Structure

16.1 Introduction Objectives

16.2 Electric Current 16.2.1 Drift Velocity and Current Density 16.2.2 Ohm's Law

16.3 Resistances in Series and Parallel 16.4 Electromotive Fame (EMF) 16.5 Kirchhoff 's Rules 16.6 RC Circuits 16.7 Energy and Power in Electric Circuits 16.8 Summary 16.9 Answers to SAQs

16.1 INTRODUCTION

In the preceding units of this Block, we have dealt with Electrostatics, which deals with the effects of stationary electric charges. You have learnt that, under static conditions, the electric field inside a metal is zero. If, however, we maintain a nonzero field in a conductor, say, by connecting two lead wires from the conductor to the terminals of a battery, the charges in the conductor becomes mobile, and, an electric current is established in the conductor. We need not emphasize the role the electric current plays in our life, as you are quite familiar with it.

In this unit, we shall discuss electric circuits, in which mobile charges exist, giving rise to a current. We will see that charges do not flow inside a material unimpeded, but the material offers a certain resistance to their flow. The amount of resistance offered varies from material to material and depends on the dimensions of the specimen. Circuits with currents flowing through resistances are most common ones in practice. We will restrict ourselves to the case where the sense of the current is continuous along one direction. Such currents, called direct current (dc) are supplied by batteries, generators etc. Circuits in which the sense of current oscillates back and forth are called alternating current circuits will be briefly talked about in the Block 5. Later in this unit, we will consider circuits with both resistances and capacitors.

After reading this unit, you should be able to :

explain the concepts of current density and drift velocity (SAQs 1-3),

use Ohm's law to calculate the resistance of conductors (SAQs 4-6),

explain what is electromotive force and obtain current distribution in a circuit containing resistances and Emfs (SAQs 7-9)'

solve problems on electrical circuits containing capacitors and resistances (SAQ l l ) , and

explain how power is dissipated in a circuit containing a resistance

16.2 ELECTRIC CURRENT P

The electric current measures the rate of flow of charge through a rnaierial. In Fig. 16.1, a conducting wire is shown with positive charge carriers moving to the right. If a net amount of charge AQ passes through a cross section labelled S in a time interval At, then .the current I is defined as

Figure 16.1 Figure 16.2

You should be clear about the meaning of the word "net" in this context. We pointed out that in a conductor, free charges exist, which can move when the conductor is subjected to an electric field. These charge carriers are usually electrons, which carry negative charges. When an electric field is applied, the negative charges move in a direction opposite to the direction in which the carriers would move if they were positively charged. The electric current is not a vector quantity. Nevertheless, it is common to talk about a direction of current. Conventionally, the direction of the electric current is taken to be the direction in which a positive charge carrier moves. If the charge carriers are negative (which is usually the case with metals), the direction of the electric current is opposite to the direction of charge flow. The word "net" in our definition, takes care of charge carriers accounting for their signs. For example, if equal amount of positive and negative charge flow through an area in the same direction in a time At, the net flow of charge is zero. On the other hand, if an amount AQ of negative charge flows through a cross section to the left, and an equal amount of positive charge cross the same cross section simultaneously to the right, the net flow of charge is 24Q towards right. Clearly, if an electrically neutral at0111 passes throgh a given cross section, it does not contribute to current, because i t carries equaa amount of positive and negative charge which pass through the section siniultaneously. Figure 16.2 shows the direction of current in a wire in which the charge carriers are electrons.

The SI unit of current is Ampere which is defined as a rate of flow of one Coulomb of charge through a cross section per second.

Example 1: A current of 10 A exists in a conductor. Assuming that this current is entirely due to the flow of electrons, (a) find the number of electron crossing the area of cross section per second, (b) if such a current is maintained for 1 hour, find the net flow of the charge.

Solution : (a) A current of 10 Amperes means a flow of 10 C of charge per second. As

the charge on electron is 1.6 x ~ O - ' ~ C , there is a net flow of 1011.6 x = 6.25 x 10" electrons per second through a given cross section of the conductor. (b) If a constant current of 10 A is maintained for an hour, the net charge flowing through the cross sectiorl would be 10 x 3600 = 3.6 x lo4 Coulomb.

Suppose we can charge a sphere of radius 0.2 m by supplying it a constant current of 1 micro ampere, the supplied charge being accuml~lated inside th sphere. Find the time taken for the sphere to be raised through a potential difference of 500 V.

