source and is active. The power is

12V 12V 12V 4VP V i 12V ( 4.6A) 55.2W P Active= = ⋅ − = − = Why do high voltage transmission lines have such high voltage?