Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 33

Now, imagine a positive charge dQ flowing clockwise from point a through the battery and the resistor, and back to the same point a.. A battery is often called a source of electromotive

24.2 Ohm’s Law and Electric Resistance 823

I =V

2.4 × 103= 5 × 10−3A= 5 mA

(c) At the inner and outer faces of the silicon, namely 2π a L and 2 π b L,

respectively, we use Eq.24.7to find the corresponding current density as follows:

trans-of the filament It is important to calculate the rate trans-of this energy transfer

Figure24.10shows a battery of potential differenceV connected to a simple circuit (our system) containing a resistor of resistance R The resistor is usually rep-

resented by the symbol Unless noted otherwise, we assume that the connectingwires have zero resistance

Fig 24.10 A simple circuit

containing one battery and one

V

R

S

Now, imagine a positive charge dQ flowing clockwise from point a through the battery and the resistor, and back to the same point a In a time interval dt a quantity

of charge dQ enters point a, and an equal quantity leaves point b Thus, the electric potential energy of the system increases by the amount dU = dQ V , see Eq.22.18,

while the stored chemical potential energy of the battery decreases by the same amount On the other hand, as the charge enters the resistor at b and an equal

quantity leaves a(which is identical to a) over the same time dt, the system losesthis energy through collisions with the molecules of the resistor The net result isthat some of the chemical energy of the battery has been delivered to the resistor asinternal energy associated with molecular vibration (rise in temperature) This rise intemperature will ultimately transfer to the surroundings through thermal radiation.The rate at which the system loses energy as the charges pass through theresistor is:

Using the relationV = I R for a resistor of resistance R, the electric power P

deliv-ered in the resistor can be written in the following form:

P = I V = I2R= (V )2

Because P = I V , the same amount of power P can be transported either at high I

and lowV , or at low I and high V

We previously introduced the battery as a device that produces a potential difference

and causes charges to move In fact, it is a device that works as an energy converter.

A battery is often called a source of electromotive force or, a source of emf (this

unfortunate historical name describes a potential difference in volts, but not a force).Spotlight

The emfE of a battery is the maximum possible potential difference that the

battery can provide between its terminals, usually the voltage at zero current

Figure24.11a shows a device (a battery) with an emfE that is used in a simple

circuit containing a resistor of resistance R The battery keeps one terminal (labeled

with the sign+) at a higher electric potential than the other (labeled with the sign

−) Therefore, within the battery, the conventional positive charge carriers move

from a region of low electric potential (at the negative terminal) to a region of higherelectric potential (at the positive terminal)

Because a real battery is made of matter, there is a resistance against the flow of

charge within the battery This resistance is called the battery’s internal resistance

and is usually denoted by r For an ideal battery with zero internal resistance, the

I I

I

V

Fig 24.11 (a) A simple circuit containing a resistor connected to a battery (b) A circuit diagram of a source of emfE(the battery) of internal resistance r, connected to a resistor of resistance R (c) Graphical

representation of the electric potential at different points

potential difference between its terminals is equal to its emfE (directed from the −

terminal to the + terminal) For real batteries, this is not the case.

We now consider the circuit diagram in Fig.24.11b, which is the same as the real

emf device of Fig.24.11a, except we represent the battery with a dashed rectangular

box containing an ideal emf E in series with an internal resistance r Let us start at

point a (where the potential is V a ), and move clockwise to point b (where the potential

is V b), and measure the electric potential at different locations When we move from

the negative terminal to the positive terminal, the potential increases by the amount

of the emfE However, as we move through the internal resistance r in the direction

of the current I, the potential drops by an amount Ir Thus, the potential difference

between the terminals of the batteryV = V b − Vais:

We always assume that the wires in the circuit have no resistance, unless otherwise

indicated This means that the potentials of points a and aare the same The same

applies to points b and b Thus:

But according to Ohm’s law, given by Eq.24.16, V b − Va must equal IR Thus,

Vb − Va = Vb − Va= IR Combining this expression with Eq.24.25, we find that:

24.4 Electromotive Force 827Solving for the current, we get:

I = E

Note that the current I depends on the resistance R of the external resistor (which

is called the load) and the internal resistance r of the battery Since R  r in most circuits, we can usually neglect r.

Example 24.7

A device is connected to a battery that has an emfE = 9 V and internal resistance

r = 0.02  Find the current in the circuit and the terminal voltage of the battery when the device is a: (a) light bulb that has a resistance R = 4 , see Fig.24.12a.(b) conducting wire having zero resistance, i.e the battery is short circuited bythis conductor, see Fig.24.12b

V = E − Ir = 9 V − (450 A)(0.02 ) = 0

Such large values for the current I would result in a very quick depletion of the

battery as all of its stored energy would be quickly transferred to the conductingwire in the form of heat energy The term “short circuit” is applied to such cases,and they can cause fire or burns

Example 24.8

A battery that has an emfE1= 9 V and internal resistance r1= 0.02  is connected

to a second battery ofE2= 12 V and r2= 0.04 , such that their like terminals are

connected, see Fig.24.13 Find the current in the circuit and the terminal voltageacross each battery