16.2.1 Drift VeIscity and Current Density

When a conductfi7r is subjected to m electric field, the charge carriers - experience a force. I he charges that constitute current can be of many types. In ionised gases tiley can be positively charged ions and electrons. In semiconducto~s, they can be electrons or holes (the absence of an electron in an energv level, is called a hole, which effectively behaves like a positively charged particle). Elowever, the most important class of material in which we are interested, as far as the current is concerned, is metal. In most of the metals: at least those which are used for making wires to carry current, the current is exdu5ively carried by electrons only, though there are some metals in which electron as well as Roles contribute to current. In these metals some of the electrons of the atoms (usually the outer shell electrons) are comparatively free and are mobile inside the metal. Even when no external electric field is applied these electrons keep on moving inside the metals and their speed can be as high as lo6 m/s. However, these electrons do not constitute current because these electrons move with equal probability in all directions, and therefore, there is rro net flow of electrorrs through any given cross section in any particular direction.

When an external electric field 2 is applied, the electrons tend to accelerate in a direction opposite to the electric field and hence their velocity in this direction increases. If there were no other force on the carriers, they would have a constant acceleration and their velocity would go on increasir~g indefinitely. However, the charge carriers interact with other particles in the material, colliding against ions, defects in the materids etc. The net effect of such interactions is that, in an electric field, the charge carriers move with a constant average velocity called the drift velocity. An analogy can be made with the gas molecules in air. In the absence of any breeze, the gas molecules keep on moving in all directions randomly. In the presence of a breeze, on the other hand, there is a net drift of molecules in a particular direction.

Consider a wire with an area of cross section A, with a charge density of n per unit volume. If vd is the magnitude of the drift velocity of the carriers, the net flow of the charges AQ through a cross section, in time At, is (vdAtA)ne. The current I is given by

AQ I = - = nevdA At

(16.2)

Electrlc Circuits

Remember that if the charge carriers are negative, the direction of vd will be opposite to the direction of electric field, but as e also is negative, I will be in the same direction as the electric field. We see from Eqn.(16.2) that, except for area A, I depends on intrinsic properties of the conductor, Like n and vd. This leads us to define a more fundamental concept, called the current density, .f Current density is a vector, the direction of which is same as that of electric field which causes the current. The magnitude of the current density is given by

- I I J (= ;i = nevd

assuming that the. current is uniformly distributed across the cross section. The above equation can also be written in the vector form as

1

J = neGd (16.3j -

If the charge carriers in Eqn.(16.3) are negative, we shall take a negative sign for e, which will lead to a sign of fopPosite to Cd, but same as that of electric field. You may note that the current density is a microscopic concept, while the current is a macroscopic concept.

If the drift velocity of the carriers varies from point to point within the material, the current density will also vary. In such a case, the current density can be found by calculating the surface integral of the current density,

where integral is taken over the surface.

Example 2 A current of 1 A passes through a copper wire of radius of cross section 1 mm. Find the current density and the drift velocity of electron.

Solution :

In order to find the drift velocity using Eqn.(16.3), we need to know the number of electrons per unit volume. For this, we assume that each Cu atom contributes one free electron for conduction purposes. The density of Cu is 8.96 glcc while its atomic weight is 63.55. You know that 63.55 kg of Cu contains Avogadro's number, i.e., 6.022 x atoms. A unit volume of Cu thus contains 6.022 x x 8960163.55 = 8.5 x atoms. Substituting this value for n, we get the drift velocity to be given by

Notice that the magnitude of the drift velocity is many orders of magnitude smaller than that of the random velocity of the charge carriers. This does not mean that when we switch on electricity, the time taken for a bulb to glow will be determined by this drift velocity. The time lag between the switching on and the glow of a bulb is determined by the time taken for the information on the change of electric field to travel between the two points, and is practically instantaneous, since this speed is much higher and approaches that of light.

SAQ 2 : Silver has 5.9 x lo2' number of free electrons per unit volume. If the current density is lo5 A/m2 in a silver wire, what is the drift velocity of the electron? What is the radius of the wire which carries a current of 10 A ?

SAQ 3 : A salt solution kept in a long glass tube of cross section 100 mm2, carries a current of 1 A. The carriers in the solution are singly charged positive and negative ions with equal densities n+ = n- = 6 x loZS ions/m3. If the drift velocity of the positive ions is four times that of the negative ions, find the contribution to the current, the drift velocity and the Zurrent density by the two types of carriers separately.

16.2.2 Ohm's Law

We have seen that the application of electric field produces a current in a conductor. It is found that the current density is proportional to the electric field that is applied across the conductor, i.e.,

Here, a is the constant of proportionality called the wnductivity of the material. Experimentally, the conductivity is found to depend on the material used, the temperature, the purity of the sample etc. However, it does not depend on the shape of the material. The S.I. units of conductivity is (ohm m)-', which is also sometimes written as mho/m. The law represented

I- by Eqn.(16.5) is called Ohm's law.

It follows from Eqn.(16.5) that if the conductivity of the material is large, it is a better conductor, because for the same electric field, it gives a larger current density. Actually, the conductivity of material is related to the average time interval between two successive collisions of the conduction electrons with the ions or the other particles of the medium mentioned earlier. We define a quantity p, called the resistivity of the material, which is inverse of conductivity.

1 P = ;

The S.I. units of resistivity is ohm m.

Most of the time, Ohm's law is written in slightly different form, which is

Electric Clrcults

found to be more useful. Experimentally, we measure directly the current I through a wire when a potential difference V is applied across its two ends. If the area of the cross section of the wire is A, and its length I, we have, E = V/l and J = IIA. Substituting these in Eqn.(16.5), we get

Making use of Eqn.(16.6), we can write the above equation as

1 v = ( ~ ~ 1 '

The quantity in the bracket pl/A is known as the resistance R of the material. This yields the more familiar form of Ohm's law

V = I R

The resistance of a material depends on the length and area of cross section of the wire used. The S.I. unit of resistance is ohm, which is usually denoted by 51. In a circuit the symbol for the resistance is -WW[IVL.

Example 3 A wire of length 1 m, and a cross section area of 1 mm2 carries a current 1.5 A when a potential difference of 3V is applied between its two ends. (a) Find the conductivity of the wire. (b) If the same wire is redrawn so that now its length is 2 m, find the resistance of the wire, assuming its density to remain unchanged.

Solution: (a) The resistance of the wire is V/I = 2 ohms. The conductivity a = 1/RA= 5 x lo7 (51 m)-'. (b) The volume of the wire remains the same on being redrawn. Since the new length is twice the original length, the area of cross section is halved: Therefore the ratio of the length to the area is four times the original. Since the conductivity remains the same, the resistance becomes 8 51.

SAQ 4 : Two wires A and B are made of a material which has a resistivity of 5 x lo-* ohm m. The length of the wire A is 100 m while that of B is 200 m. The area of cross sections of the wire B is 2 mm2 while that of A is 1 mm2. These two wires are welded together to form a single wire of length 300m. This composite wire carries a current of 1 A. Find the potential difference, the electric field across the wires and determine the current densities.

SAQ 5 : The dimensions of a rectangular block of a metal are 3 cm x 4 cm x 5 cm. To

I \

which set of parallel faces should an electric current be applied, in order to get least resistance ?

16.3 RESISTANCES IN SERIES AND PARALLEL

ElecMc Circuit 3

In Unit 15 you have seen two limiting cases of combinations of capacitors, viz., the series and the parallel combinations. We will now discuss similar combinations with resistances. It is worth mentioning here that often, we deliberately introduce standard resistances into the circuits. Such resistances, called resistors, are commercially available and their resistances are known. In this section we shall assume that we have a series or a parallel combination of resistors, and the resistances of connecting wires are negligible in comparison to these resistors. Figure 16.3 shows a series combination of three resistors which have been connected to a battery, which provides a potential difference V across the combination. The first thing that you should note is that, the potential difference across each of the resistance, in general, are different, but v the current through all of them is the same. This is because, under steady state conditions, we do not expect any accumulation of charge. If I is the current in the circuit, then V,, the potential difference across ith resistor is given by Ohm's law

V; = R;I

Now, the total potential difference across the combination is the sum of the potential differences across each resistor. We have, therefore,

Fipre 16.3

The left hand side of above equation is the net potential difference across the R,

combination, divided by the current in the combination. Hence, this is the - effective resistance R of the combination. We therefeore have, for a series I - I combination,

Figure 16.4 shows a parallel combination of three resistors. In this case, the potential difference across each resistor is the same and equal to V, the potential difference provided by the battery. However, current is different in the three resistors . If I is the current through the battery, we have v

I = ~ I ; Figure 16.4

where Ii is the current through ith resistance. Using Ohm's law, we get.

As before we can write VII as R, the effective resistance of the combination. Thus, for a parallel combination,

,-

Example 4 In Fig. 16.5, determine the value of the unknown resistance R, so that the effective resistance between the points a and b is 10 ohms ?

Solution : The 30 resistance is in parallel with R. Their effective resistance Re is

which gives 3R

Re = - 3 + R

This effective resistance R, is in series with 1 0 and 7 0 resistances. Thus, i r i

order that the effective resistance between the points a and b is 10 R , we 111uht

have 3R

Fignre 16.6 + 1 + 7 = 1 0

3 + R I which gives the value of the unknown resistance R to be 60. 7

SAQ 6 : In Example 4, find the current through each of the resistances, if the potcnti;~l difference between the points a and b is maintained at 10 V.

Example 5: Find the resistance of a spherical shell of outer radius rl and inner radius r2. The shell is made of a material of conductivity a .

Solution : Consider an element of the shell, in the form of a concentric thin shell of radius r and thickness dr, as shown in Fig. 16.6 but concentric with the conducting shell. The resistance of this element is given by

1 1 1 d r dR = -- = -- a a a4.rrr2

As the entire medium can be thought of as a series combination of resistances offered by such elements, and using the fact that the resistances in series add

Figure 16.6 up, we can write the total resistance R of the spherical shell as

16.4 ELECTROMOTIVE FORCE (EMF)

As discussed earlier, the electric current is produced in a conductor due to thc. presence of an electric field, which exerts a force on the charges which move from a higher potential to a lower potential. Clearly, this cannot go on indefinitely, for, if we are to have a continuous current, there must be a constant supply of charges at the higher potentid end. In any closed circuit,

this means that the charges which E z ~ e reached the lower end of the potential energy, must be pumped to higher ptential end by some device. The situation is somewhat analogous to a wat'er fountain. If one has to continuously run such a foantain, the water which has fallen to gr&nd (lower potential energy) must be pumped back to a height by an external agency. An agency which continuously pumps charges from low potential energy to high potential energy and thus help maintain a current is called a source of electmmotive force. The more common terminology is to call it a source of emf. A battery, a generator, a solar cell, a thermocouples are examples of sources of emf. In a battery, chemical reactions take place, which performs the work of pumping charges from a lower potential to a higher potential. In other words, it helps to maintain a constant potential difference between its electrodes. If a work dW is done on a charge dQ, to pump it from an electrode at a lower potential to an electrode at a higher potential, the emf & is defined as

Note that & does not have the dimensions of force. In fact, it has the same units as the potential, viz., volt.

You are familiar with the 1.5 V cells that are used in torches, radio - transistors and in many other consumer electronic items. What does such a cell do ? If it is an ideal cell, it maintains a potential difference of 1.5 V between its two terminals. If this battery is connected to a circuit, a current is established in the circuit and a charge (say of magnitude Q, assumed positive by convention) loses a total energy of 1.5Q J as it travels from the positive to the negative terminal of the battery. The battery again increases the energy of the charge by 1.54 J by doing work. Hence, the total work done by the battery per charge is 1.5 V. Thus the emf of the cell is 1.5 V.

What we said in the last paragraph is strictly true only for an ideal battery or cell. In practice, all batteries and cells possess what is called an internal resistance. A more realistic picture of a battery or a cell is, therefore, a combination of source of emf and a resistance (Fig. 16.7). Whenever there is a current in the circuit, the emf has to do extra work because of the presence of the internal resistance. Hence the potential difference between the two terminals is smaller than emf in the presence of a current. If R is the resistance of the external circuit that is connected to the battery and r is the internal resistance of the battery, the actual potential difference V between two electrodes of the battery will be given by

where I is the current in the circuit. As V appears solely across the external circuit, we have

V = I R (16.12)

Equations (16.12) and (16.13) determine the net current in the circuit

I

Elcetrielty Equation (16.12) reminds, that if the current in the circuit is zero (or if the battery is not connected which is called an open circuit condition), the , (

i potential differences between the terminals of a battery is equal to the emf. In I

the presence of a current, on the other hand, the potential difference is'smaller 1 than the emf. Further, a better battery is one which has a smaller internal resistance, as, for such a battery, the potential difference will be closer to its

1 j

emf even when a large current is drawn from it. The internal resistance of a car battery in good condition is 0.005 52, while the internal resistance of a

1 torch light battery is about 0.01 52. In most cases, the internal resistance is quite low compared to the rest of the resistance in the circuit, and the internal resistance can be neglected.

Example 6: A battery of emf 15 V and an internal resistance of 1 ohm is connected to a resistance R. Find the voltage across its terminals and the resistance, if the I current in the circuit is 1 A.

Solution : !

Equation (16.12) gives the voltage, I

and Eqn.(16.3) gives the current

SAQ 7 : A battery, when connected to a resistance of 552 gives a current of 1 A. The same battery when connected to a resistance of 1052 gives 0.6 A current. Find the emf and the internal resistance of the battery.

SAQ 8 : How much work does a source of 1.5 V emf do, in maintaining a constant current of 0.01 A for 2 seconds ? If the internal resistance of the source is 1 ohm, how much is the gain in the potential energy of the carriers as they are pumped by the source ?

16.5 KIRCHHOFF'S RULES

In Unit 15, you have seen that for circuits containing capacitors, every case cannot be reduced to a series or a parallel combination. You have also seen that we can find quantities like equivalent capacitance, charges, potential differences etc. for such circuits, if we work from basic principles. Similar situation exists in case of circuits with resistances as well. Kirchhoff's rules help us in analysing complicated circuits consisting of resistances and sources of emf. Kirchhoff's rules consist of a point rule and a loop rule.

(a) Point Rule : The point rule states that the sum of currents appmching towards a junction

4s2 b c.

of t h m or mow conductors is equal to the sum of cumnts leaving away fmm the same junction. If the currents that approach a junction are assigned a sign opposite to those which leave the junction, the point rule can be restated in 'the form that the algebraic sum of all the curnents at a junction b zem.

1 Kirchhoff's point rule is essentially a statement of conservation of charge, since charge does not accumulate at any point along the connecting wires. For I,-: 2 I

example, the point rule applied to a point a in the circuit in Fig. 16.8 gives Figure 16.8

because il and i2 are toward a while i is away from it.

(b) Loop Rule : The loop rule states that the sum of potential differences enwuntered by a charge in a mund trip amund any closed loop is %em. The loop rule is basically a statement on the conservation of energy. We can write the loop rule as

In using the loop rule, we must use a consistent convention for the signs of the current and the potential difference. Applying this to the loop abcd of the

.circuit of Fig. 16.8, we have

The potential difference from a to b is V, - fi = +2 V, or, if the internal resistance of the battery is not negligible, it is 2 - ir , where i is the current along the branch from a to b. In using the latter expression, care has to be taken to use the numerical value of the current with a proper sign. Thus the following two points are to be remembered while assigning signs to various quantities :

(i) In traversing a resistance R along the sense of the current i , the potential difference across the current is -iR, while if the resistance is traversed opposite to the direction of the current, a potential difference of + i R is assigned.

(ii) In traversing a source of emf along the sense of emf, (i.e., when a source is encountered, the positive terminal comes first) the potential difference across the source is entered as &. If the source is traversed in the opposite direction the potential difference is taken as -&.

The following examples and SAQs will help you to appreciate the use of Kirchhoff's rules in solving problems on electrical circuits. . ~5

Example 7 Find the current in various branches and the equivalent resistance of the circuit shown in Fig. 16.9. Assume that the internal resistance of the battery is negligible. Take the values of the resistances to be RI = 500, Rz = 250, & = 200, & = 100 and Rs = 350. The emf of the

Figure 16.9

Solution : Let the currents in various branches be as shown in the figure. At this stage we are not sure of the direction of current in any particular branch. We have; therefore chosen the directions arbitrarily. If the numerical value for any current turns out to be negative, it simply means that the current in that branch is opposite to the direction previously assumed. Applying Kirchhoff's point rule at the junction points a,b,c and d, we get following equations :

0 )

(ii) (iii )

(iv)

You can see yourself that only three of above equations are independent, as the first and the last equations are identical in view of the other two. Applying Kirchhoff's loop rule to loops abca, bcdb and acdefa, we get the following equations :

You may note that as per the directions of the current which we have assumed, the potentials at different junction points follow the following relationship Vh > I-$ > V'; d; > > > V' and fi > >. These inequalities have been taken care of while writing above equations. Substituting the values of the resistances, we get

Using equations (ii) and (iii), (v) can be reduced to

Multiplying (vi) by 4, we get

( 4

( 4 (vii)

Thus, i2 = is. From (vi), we then get i3 = -0.5i5 Substituting these in (ii), il = 0.5is. Similarly, (iii) and (i) give i4 = 1.5i5 and i = 2i5. Thus all the currents are expressed in terms of one current is. From (vii), we therefore get is = 0.5 A. The cyrrents in various branches can now be written down.

I Note that the current i3 is negative indicating that the actual direction of the

i current is from c to b and not from b to c as has been assumed a pn'ori. As the total current drawn from the battery is 1 A, the equivalent resistance of the circuit is R = V/I = 10Q

SAQ 9 : In Fig. 16.9, find the condition for which i3, i.e. the current through R3 is zero.

Example 8 Find the current in e&h branch of the circuit of Fig. 16.10, and determine the potential difference between the points a and b. Neglect the internal resistances of the cells. Take the values R1 = 10/3Q, R2 = 20Q, R3 = 10Q, X Vl = 25V and V2 = 10V. r Solution : fD b

Let us assume the currents in the various branches to be along the directions b u m 16.10

shown in the figure. ~ ~ ~ l ~ i n ~ the point rule, we have

Applying loop rule t o the loop abefa

r Vl - ilR2 - iZRl = 0 (ii) t

Remember that as per directions of current assumed, if a charge has to traverse through the loop abefa, the passage through Vl will increase its potential while passage through botp R~ and Rl will decrease the same, because the potential at the entry point ofcurrent into a resistor is higher than a t the exit point. Similarly, applying the loop rule t o abcda

Vl - ilR2 -i3R3- V2 = 0 (iii)

Substituting the numerical values we get, on solving (i), (ii) and (iii) it = 1 A, iz = 1.5 A and is = -0.5 A. The potential difference between the points b and a is 25 - il R2 = 5V.

M t Y SAQ 10 : Find the potential difference between points a and b in the circuit shown in Fig. 16.11. Neglect the internal resistances of the cells.

1

I Figure 16.11

16.6 RC CIRCUITS

Figure. 16.12

So far we considered only those circuits which contained sources of emf and resistances. We had implicitly assumed that the emf is constant as a function of time and therefore the current also is time independent. We shall now consider a circuit in which, besides a resistor, a capacitor is also included as a circuit element. We shall see that because of charging and discharging of the capacitor, the current, in general, will be time dependent.

Figure 16.12 shows a simple circuit. The point d can either be connected to the point a or to the point b. Let us first consider the case when capacitor C does not have any excess charge on its plates. Assume that at t = 0, we connect the point d to the point a. The battery V therefore gets included in the circuit. As the positive terminal of the battery (point k) is at a higher potential than the point m, the charges will flow out of the battery, and there will be a current in the circuit. These charges will keep on accumulating at the plates of the capacitor and the potential difference between the positive terminal of the battery and the point m will decrease, causing the current to reduce. Finally, if we wait long enough, the potential difference between points k and m will be essentially zero and the current in the circuit will also be nearly zero. Our interest is to find out how the current varies as a function of time.

Let Q be the charge on the plates of the capacitor at any given time t. The potential difference between the plates of the capacitor at this instant of time is Vm, = Q/C. If the current is i , then the potential difference across the resistor is Vkm = iR. As the total potential difference has to be V, the one that is maintained by the battery, we get

Using i = dQ/dt, we get the following differential equation :

The equation (16.16) can be written as

dQ - -- dt VC-Q RC

putting VC - Q = 2, so that -dQ = dx,we get

This can be integrated to give z = Ke-'lRC where K is a. constant of integration. Substituting back, we get

The constant K may be determined by imposing conditions on the charge. At 9: = O,Q = 0, which gives 0 = VC - K or K = VC. Therefore, we get

An expression for the current can be obtained by differentiating the above,

As at t = 0,i = V/R, we introduce V/R = i, and can write the expression for the current as

i = i,e ' -~/RC

It clear from Eqns(16.16) and (16.17) that both the charge on the capacitor and the current in the circuit are exponential functions of time. The factor R C in these two equations has the dimension of time and determines how fast the capacitor will get charged or how fast will the current reduce to zero. This factor is therefore called the time constant of the circuit and is usually denoted by T . The time constant can be interpreted as the time taken for the current to reduce to l / e x 0.37 of its original value. In terms of the charge on the capacitor, it can be interpreted as the time in which the charge on the capacitor increases to (1 - l/e) x 0.63 times its final value. For a circuit with a large time constant, the time taken for charging the capacitor is longer.

Suppose the capacitor is fully charged. We connect the point d to point b in Fig. 16.12 by changing the position of the switch. Let us now measure time from the instant d is connected to b. Obviously, the battery has been thrown out of the circuit, and, therefore, there is nothing in circuit to maintain the

r potential difference. Hence, the flow of charges will start from a higher potential to a lower potential. This flow will reduce the charges on the plates of the capacitor and thereby the potential difference between them. The

I current will reduce as a function of time until the excess charges on the plates of the capacitor become zero. This situation will correspond to the potential becoming the same everywhere in the circuit. Again, as before, our interest is to find the charge Q on the capacitor and the current i as a function of time.

We can write an equation similar to Eqn.(16.15), except that now there is no battery providing V in the circuit. The differential equation in the case of discharging will thus become

This can be written as dQ -- dt - -- Q RC

Integrating the above, we get

If at t = 0, & - Q, which is same as VC, we have

Electric Circuits

Differentiating the above, we obtain, for the current,

where i, = Q,/ RC. The negative sign indicates that the direction of the current is now opposite to the original direction. You now see that both the charge and the magnitude of the current decay with the same time constant r given by r = 1/RC. t

1 Example 9 In the circuit shown in Fig. 16.13, two capacitors are connected to a single battery. At t = 0, the capacitors are completely uncharged, and the battery is connected. Take the numerical values of the circuit elements as follows : C1 = 10pF; C2 = 30pF; R = 200 and V = 50 V.

JA (a) Find the time constant of the charging.

I v

Figure 16.13

(b) After s, what are the potential differences across the capacitors and the resistor ?

Solution : If Vl,V2 and VR are the potential differences across the capacitors C1,C2 and the resistor R respectively, we can write

As the capacitors are in series one, the charge Q on the plates of the two capacitors will he identical at any given time t. We can thus write

which can be written as

You can see the above equation is identical to Eqn.(16.16) with 1/C = l/C1 + 1/C2. This is a result which we expected, because we could have just taken an equivalent capacitance of two capacitors in series.

The time constant is given by r = R( &- + $-)-I. Substituting numerical values we get r = 1.5 X ' I O - ~ s. The charge and the current at t = s, can be obtained by direct use of equations (16.17) and (16.18) to be 182.5 pC and 1.2835 A respectively. The potential differences are given by Vl = Q/Cl = 18.25 V, V2 = Q/C2 = 6.08 V anu VR = iR = 25.67 V.

SAQ 11 : Consider the circuit of Fig. 18.1 1. w i t h Vl = 5 V, li = 20 V, R1 = 5Q, Rz = 10R and C = 2pT

- 1"; t h ~ hey S is kept open and the capacitor is

aI!~,,cd i~ rl 2~~ ' ,- ,, , , :I + r ! 't1.1t t l ~ r c h a ~ g e on C is constant. What ~s t h t ~ ~h.,~, , t I '.1gt5 ? The key S is now closed. What *.ll u l I . * I .. I I , : , L ~ C r r after a sufficiently long time elapses ?

16.7 ENERGY AND POWER IN ELECTRIC CIRCUITS

We have seen that when a current i flows through a resistance R, a potential difference of V = iR exists between the two ends of the resistance. We have seen earlier that this implies that a charge q entering the resistance a t the higher potential side and leaving it at the lower potential side suffers a loss of energy equal to qV. This loss of energy appears as heat in the resistor. Physically, this can be understood as follows. As the charge passes through a resistive medium, it collides with other particles of the medium. These scatterings cause, in general, the motion of the bound atoms in the solid to increase, which manifests in the form of increased temperature. The situation

1 can be compared with a ball falling from a height to the ground. Though the I drop causes a decrease in potential energy, this decrease is compensated by an I increase in the kinetic energy. In the case of charges drifting through a

resistor, the overall current (or drift velocity) does not increase even though there is a fall in the potential energy. This decrease in potential energy is compensated by an increase in the kinetic energy of the atoms of the solids.

Consider a time interval dt. In this time interval, a charge dq = idt flows into the resistor froin highest potential end to a slightly lower potential, and, a charge dq leaves the lowest potential and a slightly higher potential. Other conduction charges also move to lower potential side in this time within the resistor. The net decrease of potential energy is equal to dqV. This is because, effectively, the charge dq has left the highest potential end and reappeared at the lowest potential end. The energy dW transferred from conduction carriers to the atoms of the solid in this time would, therefore, be given by

The rate P of energy transfer (Power) is thus given by

p = - = dW i~ dt

Since, for a resistor V = iR, we can write

The S.I. unit of P is Watt (W), one unit of which is equal to 1 J/s. You should appreciate the fact that Eqn.(6.21) is correct even when we consider some other circuit component like a battery, between the twr, ierminals of which, there is a potential difference V. Equation (6.22), on the other hand, is correct only for a resistor, because it is only here that we have assumed V = iR, a condition true for a resistor.

Example 10 Consider a 60 W, 240 V d.c. bulb. What is the resistance of the wire? What will be the current in it, if it is connected to 240 V d.c. supply? What will be the power consumed if it is attached to 210 V d.c. supply? What is the value of the current in this case ? Assume that the resistance of the bulb wire does not change in the two cases.

Solution : A 60 W, 240 V bulb consumes 60 W power when connected to 240 V supply. Thus y 1 / R = 60, which gives R = 980R. The current in this case is i = V/R = 0.25 A. When the same bulb is connected to a 210 V supply, I

P = V ~ / R = 45.94 W. The current in this case is i = V/R = 0.22 A. 0

SAQ 12 : A 15 V battery with an internal resistance of 1R is connected to a resistance of 9 ohm. What is the rate at which the internal energy of battery is getting converted to electrical energy ? Find the power consumed within the battery and by the external resistance.

16.8 SUMMARY

In this unit, we introduced the concept of electric current, which can exist whenever an external electric field is maintained across the ends of a conductor. We found that in the presence of such a field the average velocity of the charge carriers is no longer zero, but has a small value, which is in the direction of the applied field for positive carriers. This is called the drift velocity. The electric current is a scalar quantity, but we have a quantity called the current density, which depends on the drift velocity as well as the carrier density, which is a vector.

It was seen that the charge carriers encounter opposition to their free movement through a conductor because of scattering with other particles of the medium. This provides resistance to the current. Circuits which contain resistances and sources of energy to drive the electric current were solved using the two rules due to Kirchhoff. While the current in such circuits is constant, in the presence of a capacitative element in the circuit, the charge as well as the current become time dependent. Finally, we found that power is dissipated in circuits containing resistances.

16.9 ANSWERS TO SAQs

1. As the current is 1 pA, it supplies 1 pC charge per second. As the capacitance of the sphere is 47r/c,R, in order to raise the potential of the sphere by 500 V, we require a total charge Q given by

The charge Q in time t is 1 x 10'~t and the radius R is given as 0.2.m. Substituting these in the above equation, we get the time to be 1.1 x 10'~ s.

2. ud = 1.1 x lo:', and t = 5.6 mm.

3. Recalling that the positive and the negative charges contribute to the current in the same sense, the current is proportional to the drift velocity. Thus the positive charges contribute to 80% of the total current, i.e., 0.8 A. The remaining 0.2 A current is due to the negative charges. Current density I / A are respectively 8000 A / & ~ and 2000 ~ / r n ~ . The drift speed for the positive charge is J+/nq = 8.3 mm/s.

4. The resistances of each of the wires is 5 52. The voltage drop across each wbre is therefore the same and is 5 V. The electric field across A is 51100 .=: 0.05 V/m, while that across B is Q.025 V/m. The current density in A is lo6 A/m2 and in B is lo5 A/m2.

5 . The resistance will be lowest when the ratio IIA is the least. The block has three sets of parallel faces; first set has an area of cross section 12 cm2 and a length 5 cm; second has an area of 15 cm2 and a length 4 cm and the third has an &ea of 20 cm2 and a length of 3 cm. Comparing the

i ratio of the area to the length', the third face offers the least resistance.

6 . Assume V, > Vb. Since the effective resistance of the circuit is 10 52, the total current entering the point a is 10/10 = 1 A. Clearly, the full current of 1 A passes through 152 and 752 resistances. However, at the point c, the current gets divided between the 352 and 652 resistances. The potential drop across 152 and 752 resistances are 1 V and 7 V respectively. Hence the potential drop between the points c and d is (10 - 7 - 1 ) = 2 V. This 2 V is the potential difference across both 352 and 652 resistances. Hence, the current in them are 213 A and 113 A, respectively. We can see that the total current in these branches is 213 + 113 = I A, as expected.

i i 7 . The potential drop across the 552 resistance is 5 V while that across the i 1052 resistance is 6 V. Using Eqn.(16.12)

6 = & - 0 . 6 ~

5 = & - - r

Solving, we get r = 2.552 and & = 7.5 V.

8. 0.01 A current for 2 seconds implies a total transfer of charge of 0.02 C. The work W done by, the source of emf on these charges, by using Eqn.(l6.11) is W = 1.5 x 0.02 = 0.03 J . This work is not fully utilised to increase the energy of the charge carriers, as, a part of the energy is lost

P due to the internal resistance of the source. The potential difference across the internal resistance is 0.01 x I = 0.01 V. The total drop in the energy of the carriers, in passing through the internal resistance is

r 0.01 x 0.02 = 2 x J. Hence, the net increase in the energy of the charge carriers is 0.03 - 2 x = 0.0298 J .

9. If ig = 0 , i l = il and i4 = is. Moreover, the potential difference between the points b and c should be zero. This implies that the potential drop across R1 should be same as that across R4. Thus, il Rl = is R4. Similarly, the potential drop between the points a and d is given by i l ( R 1 + R 2 ) = i4(R4 + R5) . Combining these, we get

A circuit of this type is called a wheatstone bridge which is used to measure an unknown resistance, normally put in place of Rs.

10. If there is current i in the loop bcefb, we get, applying the loop rule, 25 - 7i - 8i - 10 = 0, i.e., i = 1 A. The poiential differences between the points b and e, is thus 25 - 7 x 1 = 18 V. As a and b are not connected, there is no current through the 1052 resistor. Hence, there is no voltage drop across this resistor, and therefore, we have & - V a = 8 x 1 + 1 0 - 8 = 1 0 V e

11. When the key S is open, the part of the circuit to the ieft is ineffective. Hence it is a simple RC circuit with R = R2 = 25R and C = 2pF. The final charge on the capacitor is therefore, VC = 40 x C. When the key S is closed, the part of the circuit to the left becomes effective and redistribution of charges take place. But once the charges or1 the capacitor stabilizes, the current in the arm containing the capnc.itor is zero. However, there will be a current in rest of the circuit as there is a net voltage due to the emf. Applying loop rule to abcdefa, we get i(10 + 5) - 5 = 20, which gives i = 1 A. The potential difference he between b and e is therefore given by %, = fi - iR2 = 20 - 10 = 10V. The new charge on the capacitor is 20 x C.

12. The total current in the circuit is 15/(9 + 1) = 1.5 A. The rate of ene transfer within the battery as given by Eqn.(6.21) is 15 x 1.5 = 22.5 7 Power consumed by the internal resistance = 1.5 x 1.5 x 1 = 2.25W. Power consumed by the external resistance = 1.5 x 1.5 x 9 = 20.25W.

SUGGESTED FURTHER READINGS

Mahajan, A. S. and Rangwala, A. A., Electricity and Magnetism , 1993 , Tata McGraw-Hill Publishing Company, New Delhi.

Purcell, E. M., Berkeley Physics Course, Vo1.2. International Student Edn. 1985 McGraw-Hill Book Co. Singapore

Halliday, D. and Resnick, R., Fundamentals of Physics, Third Extended P Edn., John Wiley, New York, 1988.

Plonsey, R. and Collin, R. E., Principles and Applications of Electromagnetic Fields. Tata McGraw-Hi9 Publishing Company, New Delhi.