Solution: The two batteries are oppositely directed around the circuit Since E2>

E1, then the net emfEnetin this circuit will be in the counterclockwise direction,i.e.:

Enet= E2− E1= 12 − 9 V = 3 V (Counterclockwise direction)

Consequently, the current I in this circuit will also be in the counterclockwise

direction as indicated in Fig.24.13 This current is opposite to the discharging

current that theE1= 9 V battery should produce when connected to circuits

con-taining only resistors Actually, this current will charge theE1= 9 V battery.The total resistance of this circuit is only due to the presence of the internal

resistances r1and r2of the two batteries Therefore, Eq.24.28gives us the value

of the current as follows:

24.4 Electromotive Force 829Depending on the direction of the current in each battery, the terminal voltagesacross the batteries are:

V = V b − Va= E1+ Ir1 = 9 V + (50 A)(0.02 )

= 10 V (Gain from a to b)

V = V b− Va = E2− Ir2 = 12 V − (50 A)(0.04 )

= 10 V (Drop from ato b)

24.5 Resistors in Series and Parallel

Resistors in a circuit may be used in different combinations, and we can sometimes

replace a combination of resistors with one equivalent resistor In this section, we

introduce two basic combinations of resistors that allow such a replacement

Resistors in a Series Combination

Figure24.14a shows two resistors R1 and R2 that are connected in series with a

battery B Figure24.14b shows a circuit diagram for this combination of resistors

When the circuit is connected, the amount of charge that passes through R1must

also pass through R2in the same time interval Otherwise, charge will accumulate on

the wire between resistors Thus, for series combination of resistors, the current I is

the same in both resistors Figure24.14c shows a single resistor Reqthat is equivalent

to this combination and has the same effect on the circuit This means that when the

potential differenceV is applied across the equivalent resistor, it must produce the same current I as in the series combination.

The potential differenceV is divided to V1andV2across the resistors R1

and R2, respectively Thus:

For the two resistors in Fig.24.14b, we have:

V1= Vc − Vb = IR1 and V2= Vb − Va = IR2 (24.30)Substituting in Eq.24.29, we get:

The equivalent resistor Reqhas the same applied potential differenceV and the same circuit current I flowing through it; thus:

Canceling I, we arrive at the following relationship:

We can extend this treatment to n resistors connected in series as:

Req = R1 + R2 + · · · + Rn (Series combination) (24.34)Thus, the equivalent resistor of a series combination of resistors is simply the alge-braic sum of the individual resistances and will always be greater than any one ofthem

Example 24.9

In Fig.24.14, let R1= 6 , R2 = 3 , and V = 18 V Find I in the circuit and the

potential differencesV1andV2

Solution: The equivalent resistance of the series combination is:

Req = R1 + R2 = 6  + 3  = 9 

24.5 Resistors in Series and Parallel 831Using Ohm’s law, given by Eq.24.16, we find:

Resistors in a Parallel Combination

Figure24.15a shows two resistors of resistances R1 and R2 that are connected in

parallel with a battery B Figure24.15b shows a circuit diagram for this combination

of resistors The potential differenceV between the battery’s terminals is the same as

the potential difference across each resistor Figure24.15c shows a single resistance

Reqthat is equivalent to this combination and has the same effect on the circuit

Req replacing the parallel combination

When the current I reaches junction b, it will split into two parts, I1in R1and I2

in R2 Because electric charge is conserved, the current I that enters junction b must

equal the total current leaving that junction; that is:

Because the potential difference V across the resistors is the same, then from

Fig.24.15b, we have:

V = I1R1 and V = I2R2 (24.36)Substituting into Eq.24.35, we get:

I= V

R1 +V

R2 =

1

R1+ 1

R2



An equivalent resistor with the same applied potential differenceV and total current

I has a resistance Reqgiven byV = I Req Thus:

24.5 Resistors in Series and Parallel 833

Example 24.11

In Fig.24.16, let R1= 3 , R2 = 6 , R3 = 1 , R4 = 7 , and Vda = Va − Vd=

30 V (a) What is the equivalent resistance between points a and d? (b) Evaluate

the current passing through each resistor

Solution: (a) We can simplify the circuit by the rule of adding resistances in series

and in parallel in steps The resistors R1and R2are in parallel and their equivalent

resistance R12between b and c is:

I =V da

Req = 30 V

10 = 3 A (Current through the battery, R3 and R4)

SinceV cb = IR12 = I1R1 = I2R2 , then we find I1and I2as follows:

I1= IR12

R1 = (3 A)(2 )

3 = 2 A and I2=

IR12 R2 = (3 A)(2 )

6 = 1 A

24.6 Kirchhoff’s Rules

Not all circuits can be reduced to simple series and parallel combinations A techniquethat is applied to loops in complicated circuits consists of two principles calledKirchhoff’s Rules

Kirchhoff’s Rules:

1 Junction rule

At any junction in a circuit, the sum of the ingoing currents must equal the sum

of the outgoing currents That is:

When we apply Kirchhoff’s second rule to a loop, we should note the followingsign conventions:

(1) When a resistor is traversed in the direction of the current, the potential enceV is −IR (Fig.24.17a)

differ-(2) When a resistor is traversed in the direction opposite the current, the potentialdifferenceV is +IR (Fig.24.17b)

(3) When a source of emf is traversed in the direction of its emf (from− to +), thepotential differenceV is +E (Fig.24.17c)

(4) When a source of emf is traversed in a direction opposite to its emf (from+

to−), the potential difference V is −E (Fig.24.17d)

Fig 24.17 The potential differencesV = V b − V a across a resistor of resistance R and a battery of emf

E(assumed to have zero internal resistance), when each element is traversed from a to b

Of course, we do not need Kirchhoff’s rules to solve this simple loop circuit

We are just using it to practice applying the loop rule

Example 24.13

In Fig.24.18, let R1= 2 , R2 = 6 , R3 = 4 , E1= 10 V, and E2= 14 V Find

the currents I1, I2, and I3in the circuit

Solution: We cannot simplify the circuit by the rule of adding resistances in series

and in parallel Thus, we must use Kirchhoff’s rules By applying Kirchhoff’s

junction rule to the junction f, we get:

We have three loops in this circuit, but we need only two loop equations todetermine the three unknown currents Applying Kirchhoff’s loop rule to the loops

abcfa and fcdef and traversing these loops clockwise, we obtain the following

equations (after temporarily omitting the units, since they are all consistent SIunits):

2

(2) Loop abcfa: I1R1 + I2R2− E1= 0 ⇒ 2I1 + 6I2− 10 = 0

(3) Loop fcdef : E1− I2 R2 + I3 R3+ E2= 0 ⇒ 24 − 6 I2 + 4 I3= 0Substituting Eqs (1) into (3) gives:

24− 10 I2 + 4 I1= 0Dividing this equation by 2 gives:

I3 = I1 − I2= −1 A − 2 A = −3 AThus (I = −1A, I2 = 2A, I3 = −3A)

24.6 Kirchhoff’s Rules 837

We notice that I1and I2are both negative This means that the currents are opposite

to the direction we chose However, the numerical values are correct

Using this value of I2in Eq (4) gives I1a value of:

−4 I1 + 8(0.5) − 3 = 0 ⇒ I1 = 0.25 A

Finally, from Eq (1) we have:

I3 = I1 − I2 = 0.5 − 0.25 A = 0.25 A Applying Kirchhoff’s loop rule to the loop cdefc gives:

Loop cdefc: Q /C − E2+ E1= 0 ⇒ Q = (E2− E1)C = − 6 µC

the switch S is closed at time t = 0, charge begins to flow, setting up a current I in

the circuit, until the capacitor is fully charged and the current becomes zero

- q

t > 0

Fig 24.20 (a) A capacitor in series with a resistor, switch, and battery (b) The circuit diagram before

the switch is closed (t < 0) (c) The circuit diagram at time t > 0 after the switch is closed at t = 0

Assume that the current in the circuit at time t > 0 is I and the magnitude of the charge on the capacitor is q (Fig.24.20c) Applying Kirchhoff’s loop rule andtraversing the circuit clockwise, we get:

At t = 0, the potential difference across the battery appears entirely on the resistor.

At t = ∞, when the capacitor is fully charged to its maximum value Q, the current is

zero and the potential difference across the battery appears entirely on the capacitor.Therefore, Eq.24.43gives:

q − C E = −

dt

R C Substituting with C E = Q in this expression and integrating from the initial charge

q = 0 at t = 0 to an arbitrary charge q at time t, we get:

q

0

dq

q − Q = −

t

0

The quantity R C in the exponents of Eqs.24.46and24.47is called the time

con-stantτ of the circuit Therefore, the quantity τ = R C represents the time interval ing which the charge on the capacitor increases to Q (1 − e−1) = 0.632 Q, i.e., ∼ 63%

dur-increase Similarly, after a time intervalτ, the current decreases to 1/e of its initial value; that is, I = e−1I◦= 0.368I◦(∼37% decrease).

Figure24.21shows the variation of the capacitor charge q and the circuit current

I as a function of time.

0 1 2 3 4 5 6

Fig 24.21 (a) A plot of the charge q on the capacitor of Fig.24.20versus time t (b) A plot of the current

I in the same figure versus time t The two curves are for R = 2 k, C = 1 µF, andE= 10 V

Discharging a Capacitor

Let us consider the circuit shown in Fig.24.22a, in which we have a capacitor of

capacitance C carrying an initial charge Q, a resistor of resistance R, and an open switch S When the switch is closed at time t = 0, the capacitor begins to discharge through the resistor If the current in the circuit at time t > 0 is I and the magnitude

of the charge on the capacitor is q (Fig.24.22b), then by applying Kirchhoff’s looprule and traversing the circuit clockwise, we get:

+IR− q

To find the charge as a function of time, we substitute with I = −dq/dt (the rate

of decrease of charge on the capacitor) into Eq.24.48and rearrange the equation asfollows: